$s$ and $t$ are non-symplectic involutions, Fix$(s)$ and Fix$(t)$ are a set of
curves and Fix$(s\circ t)$ is a set of 8 isolated points. Since $(B_{2,3}\cdot
B_{0,1})=2$, $|p^{-1}(B_{2,3})\cap p^{-1}(B^{1}_{1,1})|=4$. Since Fix$(s\circ
t)\supset p^{-1}(B_{2,3})\cap p^{-1}(B_{0,1})$, $X/(G_{2,3}\oplus G_{0,1})$
has 2 singular points, however, since the branch divisor of the double cover
$X/(G_{2,3}\oplus G_{0,1})\rightarrow X/G$ is $2B^{1}_{1,1}+2B^{2}_{1,1}$ and
$(B^{1}_{1,1}\cdot B^{2}_{1,1})=1$, the number of singular points of
$X/(G_{2,3}\oplus G_{0,1})$ must be 1. This is a contradiction.

As for the case of (121), the numerical class of $B$ is not one of (191,192)
of the list in section 6.

We assume that the numerical class of $B$ is (123) of the list in section 6.
We denote $B$ by $2B_{2,2}+2B_{1,2}+2B_{1,1}+2B_{0,1}$. By Theorem 2.8,
$G_{2,2}\cong G_{1,2}\cong G_{1,1}\cong\mathbb{Z}/2\mathbb{Z}$. Since an
intersection of two of $B_{2,2},B_{1,2},B_{1,1}$ is not an empty set, by
Theorem 2.8, $G=G_{2,2}\oplus G_{1,2}\oplus G_{1,1}$. Since $(B_{2,2}\cdot
B_{1,2})=4$, $X/(G_{2,2}\oplus G_{1,2})$ is smooth. Then there is a double
cover $X/G_{2,2}\oplus G_{1,2}\rightarrow X/G\cong{\mathbb{F}}_{1}$ whose
branch divisor is $2B_{1,1}+2B_{0,1}$. Since
$\frac{B_{1,1}+B_{0,1}}{2}\not\in$Pic$({\mathbb{F}}_{1})$, by Theorem 3.5,
this is a contradiction.

We assume that the numerical class of $B$ is (125) of the list in section 6.
We denote $B$ by $2B^{1}_{1,2}+2B^{2}_{1,2}+2B^{1}_{1,1}+2B^{2}_{1,1}$. By
Theorem 2.8, $G^{i}_{1,2}\cong G^{i}_{1,1}\cong\mathbb{Z}/2\mathbb{Z}$ for
$i=1,2$. Since an intersection of two of
$B^{1}_{1,2},B^{2}_{1,2},B^{1}_{1,1},B^{2}_{1,1}$ is not an empty set, by
Theorem 2.8, $G=G^{1}_{1,2}\oplus G^{2}_{1,2}\oplus G^{1}_{1,1}\oplus
G^{2}_{1,1}$ or $G=G^{1}_{1,2}\oplus G^{2}_{1,2}\oplus G^{1}_{1,1}$. We assume
that $G=G^{1}_{1,2}\oplus G^{2}_{1,2}\oplus G^{1}_{1,1}\oplus G^{2}_{1,1}$.
Since $|G|=16$ and $(B^{1}_{1,2}\cdot B^{2}_{1,2})=3$,
$|p^{-1}(B^{1}_{1,2}\cap B^{2}_{1,2})|\geq 12$. Since the number of isolated
points of symplectic involution is 8, this is a contradiction. Therefore,
$G=G^{1}_{1,2}\oplus G^{2}_{1,2}\oplus G^{1}_{1,1}$.

By Theorem 3.5, there are the Galois covers
$p_{1}:Y_{1}\rightarrow{\mathbb{F}}_{1}$ and
$p_{2}:Y_{2}\rightarrow{\mathbb{F}}_{1}$ such that the branch divisor of
$p_{1}$ is $2B^{1}_{1,2}+2B^{2}_{1,2}$ and that of $p_{2}$ is
$2B^{1}_{1,1}+2B^{2}_{1,1}$. Let
$X^{\prime}:=Y_{1}\times_{{\mathbb{F}}_{1}}Y_{2}$. Then there is the Galois
cover $q:X^{\prime}\rightarrow{\mathbb{F}}_{1}$ whose branch divisor is
$2B^{1}_{1,2}+2B^{2}_{1,2}+2B^{1}_{1,1}+2B^{2}_{1,1}$ and Galois group is
isomorphic to ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ as a group. By Theorem
2.3, there is a symplectic automorphism of order 2 $s\in G$ such that
$X^{\prime}=X/\langle s\rangle$. Since $s$ is symplectic, the minimal
resolution $f:X^{\prime}_{m}\rightarrow X^{\prime}$ is a $K3$ surface. Let
$e_{1},\ldots,e_{8}$ be the exceptional divisors of $f$. We set
$\\{p_{1},p_{2},p_{3}\\}:=B^{1}_{1,2}\cap B^{2}_{1,2}$, and
$\\{p_{4}\\}:=B^{1}_{1,1}\cap B^{2}_{1,1}$. Let $\pi:{\rm
Blow}_{\\{p_{1},p_{2},p_{3},p_{4}\\}}{\mathbb{F}}_{1}\rightarrow{\mathbb{F}}_{1}$
be the blow-up of ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ at points
$p_{1},p_{2},p_{3},p_{4}$, and $E_{i}:=\pi^{-1}(p_{i})$ be an exceptional
divisor of $\pi$ for $i=1,2,3,4$. Since the support of $B$ is simple normal
crossing, in the same way of Proposition 3.15, there is a Galois cover
$q:X^{\prime}_{m}\rightarrow{\rm
Blow}_{\\{p_{1},p_{2},p_{3},p_{4}\\}}{\mathbb{F}}_{1}$ whose branch divisor is
$2C^{1}_{1,2}+2C^{2}_{1,2}+2C^{1}_{1,1}+2C^{2}_{1,1}$ and Galois group is
isomorphic to ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ as a group, where
$C^{1}_{1,2},C^{2}_{1,2},C^{1}_{1,1},C^{2}_{1,1}$ are proper transform of
$B^{1}_{1,2},B^{2}_{1,2},B^{1}_{1,1},B^{2}_{1,1}$ by the birational map
$\pi^{-1}$ in order. Notice that
$q^{\ast}(\sum_{i=1}^{4}E_{i})=\sum_{j=1}^{8}e_{j}$ and there is the
commutative diagram:

$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathbb{F}}_{1}}$$\textstyle{X^{\prime}_{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{q}$$\textstyle{{\rm
Blow}_{\\{p_{1},p_{2},p_{3},p_{4}\\}}{\mathbb{F}}_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces.}$$\scriptstyle{\pi}$

Furthermore, we put $\\{x_{1},\ldots,x_{8}\\}:=$Fix$(s)$. Then ${\rm
Blow}_{\\{x_{1},\ldots,x_{8\\}}}X/\langle s\rangle=X^{\prime}_{m}$, the branch
divisor of the double cover ${\rm Blow}_{\\{x_{1},\ldots,x_{8}\\}}X\rightarrow
X^{\prime}_{m}$ is $\sum_{j=1}^{8}e_{j}$, and there is the commutative
diagram:

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{\prime}}$$\textstyle{{\rm
Blow}_{\\{x_{1},\ldots,x_{8}\\}}X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{\prime}_{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

In the same way of Proposition 3.15, we get that

$\sum_{i=1}^{4}q^{\ast}E_{i}=2(\pi\circ
q)^{\ast}(C+\frac{3}{2}F)-2C^{1}_{1,2}-2C^{1}_{1,1}\ {\rm in\
Pic}(X^{\prime}_{m}).$

Since ${\rm Blow}_{\\{x_{1},\ldots,x_{8}\\}}X$ and $X^{\prime}_{m}$ are
smooth, and $q^{\ast}(\sum_{i=1}^{4}E_{i})=\sum_{j=1}^{8}e_{j}$, we get that
$\frac{\sum_{i=1}^{4}q^{\ast}E_{i}}{2}\in$Pix$(X^{\prime}_{m})$, and hence
$\frac{F}{2}\in$Pic$(X^{\prime}_{m})$.

Since $C^{1}_{1,2}\cap C^{2}_{1,2}$ is an empty set and
$\frac{C^{1}_{1,2}+C^{2}_{1,2}}{2}\in$Pic$({\rm
Blow}_{\\{p_{1},p_{2},p_{3},p_{4}\\}}{\mathbb{F}}_{1})$, by Theorem 3.5, there
is the Galois cover $g:Z\rightarrow{\rm
Blow}_{\\{p_{1},p_{2},p_{3},p_{4}\\}}{\mathbb{F}}_{1}$ such that $Z$ is
smooth, the branch divisor is $2C^{1}_{1,2}+2C^{2}_{1,2}$, and the Galois
group is isomorphic to ${\mathbb{Z}}/2{\mathbb{Z}}$ as a group. By Theorem
2.3, there is a non-symplectic automorphism of order 2 $\rho$ of
$X^{\prime}_{m}$ such that $X^{\prime}_{m}/\langle\rho\rangle=Z$. Let
$h:X^{\prime}_{m}\rightarrow Z$ be the quotient map. Then $q=g\circ h$, and
hence $\frac{F}{2}\in$Pic$(X^{\prime}_{m})^{\rho}$. Since the degree of $g$ is
2 and $(C^{1}_{1,2}\cdot\frac{F}{2})=\frac{1}{2}$ and
$\frac{g^{\ast}C^{1}_{1,2}}{2}\in$Pic$(Z)$, we get that
$g^{\ast}\frac{F}{2}\not\in$Pic$(Z)$. Recall that $C^{i}_{1,1}=C+F-e_{4}$ in
Pic$({\rm Blow}_{\\{p_{1},p_{2},p_{3},p_{4}\\}}{\mathbb{F}}_{1})$ for $i=1,2$.
Since the branch divisor of $h$ is
$2g^{\ast}C^{1}_{1,1}+2g^{\ast}C^{2}_{1,1}$, we get that
$q^{\ast}(\frac{1}{2}C+\frac{1}{2}F-e_{4})\in$Pic$(X^{\prime}_{m})$. By
[References], Pic$(X^{\prime}_{m})^{\rho}$ is generated by $h^{\ast}{\rm
Pic}(Z)$ and $q^{\ast}(\frac{1}{2}C+\frac{1}{2}F-e_{4})$ over ${\mathbb{Z}}$.
Since $g^{\ast}\frac{F}{2}\not\in$Pic$(Z)$ and
$2q^{\ast}(\frac{1}{2}C+\frac{1}{2}F-e_{4})\in h^{\ast}{\rm Pic}(Z)$, we may
assume that there is $\alpha\in{\rm Pic}(Z)$ such that

$q^{\ast}\frac{F}{2}=h^{\ast}\alpha+q^{\ast}(\frac{1}{2}C+\frac{1}{2}F-e_{4}).$

Then $g^{\ast}(\frac{-1}{2}C+e_{4})\in$Pic$(Z)$. Since the degree of $g$ is 2
and $(C^{1}_{1,2}\cdot\frac{-1}{2}C+e_{4})=\frac{3}{2}$ and
$\frac{g^{\ast}C^{1}_{1,2}}{2}\in$Pic$(Z)$, we get that
$(\frac{g^{\ast}C^{1}_{1,2}}{2}\cdot
g^{\ast}(\frac{-1}{2}C+e_{4}))=\frac{3}{2}$. By the assumption that
$\frac{g^{\ast}C^{1}_{1,2}}{2}\in$Pic$(Z)$ and
$g^{\ast}(\frac{-1}{2}C+e_{4})\in$Pic$(Z)$, this is a contradiction.
Therefore, the numerical class of $B$ is not (125) of the list in section 6.

We assume that the numerical class of $B$ is (126) of the list in section 6.
We denote $B$ by $2B_{1,2}+2B^{1}_{1,1}+2B^{2}_{1,1}+2B^{3}_{1,1}+2B_{0,1}$.
By Theorem 2.8, $G_{1,2}\cong G^{i}_{1,1}\cong
G_{0,1}\cong\mathbb{Z}/2\mathbb{Z}$ where $i=1,2,3$. Since an intersection of
two of $B_{1,2},B^{1}_{1,1},B^{2}_{1,1},B^{3}_{1,1},B_{0,1}$ is not an empty
set, by Theorem 2.8, $G=G_{1,2}\oplus G^{1}_{1,1}\oplus G^{2}_{1,1}\oplus
G^{3}_{1,1}$. Let $G_{s}$ be the subgroup of $G$ which consists of symplectic
automorphisms of $G$. Then $G_{s}\cong\mathbb{Z}/2\mathbb{Z}^{\oplus 3}$. By
[References], the number of singular points of $X/G_{s}$ is 14, however, since
the branch divisor of the double cover $X/G_{s}\rightarrow X/G$ is
$B=2B_{1,2}+2B^{1}_{1,1}+2B^{2}_{1,1}+2B^{3}_{1,1}+2B_{0,1}$ and the support
of $B$ is simple normal crossing, the number of singular points of $X/G_{s}$
is 13. This is a contradiction. Therefore, the numerical class of $B$ is not
(126) of the list in section 6.

######  Lemma 3.38.

If $X/G\cong{\mathbb{F}}_{n}$ where $n\geq 1$, then
$B\not=aC+bB_{s,t}+cB_{u,v}$ where $a,b,c>0$ are even integers, and $(C\cdot
B_{s,t})\not=0$ and $(C\cdot B_{u,v})\not=0$, i.e. $s\not=t$ or $u\not=v$.

###### Proof.

We assume that $B=aC+bB_{s,t}+cB_{u,v}$ where $a,b,c>0$ are even integers, and
$(C\cdot B_{s,t})\not=0$ and $(C\cdot B_{u,v})\not=0$ By Theorem 2.8 and
$(B_{s,t}\cdot B_{u,v})\not=0$, $G=G_{s,t}\oplus G_{u,v}$. Then the number of
non-symplectic involutions of $G$ is 2. Since $(C\cdot B_{s,t})\not=0$ and
$(C\cdot B_{u,v})\not=0$, $G$ must have at least 3 non-symplectic involutions.
This is a contradiction. ∎

By Lemma 3.38, the numerical class of $B$ is not one of (139,181,182,244) of
the list in section 6.

We assume that the numerical class of $B$ is (189) of the list in section 6.
We denote $B$ by $2B_{1,0}+2B_{1,4}+2B^{1}_{1,1}+2B^{2}_{1,1}$. By Theorem
2.8, $G_{1,0}\cong G_{1,4}\cong G^{i}_{1,1}\cong\mathbb{Z}/2\mathbb{Z}$ where
$i=1,2$. Since $(B_{1,4}\cdot B^{i}_{1,1})\not=0$ for $i=1,2$, by Theorem 2.8,
$G=G_{1,4}\oplus G^{1}_{1,1}\oplus G^{2}_{1,1}$. Let $s\in G^{1}_{1,1}$ and
$t\in G^{2}_{1,1}$ be generators. Since the number of non-symplectic
automorphisms of order 2 of $G$ is 4 and Theorem 2.8, we may assume that
Fix$(s)$ is the support of $p^{\ast}B^{1}_{1,1}$. Since the support of $B$ is
simple normal crossing and $(B_{1,4}\cdot B^{1}_{1,1})=4$, $X/(G_{1,4}\oplus
G^{1}_{1,1})$ is smooth. Then there is the Galois cover $X/G_{1,4}\oplus
G^{1}_{1,1}\rightarrow{\mathbb{F}}_{1}$ such that the branch divisor is
$2B_{1,0}+2B^{2}_{1,1}$ and the Galois group is isomorphic to
${\mathbb{Z}}/2{\mathbb{Z}}$ as a group. Since
$\frac{B_{1,0}+B^{2}_{1,1}}{2}\not\in$Pic$({\mathbb{F}}_{1})$, this is a
contradiction.

As for the case of (189), the numerical class of $B$ does not (190) of the
list in section 6.

We assume that the numerical class of $B$ is (228) of the list in section 6.
We denote $B$ by $3B_{1,0}+3B_{1,2}+3B_{1,4}$. By Theorem 2.8 and
$(B_{1,2}\cdot B_{1,4})\not=0$, $G=G_{1,2}\oplus G_{1,4}$. Let $s\in G_{1,4}$
be a generator of $G_{1,4}$. Then the only curve of Fix$(s)$ is $C_{1,4}$.
Since $(B_{1,4}\cdot B_{1,4})=6$, the genus of $C_{1,4}$ is 4. By
[References,References], Fix$(s)$ does not have isolated points, and hence
$X/G_{1,4}$ is smooth. Let $q:X/G_{1,4}\rightarrow X/G$ be the quotient map.
Then the degree of $q$ is 3, and the branch divisor of $q$ is
$3B_{1,0}+3B_{1,2}$. Since the degree of $q$ is 3 and $X/G_{1,4}$ is smooth,
$\frac{3}{3^{2}}(B_{1,0}\cdot B_{1,0})$ is an integer. Since $(B_{1,0}\cdot
B_{1,0})=-2$, $\frac{3}{3^{2}}(B_{1,0}\cdot B_{1,0})=-\frac{2}{3}$. This is a
contradiction.

We assume that the numerical class of $B$ is (229) of the list in section 6.
We denote $B$ by $3B_{1,0}+3B_{1,2}+3B_{1,3}+3B_{0,1}$. By Theorem 2.8,
$G_{1,0}\cong G_{1,2}\cong G_{1,3}\cong G_{0,1}\cong\mathbb{Z}/3\mathbb{Z}$.
Since $(B_{1,2}\cdot B_{1,3})\not=0$, by Theorem 2.8, $G=G_{1,2}\oplus
G_{1,3}$. Let $s,t\in G$ be generators of $G_{1,2}$ and $G_{1,3}$ respectively
such that $s\circ t$ is a non-symplectic automorphism of order 3. Since
$G=G_{1,2}\oplus G_{1,3}$, the number of subgroups of $G$ which are generated
by a non-symplectic automorphism of order 3 is 3. Since $(B_{1,2}\cdot
B_{0,1})\not=0$ and $(B_{1,3}\cdot B_{0,1})\not=0$, we get that
$p^{-1}(B_{0,1})$ is contained in Fix$(s\circ t)$, and hence $p^{-1}B_{1,0}$
is contained in Fix$(s)$. Since $|G|=9$, there is an elliptic curve $C_{0,1}$
on $X$ such that $p^{\ast}B_{0,1}=3C_{0,1}$. By [References,References], the
number of isolated points of Fix$(s\circ t)$ is 3. Since $(B_{1,0}\cdot
B_{1,3})=1$ and $(B_{1,2}\cdot B_{1,3})=3$, we have $|p^{-1}(B_{1,0}\cup
B_{1,2})\cap p^{-1}(B_{1,3})|=4$. Since $p^{-1}(B_{1,0}\cup
B_{1,2})\subset$Fix$(s)$ and $p^{-1}(B_{1,3})\subset$Fix$(t)$, we get that
$p^{-1}(B_{1,0}\cup B_{1,2})\cap p^{-1}(B_{1,3})\subset$Fix$(s\circ t)$. By
the fact that the number of isolated points of Fix$(s\circ t)$ is 3, this is a
contradiction.

We assume that the numerical class of $B$ is (243) of the list in section 6.
We denote $B$ by $2B_{1,0}+2B_{1,4}+2B_{2,4}$. By Theorem 2.8,
$G=G_{1,4}\oplus G_{2,4}$. Let $s\in G$ be a generator of $G_{1,4}$. Since
$(B_{1,0}\cdot B_{1,4})\not=0$ and $(B_{1,4}\cdot B_{2,4})\not=0$, the only
curve of Fix$(s)$ is $C_{1,4}$. Since the fixed locus of a non-symplectic
involution does not have isolated points, $X/G_{1,4}$ is smooth. Let
$q:X/G_{1,4}\rightarrow X/G\cong{\mathbb{F}}_{2}$ be the quotient map. The
degree of $q$ is 2 and the branch divisor of $q$ is $2B_{1,0}+2B_{2,2}$. Since
$\frac{B_{1,0}+B_{2,2}}{2}\not\in$Pic$({\mathbb{F}}_{2})$, by Theorem 3.5,
this is a contradiction.

We assume that the numerical class of $B$ is (249) of the list in section 6.
We denote $B$ by $2B_{1,0}+2B^{1}_{1,3}+2B^{2}_{1,3}+2B_{1,2}$. By Theorem
2.8, $G^{i}_{1,3}\cong G_{1,2}\cong\mathbb{Z}/2\mathbb{Z}$ where $i=1,2$.
Since an intersection of two of $B^{1}_{1,3},B^{2}_{1,3},B_{1,2}$ is not an
empty set, $G=G^{1}_{1,3}\oplus G^{2}_{1,3}\oplus G_{1,2}$. Since $|G|=8$ and
$(B^{1}_{1,3}\cdot B^{2}_{1,3})=4$, $Y:=X/(G^{1}_{1,3}\oplus G^{2}_{1,3})$ is
smooth. Then there is the Galois cover $q:Y\rightarrow X/G$ such that the
branch divisor is $2B_{1,0}+2B_{1,2}$, and the Galois group is
${\mathbb{Z}}/2{\mathbb{Z}}$ as a group. Since the fixed locus of a non-
symplectic automorphism of order 2 does not have isolated points,
$X/G^{1}_{1,3}$ is smooth, and there is the Galois cover
$q^{\prime\prime}:X/G^{1}_{1,3}\rightarrow Y$ such that the branch divisor of
$q^{\prime\prime}$ is $2q^{\ast}B^{1}_{1,3}$ and the Galois group of
$q^{\prime\prime}$ is ${\mathbb{Z}}/2{\mathbb{Z}}$ as a group. Since $Y$ and
$X/G^{1}_{1,3}$ are smooth, and the degree of $q^{\prime\prime}$ is two, we
get that $\frac{q^{\ast}B^{1}_{1,3}}{2}\in$Pic$(Y)$. Recall that the branch
divisor of $q$ is $2B_{1,0}+2B_{1,2}$, and the degree of $q$ is two. Since
$\frac{q^{\ast}B_{1,2}}{2}\in$Pic$(Y)$, we get that
$\frac{q^{\ast}F}{2}=\frac{q^{\ast}B^{1}_{1,3}}{2}-\frac{q^{\ast}B_{1,2}}{2}\in$Pic$(Y)$.
Since $(B_{1,0}\cdot F)=1$, we get that
$(\frac{q^{\ast}B_{1,0}}{2}\cdot\frac{q^{\ast}F}{2})=\frac{1}{2}$. Since
$\frac{q^{\ast}B_{1,0}}{2}\in$Pic$(Y)$ and $\frac{q^{\ast}F}{2}\in$Pic$(Y)$,
this is a contradiction. Therefore, the numerical class of $B$ is not (249).

We assume that the numerical class of $B$ is (250) of the list in section 6.
We denote $B$ by $2B_{1,0}+2B_{1,3}+2B^{1}_{1,2}+2B^{2}_{1,2}+2B_{0,1}$. By
Theorem 2.8, $G_{1,3}\cong G^{i}_{1,2}\cong\mathbb{Z}/2\mathbb{Z}$ where
$i=1,2$. Since an intersection of two of $B_{1,3},B^{1}_{1,2},B^{2}_{1,2}$ is
not an empty set, by Theorem 2.8, $G=G_{1,3}\oplus G^{1}_{1,2}\oplus
G^{2}_{1,2}$. Let $s\in G^{1}_{1,2}$ be a generator. Since the number of non-
symplectic automorphisms of order 2 of $G$ is 4 and Theorem 2.8, we may assume
that $p^{-1}(B^{1}_{1,3})$ and $p^{-1}(B_{1,0})$ are contained in Fix$(s)$.
Since the support of $B$ is simple normal crossing and $(B_{1,3}\cdot
B_{1,0}+B^{1}_{1,2})=4$, $X/(G_{1,3}\oplus G^{1}_{1,2})$ is smooth and there
is the Galois cover $X/(G_{1,3}\oplus G^{1}_{1,2})\rightarrow{\mathbb{F}}_{2}$
such that the branch divisor is $2B^{2}_{1,2}+2B_{0,1}$ and the Galois group
is ${\mathbb{Z}}/2{\mathbb{Z}}$ as a group. Since
$\frac{B^{2}_{1,2}+B_{0,1}}{2}\not\in$Pic$({\mathbb{F}}_{2})$, this is a
contradiction.

We assume that the numerical class of $B$ is (286) of the list in section 6.
We denote $B$ by $2B_{1,0}+3B^{1}_{1,4}+6B^{2}_{1,4}$. By Theorem 2.8,
$G_{1,0}\cong\mathbb{Z}/2\mathbb{Z}$,
$G^{1}_{1,4}\cong\mathbb{Z}/3\mathbb{Z}$,
$G^{2}_{1,4}\cong\mathbb{Z}/6\mathbb{Z}$, and $G=G^{1}_{1,4}\oplus
G^{2}_{1,4}$. Let $s$ be a generator of $G^{1}_{1,4}$. Since
$(B^{1}_{1,4}\cdot B^{1}_{1,4})=4$, the genus of $C^{1}_{1,4}$ is $5$ where
$p^{\ast}B^{1}_{1,4}=3C^{1}_{1,4}$. Since $G_{1,0}\cong\mathbb{Z}/2\mathbb{Z}$
and $(B^{1}_{1,4}\cdot B^{2}_{1,4})\not=0$, the only curve of Fix$(s)$ is
$C^{1}_{1,4}$. By [References,References], this is a contradiction.

We assume that the numerical class of $B$ is (287) of the list in section 6.
We denote $B$ by $2B_{1,0}+4B^{1}_{1,4}+4B^{2}_{1,4}$. By Theorem 2.8,
$G^{i}_{1,4}\cong\mathbb{Z}/4\mathbb{Z}$ for $i=1,2$. Since $(B^{1}_{1,4}\cdot
B^{2}_{1,4})\not=0$, by Theorem 2.8, $G=G^{1}_{1,4}\oplus G^{2}_{1,4}$. Let
$s\in G^{1}_{1,4}$ and $t\in G^{2}_{1,4}$ be generators. Then non-symplectic
involutions of $G$ are $s^{2}$ and $t^{2}$. By Theorem 2.8, we may assume that
Fix$(s^{2})=p^{-1}(B_{1,0})\cup p^{-1}(B^{1}_{1,4})$ and
Fix$(t^{2})=p^{-1}(B^{2}_{1,4})$. For a symplectic involution $s^{2}\circ
t^{2}$, since $X/G$ is smooth, Fix$(s^{2}\circ t^{2})\subset{\rm
Fix}(s^{2})\cap{\rm Fix}(t^{2})$. Since $(C\cdot B^{i}_{1,4})=0$ and
$(B^{1}_{1,4}\cdot B^{1}_{1,4})=4$, we get that $p^{-1}(B_{1,0}\cup
B^{1}_{1,4})\cap p^{-1}(B^{2}_{1,4})$ are 4 points. By the fact that the fixed
locus of a symplectic involution of a $K3$ surface are 8 isolated points, this
is a contradiction.

We assume that the numerical class of $B$ is (305) of the list in section 6.
We denote $B$ by $3B_{1,0}+2B^{1}_{1,6}+6B^{2}_{1,6}$. By Theorem 2.8 and
$(B^{1}_{1,6}\cdot B^{2}_{1,6})\not=0$, $G=G^{1}_{1,6}\oplus G^{2}_{1,6}$. Let
$\rho_{1},\rho_{2}\in G$ be generators of $G_{B^{1}_{1,6}}$ and
$G_{B^{2}_{1,6}}$ respectively. Then $\rho^{2}_{2}$ is a non-symplectic
automorphism of order 3 and a generator of $G_{1,0}$. Since $(C\cdot C)=-6$
and $|G|=12$, we get that $p^{\ast}C=\sum_{j=1}^{4}3C_{j}$ where $C_{j}$ is a
smooth rational curve. Then
$C_{1},\ldots,C_{4},C^{2}_{1,6}\subset$Fix$(\rho^{2}_{2})$ where
$p^{\ast}B^{2}_{1,6}=6C^{2}_{1,6}$. By [References,References], this is a
contradiction.

We assume that the type of $B$ is (45) of the list in section 6. We denote $B$
by $4B^{1}_{1,0}+4B^{2}_{1,0}+2B_{1,3}+2B_{0,1}$. We take the Galois cover
$q:{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}\rightarrow{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$
whose branch divisor is $4B^{1}_{1,0}+4B^{2}_{1,0}$. Since the support of $B$
is simple normal crossing, $q^{\ast}(2B_{1,3}+2B_{0,1})=2B_{4,3}+2B_{0,1}$. By
Theorem 2.4, there is the Galois morphism
$g:X\rightarrow{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ such that the branch
divisor is $2B_{4,3}+2B_{0,1}$ and the Galois group is abelian. Since the
numerical class of $2B_{4,3}+2B_{0,1}$ is (25), this is a contradiction.

As for the case of (45), the numerical class of $B$ is not one of
(46,55,57,59,60,65,66,
67,68,102,105, 108,111,116,124,136,138,142,144,149,153,178,186,221,
222,226,260,
267) of the list in section 6 by
(25,24,27,25,37,34,40,34,34,212,213,214,215,216,217,
286,286,287,287,305, 228, 241,243,286,287,303,305,308) in order.

Therefore, we get Theorem 1.6.

##  4\. Abelian groups of K3 surfaces with smooth quotient

In this section, first of all, we will show Theorem 1.1 and 1.2. Next, we will
show Theorem 1.5. By Section 3, we had that if $X/G$ is ${\mathbb{P}}^{2}$ or
${\mathbb{F}}_{n}$, then $G$ is one of ${\mathcal{A}G}$ as a group.

######  Proposition 4.1.

Let $X$ be a $K3$ surface and $G$ be a finite subgroup of Aut$(X)$ such that
$X/G$ is a smooth rational surface. For a birational morphism
$f:X/G\rightarrow{\mathbb{F}}_{n}$, we get that $0\leq n\leq 12$.

###### Proof.

Let $f:X/G\rightarrow{\mathbb{F}}_{n}$ be a birational morphism, $e_{i}$ be
the exceptional divisors for $i=1,\ldots,m$, and $B=\sum_{i=1}^{k}b_{i}B_{i}$
be the branch divisor. Since $X/G$ and ${\mathbb{F}}_{n}$ are smooth and $f$
is a birational morphism, we get ${\rm Pic}(X/G)=f^{\ast}{\rm
Pic}({\mathbb{F}}_{n})\bigoplus_{i=1}^{m}{\mathbb{Z}}e_{i}$ and there are
positive integers $a_{i}$ for $i=1,\ldots,m$ such that
$K_{X/G}=f^{\ast}K_{{\mathbb{F}}_{n}}+\sum_{i=1}^{m}a_{i}e_{i}$. By Theorem
2.7,

$0=f^{\ast}K_{{\mathbb{F}}_{n}}+\sum_{i=1}^{m}a_{i}e_{i}+\sum_{i=1}^{k}\frac{b_{i}-1}{b_{i}}B_{i}.$

Since ${\rm Pic}(X/G)=f^{\ast}{\rm
Pic}({\mathbb{F}}_{n})\bigoplus_{i=1}^{m}{\mathbb{Z}}e_{i}$, at least one of
$B_{1},\ldots,B_{k}$ is not an exceptional divisor of $f$. By rearranging if
necessary, we assume that $B_{i}$ is not an exceptional divisor of $f$ for
$1\leq i\leq u$, and $B_{j}$ is an exceptional divisor of $f$ for $u+1\leq
j\leq k$. Then $f_{\ast}B_{i}$ is an irreducible curve on ${\mathbb{F}}_{n}$
for $1\leq i\leq u$. Therefore, for $1\leq i\leq u$, there are non-negative
integers $c_{i},d_{i},g^{i}_{j}$ such that

$B_{i}=f^{\ast}(c_{i}C+d_{i}F)-\sum_{j=1}^{m}g^{i}_{j}e_{j}\ {\rm in\
Pic}(X/G)$

where $(c_{i},d_{i})=(1,0)$, $(0,1)$, or $d_{i}\geq c_{i}n>0$. Since
$K_{{\mathbb{F}}_{n}}=-2C-(n+2)F$ in Pic$({\mathbb{F}}_{n})$, by Theorem 2.7,
we get that $2=\sum_{i}\frac{b_{i}-1}{b_{i}}c_{i}$ and
$n+2=\sum_{i}\frac{b_{i}-1}{b_{i}}d_{i}$. In the same way as Theorem 3.4, we
get this proposition. ∎

Let $X$ be a $K3$ surface, $G$ be a finite subgroup of Aut$(X)$ such that
$X/G$ is smooth, and $f:X/G\rightarrow{\mathbb{F}}_{n}$ be a birational
morphism. By Proposition 4.1, we get $0\leq n\leq 12$. By the proof of
Proposition 4.1, the numerical class of $f_{\ast}B$ is one of the list on
Section 3. Let $B=\sum_{i=1}^{k}b_{i}B_{i}+\sum_{j=k+1}^{l}b_{j}B_{j}$ where
$B_{i}$ is not an exceptional divisor of $f$ for $i=1,\ldots,k$ and $B_{j}$ is
an exceptional divisor of $f$ for $j=k+1,\ldots,l$. Since
$(X/G)\backslash\bigcup_{j=k+1}^{l}B_{j}$ is isomorphic to
${\mathbb{F}}_{n}\backslash\bigcup_{j=k+1}^{l}f(B_{j})$ and $f(B_{j})$ is a
point for $j=k+1,\ldots l$, $(X/G)\backslash\bigcup_{j=k+1}^{l}B_{j}$ is
simply connected. By Theorem 2.8, $G$ is generated by $G_{1},\ldots,G_{k}$.
Therefore, as for the case of Hirzebruch surface, we will guess $G$ from the
numerical class of $f_{\ast}B$. Recall that if $G$ is abelian, then $G_{i}$ is
a cyclic group, which is generated by a purely non-symplectic automorphism of
order $b_{i}$. If $f_{\ast}B_{1}=C$, or $F$, then $G$ is generated by
$G_{2},\ldots,G_{k}$, and If $(f_{\ast}B_{1},f_{\ast}B_{2})=(C,F)$, then $G$
is generated by $G_{3},\ldots,G_{k}$.

Recall that since $X/G$ is a smooth rational, $X/G$ is given by blowups of
${\mathbb{F}}_{n}$. Next, we will investigate the relationship between a
branch divisor and exceptional divisors of blow-ups.

######  Lemma 4.2.

Let $X$ be a $K3$ surface, and $G\subset{\rm Aut}(X)$ a finite subgroup such
that $X/G$ is a smooth rational surface, and $B$ be the branch divisor of the
quotient map $p:X\rightarrow X/G$. For a birational morphism $h:X/G\rightarrow
T$ where $T$ is a smooth projective surface, let $e_{i}$ be the exceptional
divisor of $h$ for $i=1,\ldots,m$. Then for $i=1,\ldots,m$ we have that
$h(e_{i})\in{\rm Supp}(h_{\ast}B)$.

###### Proof.

Let $e_{1},\ldots,e_{m}$ be the exceptional divisors of $h$. Since $X/G$ and
$T$ are smooth and $h$ is birational, ${\rm Pic}(X/G)=h^{\ast}{\rm
Pic}(T)\bigoplus_{j=1}^{m}{\mathbb{Z}}e_{j}$ and there are positive integers
$a_{i}$ such that

$K_{X/G}=h^{\ast}K_{T}+\sum_{i=1}^{m}a_{i}e_{i}.$

We assume that $h(e_{i})\not\in{\rm Supp}(h_{\ast}B)$ for some $1\leq i\leq
m$. For simply, we assume that $i=1$, i.e. $h(e_{1})\not\in{\rm
Supp}(h_{\ast}B)$. Let $B_{1},\ldots,B_{k}$ be irreducible components of $B$
such that $B_{j}$ is not an exceptional divisor of $h$ for $j=1,\ldots,k$.
Since $h(e_{1})\not\in{\rm Supp}(h_{\ast}B)$, there are integers $c_{j,s}$
such that $B_{j}=h^{\ast}C_{j}+\sum_{s=2}^{m}c_{j,s}e_{s}$, where $C_{j}$ is
an irreducible curve in $T$. By Theorem 2.7, we get that

$0=(h^{\ast}K_{T}+\sum_{i=1}^{m}a_{i}e_{i})+\sum_{j=1}^{k}\frac{b_{j}-1}{b_{j}}(h^{\ast}C_{j}+\sum_{s=2}^{m}c_{j,s}e_{s})+\sum_{j=1}^{m}l_{j}e_{j}\
{\rm in\ Pic}(X/G),$

where $l_{j}=0$ or $l_{j}=\frac{d_{j}-1}{d_{j}}$ for some an integer
$d_{j}\geq 2$. Since $a_{i}\geq 1$, $c_{j,1}=0$, $l_{j}\geq 0$, and ${\rm
Pic}(X/G)=h^{\ast}{\rm Pic}(T)\bigoplus_{j=1}^{m}{\mathbb{Z}}e_{j}$, this is a
contradiction. ∎

######  Proposition 4.3.

Let $X$ be a $K3$ surface, $G\subset{\rm Aut}(X)$ a finite subgroup such that
the quotient space $X/G$ is smooth, and $B$ be the branch divisor of the
quotient morphism $p:X\rightarrow X/G$. Let $f:X/G\rightarrow T$ be a
birational morphism where $T$ is a smooth surface, $e_{1},\ldots,e_{m}$ be the
exceptional divisors of $f$, and
$f_{\ast}B:=\sum_{i=1}^{u}b_{i}\widetilde{B_{i}}$ where $\widetilde{B_{i}}$ is
an irreducible curves on $U$ for $i=1,\ldots,u$. If $\widetilde{B_{i}}$ is
smooth for each $1\leq i\leq u$, then for $1\leq j\leq m$ there are $1\leq
s<t\leq u$ such that $f(e_{j})\in\widetilde{B_{s}}\cap\widetilde{B_{t}}$.

###### Proof.

We set $B=\sum_{i=1}^{u}b_{i}B_{i}+\sum_{j=u+1}^{k}b_{j}B_{j}$ where $B_{i}$
is not an exceptional divisor of $f$ for $i=1,\ldots,u$, and $B_{j}$ is an
exceptional divisor of $f$ for $j=u+1,\ldots,k$. Then
$f_{\ast}B=\sum_{i=1}^{u}b_{i}f_{\ast}B_{i}$. We assume that $f_{\ast}B_{i}$
is a smooth curve for $i=1,\ldots,u$. By Lemma 4.2, $f(e_{i})\in{\rm
supp}(f_{\ast}B)$ for $i=1,\ldots,m$. Let $S:=X/G$,
$Z:=\\{f(e_{1}),\ldots,f(e_{m})\\}:=\\{z_{1},\ldots,z_{v}\\}\subset T$ where
$v:=|\\{f(e_{1}),\ldots,f(e_{m})\\}|$, $q:{\rm Blow}_{Z}T\rightarrow T$ be the
blow-up, and $E_{i}:=q^{-1}(z_{i})$ be the exceptional divisor of $q$ for
$1\leq i\leq v$. Then there is a birational morphism $g:S\rightarrow{\rm
Blow}_{Z}T$ such that $f=q\circ g$, i.e. the following diagram is commutative:

$\textstyle{{\rm
Blow}_{Z}T\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\textstyle{T}$$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces.\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\scriptstyle{f}$

By changing the number if necessary, we assume that $g(e_{i})=E_{i}$ for
$1\leq i\leq v$. Then the exceptional divisors of $g$ are
$e_{v+1},\ldots,e_{m}$. Since ${\rm Pic}({\rm Blow}_{Z}T)=q^{\ast}{\rm
Pic}(T)\bigoplus_{j=1}^{v}{\mathbb{Z}}E_{j}$ and $f=q\circ g$,

${\rm Pic}(S)=g^{\ast}{\rm Pic}({\rm
Blow}_{Z}T)\bigoplus_{j=v+1}^{m}{\mathbb{Z}}e_{j}=\Bigl{(}f^{\ast}{\rm
Pic}(T)\bigoplus_{i=1}^{v}{\mathbb{Z}}g^{\ast}E_{i}\Bigr{)}\bigoplus_{j=v+1}^{m}{\mathbb{Z}}e_{j}.$

Since $K_{{\rm Blow}_{Z}T}=q^{\ast}K_{T}+\sum_{j=1}^{v}E_{j}$,

$K_{S}=g^{\ast}K_{{\rm
Blow}_{Z}T}+\sum_{i=v+1}^{m}a^{\prime}_{i}e_{i}=\Bigl{(}f^{\ast}K_{T}+\sum_{j=1}^{v}g^{\ast}E_{i}\Bigr{)}+\sum_{i=v+1}^{m}a^{\prime}_{i}e_{i}$

where $a^{\prime}_{i}$ is a positive integer for $i=v+1,\ldots,m$.

We assume that for some $1\leq i\leq m$, $f(e_{i})\not\in f_{\ast}B_{s}\cap
f_{\ast}B_{t}$ for each $1\leq s<t\leq u$. Since
$Z=\\{f(e_{1}),\ldots,f(e_{v})\\}$, we assume that $1\leq i\leq v$. for
simplicity, we assume that $i=1$. In addition, since $f(e_{j})\in{\rm
supp}(f_{\ast}B)$ for $j=1,\ldots,m$, by changing the number if necessary, we
assume that $f(e_{1})\in{\rm supp}(f_{\ast}B_{1})$, and $f(e_{1})\not\in{\rm
supp}(f_{\ast}B_{j})$ for $2\leq j\leq u$. Recall that the exceptional
divisors of $q$ are $E_{1},\ldots,E_{v}$, the exceptional divisors of $g$ are
$e_{v+1},\ldots,e_{m}$, and $g(e_{i})=E_{i}$ for $1\leq i\leq v$. Since
$f=q\circ g$, for $j=1,\ldots,u$ there are non-negative integers
$c_{j,s},c^{\prime}_{j,t}$ such that

$B_{j}=f^{\ast}f_{\ast}B_{j}-\sum_{s=1}^{v}c_{j,s}g^{\ast}E_{s}-\sum_{t=v+1}^{m}c^{\prime}_{j,t}e_{t}\
\ {\rm in\ Pic}(S).$

Since $f(e_{1})\not\in f_{\ast}B_{j}$ for $2\leq j\leq u$, we get that
$c_{j,1}=0$ for $2\leq j\leq u$. Since $f_{\ast}B_{1}$ is smooth, $c_{1,1}=1$.
Since
$K_{S}=f^{\ast}K_{T}+\sum_{j=1}^{v}g^{\ast}E_{i}+\sum_{i=v+1}^{m}a^{\prime}_{i}e_{i}$
and $0=K_{S}+\sum_{i=1}^{k}\frac{b_{i}-1}{b_{i}}B_{i}$ in Pic$(S)$,

$\begin{split}0=&(f^{\ast}K_{T}+\sum_{j=1}^{v}g^{\ast}E_{i}+\sum_{i=v+1}^{m}a^{\prime}_{i}e_{i})\\\
&+\sum_{i=1}^{u}\frac{b_{i}-1}{b_{i}}(f^{\ast}f_{\ast}B_{j}-\sum_{s=1}^{v}c_{j,s}g^{\ast}E_{s}-\sum_{t=v+1}^{m}c^{\prime}_{j,t}e_{t})\\\
&+\sum_{j=u+1}^{k}\frac{b_{j}-1}{b_{j}}B_{j}\ \ {\rm in\ Pic}(S).\end{split}$

From the coefficient of $g^{\ast}E_{1}$, we get that $1=\frac{b_{1}-1}{b_{1}}$
Since $b_{1}\geq 2$, this is a contradiction. ∎

Let $X$ be a $K3$ surface, $G$ be a finite subgroup of Aut$(X)$ such that
$X/G$ is a smooth rational surface, and $B$ be the branch divisor of the
quotient map $p:X\rightarrow X/G$. Let $h:X/G\rightarrow T$ be a birational
morphism where $T$ is a smooth projective surface, and $e_{1},\ldots,e_{m}$ be
the exceptional divisors of $h$. We set
$h_{\ast}B:=\sum_{j=1}^{l}b_{j}B^{\prime}_{j}$. We write
$B=\sum_{i=1}^{l}b_{i}B_{i}+\sum_{j=l+1}^{k}b_{j}B_{j}$ such that
$h_{\ast}B_{i}=B^{\prime}_{i}$ for $i=1,\ldots,l$. Then $B_{j}$ is one of the
exceptional divisor of $h$ for $j=l+1,\ldots,k$, and for $i=1,\ldots,l$ there
are non-negative integers $c_{i,1},\ldots,c_{i,m}$ such that
$B_{i}=h^{-1}_{\ast}B^{\prime}_{i}-\sum_{t=1}^{m}c_{i,t}e_{t}$.

######  Remark 4.4.

In the above situation, for $e_{u}$ and $e_{v}$ where $1\leq u<v\leq m$ and
$h(e_{u})=h(e_{v})$, we get that $c_{i,u}=0$ if and only if $c_{i,v}=0$.

######  Remark 4.5.

In the situation of Proposition 4.3, we assume that $T={\mathbb{F}}_{n}$. Then
there are positive integers $a_{1},\ldots,a_{m}$ such that
$K_{X/G}=h^{\ast}K_{{\mathbb{F}}_{n}}+\sum_{i=1}^{m}a_{i}e_{i}$. By the proof
of Proposition 4.3, we get that $a_{1}=\cdots=a_{u}=1$ and

$1+\frac{\beta_{i}-1}{\beta_{i}}=\sum_{j=1}^{k}\frac{b_{j}-1}{b_{j}}c_{i,j}\
for\ i=1,\ldots,u,$

where $\beta_{i}=1$ if $e_{i}$ is not an irreducible component of $B$, and
$\beta_{i}$ is the ramification index at $e_{i}$ if $e_{i}$ is an irreducible
component of $B$.

Furthermore, we assume that $X/G\not={\rm
Blow}_{\\{h(e_{1}),\ldots,h(e_{u})\\}}{\mathbb{F}}_{n}$. For the birational
morphism $g:X/G\rightarrow{\rm
Blow}_{\\{h(e_{1}),\ldots,h(e_{u})\\}}{\mathbb{F}}_{n}$ in the proof of
Proposition 4.3, we rearrange the order so that
$\\{g(e_{u+1}),\ldots,g(e_{u+v})\\}=\\{g(e_{u+1}),\ldots,g(e_{m})\\}$, where
$v:=|\\{g(e_{u+1}),\ldots,g(e_{m})\\}|$. Like the proof of Proposition 4.3, by
considering the blow-up of ${\rm
Blow}_{\\{h(e_{1}),\ldots,h(e_{u})\\}}{\mathbb{F}}_{n}$ at
$\\{g(e_{u+1}),\ldots,g(e_{u+v})\\}$, we get that $a_{u+1}=\cdots=a_{u+v}=2$
and

$2+\frac{\beta_{i}-1}{\beta_{i}}=\sum_{j=1}^{k}\frac{b_{j}-1}{b_{j}}c_{i,j}\
for\ i=u+1,\ldots,u+v,$

where $\beta_{i}=1$ if $e_{i}$ is not an irreducible component of $B$, and
$\beta_{i}$ is the ramification index at $e_{i}$ if $e_{i}$ is an irreducible
component of $B$.

Recall that by Theorem 2.8, $G_{B_{i}}$ is generated by a non-symplectic
automorphism of order $b_{i}$. As a corollary of Theorem 2.8 and Proposition
4.3, we get the following Theorem 4.6.

######  Theorem 4.6.

Let $X$ be a $K3$ surface, $G$ be a finite subgroup of Aut$(X)$ such that
$X/G$ is smooth, and $B$ be the branch divisor of the quotient map
$p:X\rightarrow X/G$. Let $f:X\rightarrow S$ be the birational morphism where
$S$ minimal rational surface. We put $f_{\ast}B:=\sum_{i=1}^{k}b_{i}B_{i}$
where $B_{i}$ is an irreducible curve for $i=1,\ldots,k$. We denote by $G_{s}$
the subgroup of $G$, which consists of symplectic automorphisms of $G$, and
$b$ the least common multiple of $b_{1},\ldots,b_{k}$. Then there is a purely
non-symplectic automorphism $g\in G$ of order $b$ such that $G$ is the
semidirect product $G_{s}\rtimes\langle g\rangle$ of $G_{s}$ and $\langle
g\rangle$.

###### Proof.

Since $G_{s}$ is a normal subgroup of $G$ and $G/G_{s}$ is a cyclic group, in
order to show Theorem 4.6, we only show that there is a purely non-symplectic
automorphism $g\in G$ of order $b$.

First of all, we assume that $X/G\cong{\mathbb{P}}^{2}$. We put
$B:=\sum_{i=1}^{k}b_{i}B_{i}$ where $B_{i}$ is an irreducible curve for
$i=1,\ldots,k$. By Theorem 2.7, $0=\sum_{i=1}^{k}\frac{b_{i}-1}{b_{i}}{\rm
deg}\,B_{i}+{\rm deg}\,K_{\mathbb{P}^{2}}$, in which $K_{\mathbb{P}^{2}}$ is
the canonical line bundle of $\mathbb{P}^{2}$. Since the degree of
$K_{{\mathbb{P}}^{2}}$ is $-3$ and $\frac{1}{2}\leq\frac{l-1}{l}<1$ for any
positive integer $l$, we get that $4\leq\sum_{i=1}^{k}{\rm deg}B_{i}\leq 6$.
If $\sum_{i=1}^{k}{\rm deg}B_{i}=6$, then $b_{1}=\cdots=b_{k}=2$. By Theorem
2.8, in this case the statement of theorem is established. We assume that
$\sum_{i=1}^{k}{\rm deg}B_{i}\leq 5$. By [References, Theorem 2], $b=b_{i}$
for some $1\leq i\leq k$ or $b=l.c.m(b_{i},b_{j})$ for $i<j$. By Theorem 2.8,
in the former case, we get this theorem.

For the latter, i.e. if $b\not=b_{i}$ for $1\leq i\leq k$, then $B$ is one of
(i) $3L_{1}+3L_{2}+3L_{3}+2L_{4}+2L_{5}$ where $L_{3}$ pass through the points
$L_{1}\cap L_{2}$ and $L_{4}\cap L_{5}$ (see [References, pp. 408]), (ii)
$3L_{1}+3L_{2}+3L_{3}+2Q$ where $L_{1},L_{2}$ are the tangent to $Q$ and
$L_{3}$ is in general position with respect to $L_{1}\cup L_{2}\cup Q$ (see
[References, pp. 408]), and (iii) $2L_{1}+2L_{2}+3L_{3}+3Q$ where
$L_{1},L_{2},L_{3}$ are three distinct tangent lines to $Q$ (see [References,
pp. 410]). Here, $L_{i}$ and $Q$ are smooth curves on ${\mathbb{P}}^{2}$ with
deg $L_{i}=1$ and deg $Q=2$ for $i=1,\ldots,5$. Then there are $1\leq i<j\leq
k$ such that $b=l.c.m(b_{i},b_{j})$, $B_{i}+B_{j}$ is simple normal crossing,
and $(B_{i}\cap B_{j})\backslash\cup_{s\not=i,j}B_{s}$ is not an empty set.
For clarity, we may assume that $i=1$, $j=2$. We take one point
$y\in(B_{1}\cap B_{2})\backslash\cup_{i=3}^{k}B_{i}$. Let $x\in p^{-1}(y)$. By
the assumption for $y$ and Theorem 2.3, there are open subset
$V\subset{\mathbb{P}}^{2}$ and $U\subset X$ such that $y\in V$, $x\in U$,
$p_{|U}:U\rightarrow V$ is isomorphic to $\\{z\in{\mathbb{C}}^{2}:\
|z|<1\\}\ni(z_{1},z_{2})\mapsto(z^{b_{1}}_{1},z^{b_{2}}_{2})\in\\{z\in{\mathbb{C}}^{2}:\
|z|<1\\}$, and hence $G_{x}:=\\{g\in
G\,|\,g(x)=x\\}\cong\mathbb{Z}/b_{1}\mathbb{Z}\oplus\mathbb{Z}/b_{2}\mathbb{Z}$.
Since $b=l.c.m(b_{1},b_{2})$, there is a purely non-symplectic automorphism
$g\in G$ with order $b$.

Next, we assume that $X/G\cong{\mathbb{F}}_{n}$. By the list of the numerical
class of $B$ in Section 6, if the numerical class of $B$ is not one of
(65,70,73,77,83,92,102,127,128,132,136, 143,153,154,170,235,251,252,253), then
$b=b_{i}$ for some $1\leq i\leq k$. Therefore, by Theorem 2.8, we get this
theorem. If the numerical class of $B$ is one of
(65,70,73,77,92,127,128,132,136,143, 153,154,170,235,251,252, 253), then there
are $1\leq i<j\leq k$ such that $b=l.c.m(b_{i},b_{j})$, $B_{i}+B_{j}$ is
simple normal crossing, and $(B_{i}\cap B_{j})\backslash\cup_{s\not=i,j}B_{s}$
is not an empty set. As for the case of $\mathbb{P}^{2}$, we get this theorem.

We assume that the numerical class of $B$ is (83). We write
$B=3B_{3,3}+2B_{0,1}^{1}+2B_{0,1}^{2}$. Since $B^{1}_{0,1}\cap B^{2}_{0,1}$ is
an empty set, if $B_{3,3}\cap B^{1}_{0,1}$ is not one point, then by
$(B_{3,3}\cdot B^{1}_{0,1})=3$, there is a point $y\in B_{3,3}\cap
B^{1}_{0,1}$ such that the support of $B$ is simple normal crossing at $y$.
Since $b=6$, by Theorem 2.8, we get this theorem. Therefore, we assume that
$B_{3,3}\cap B^{1}_{0,1}$ and $B_{3,3}\cap B^{2}_{0,1}$ are one point. Let
$q:X/G_{s}\rightarrow X/G$ be the quotient map. Then the singular locus of
$X/G_{s}$ is $q^{-}1(B_{3,3}\cap B^{1}_{0,1})\cup q^{-}1(B_{3,3}\cap
B^{2}_{0,1})$. Since the Galois group of $q$ is
$G/G_{s}\cong{\mathbb{Z}}/6{\mathbb{Z}}$, the branch divisor of $q$ is $B$,
and $B_{3,3}\cap B^{1}_{0,1}$ and $B_{3,3}\cap B^{2}_{0,1}$ are one point,
$X/G_{s}$ has just two singular point. By [References, Theorem 3], this is a
contradiction. Therefore, if the numerical class of $B$ is (83), then we get
this theorem. As for the case of (83), we get this theorem for (102).

Finally, we assume that $X/G$ is not $\mathbb{P}^{2}$ or $\mathbb{F}_{n}$. We
take a birational morphism $f:X/G\rightarrow{\mathbb{F}}^{n}$ where $0\leq n$.
Let $e_{1},\ldots,e_{m}$ be the exceptional divisors of $f$. In the same way
of the case where $X/G\cong\mathbb{P}^{2}$ or $\mathbb{F}_{n}$, we only
consider the case that the numerical class of $f_{\ast}B$ is one of
(65,70,73,77,83,92, 102,127,128,132,136, 143, 153,154,170,235,251,252, 253).

We assume that the numerical class of $f_{\ast}B$ is (65). By Remark 4.5,
there are positive integers $a_{1},\ldots,a_{5},b$ such that

$1+\frac{b-1}{b}=\frac{2}{3}a_{1}+\frac{5}{6}a_{2}+\frac{1}{2}a_{3}+\frac{3}{4}a_{4}+\frac{3}{4}a_{5}.$

Since the numerical class of $f_{\ast}B$ is (65), we may assume that $a_{1}$
or $a_{2}$ is 0, and either $a_{4}$ or $a_{5}$ is 0. However, there are not
such positive integers. Therefore, the numerical class of $f_{\ast}B$ is not
(65). As for the case of (65), the numerical class of $B$ is not one of
(73,77,128,132,170,235,251,253).

We assume that the numerical class of $f_{\ast}B$ is (70). By Remark 4.5,
there are positive integers $a_{1},\ldots,a_{6},b$ such that

$1+\frac{b-1}{b}=\frac{1}{2}a_{1}+\frac{2}{3}a_{2}+\frac{5}{6}a_{3}+\frac{1}{2}a_{4}+\frac{3}{4}a_{5}+\frac{3}{4}a_{6}.$

Since the numerical class of $f_{\ast}B$ is (70), we may assume that two of
$a_{1}$, $a_{2}$ and $a_{3}$ are 0, and two of $a_{4}$, $a_{5}$ and $a_{6}$
are 0. The integers satisfying the above conditions is only
$(a_{1},\ldots,a_{6},b)=(1,0,0,1,0,0,12)$. Therefore, for
$B:=\sum_{j=1}^{l}B_{j}B_{j}$, $b_{i}=12$ for some $1\leq i\leq l$. By Theorem
2.8, if the numerical class of $f_{\ast}B$ is (65), then we get this theorem.
As for the case of (70), if the numerical class of $B$ is one of (136,143),
then we get this theorem.

We assume that the numerical class of $B$ is (83). By Remark 4.5, there are
positive integers $a_{1},\ldots,a_{6},b$ such that

$1+\frac{b-1}{b}=\frac{2}{3}a_{1}+\frac{1}{2}a_{2}+\frac{1}{2}a_{3}.$

Since the numerical class of $f_{\ast}B$ is (83), we may assume that either
$a_{2}$ or $a_{3}$ is $0$. The integers satisfying the above conditions is
$(a_{1},a_{2},a_{3},b)=(2,1,0,6)$ or $(2,0,1,6)$. Therefore, we get this of
theorem. As for the case of (83), if the numerical class of $B$ is one of
(92,102,127,153,154,252), then we get this theorem. ∎

######  Theorem 4.7.

Let $X$ be a $K3$ surface and $G$ be a finite subgroup of Aut$(X)$ such that
$X/G$ is smooth. For a birational morphism $f:X/G\rightarrow{\mathbb{F}}_{n}$
where $0\leq n$, we get that $n$ is not one of $5,7,9,10,11$.

###### Proof.

Let $p:X\rightarrow X/G$ be the quotient map, and
$B:=\sum_{i=1}^{k}b_{i}B_{i}$ be the branch divisor of $p$. Let
$f:X/G\rightarrow{\mathbb{F}}_{n}$ be a birational morphism where $0\leq n$,
and $e_{1},\ldots,e_{m}$ be the exceptional divisors of $f$.

First we will show this theorem for the cases where $f$ is an isomorphism,
i.e. $X/G\cong{\mathbb{F}}_{n}$. By Theorem 2.7, $n=0,1,2,3,4,5,6,7,8,9$, or
$12$. We assume that $n=5,7,$ or $9$. Then the numerical class of $B$ is one
of (296,297,298,299,300, 301,310,314,315) of the list in section 6.

We assume that the numerical class of $B$ is (296). We denote $B$ by
$4B_{1,0}+2B_{1,5}+4B_{1,6}$. Let $p^{\ast}B_{1,0}=\sum_{i=1}^{m}4C_{i}$ where
$C_{i}$ is a smooth curve for $i=1,\ldots,m$. Since $(B_{1,0}\cdot
B_{1,0})<0$, $(C_{i}\cdot C_{i})<0$. Since $X$ is a $K3$ surface, and $C_{i}$
is irreducible, we get that $(C_{i}\cdot C_{i})=-2$. Since the degree of $p$
is $|G|$, and $(B_{1,0}\cdot B_{1,0})=-5$, we get that
$\frac{-5|G|}{16}=-2m+2\sum_{1\leq i<j\leq m}(C_{i}\cdot C_{j})$, and hence
$\frac{5|G|}{32}\leq m$. Let $p^{\ast}B_{1,6}=\sum_{j=1}^{l}4C^{\prime}_{j}$
where $C^{\prime}_{j}$ is a smooth curve for $j=1,\ldots,l$. Since
$(B_{1,0}\cdot B_{1,6})=1$,
$\frac{|G|}{16}=m(C_{1}\cdot\sum_{j=1}^{l}C^{\prime}_{j})$. Since
$(C_{1}\cdot\sum_{j=1}^{l}C^{\prime}_{j})\geq 1$, we get that
$m\leq\frac{|G|}{16}$. By $\frac{5|G|}{32}\leq m$ and $m\leq\frac{|G|}{16}$,
we get that the numerical class of $B$ is not (296). As for the case of (296),
the numerical class of $f_{\ast}B$ is not one of (297,298,299,300,
301,310,314,315). Therefore, if $X/G\cong{\mathbb{F}}_{n}$, then
$n\not=5,7,9,10,11$.

Next, we assume that $f$ is not an isomorphism, i.e. $X/G$ is not a Hirzebruch
surface $\mathbb{F}_{n}$. By Theorem 4.1, $n=0,1,2,3,4,5,6,7,8,9$, or $12$. We
assume that $n=5,7,$ or $9$. The numerical class of $f_{\ast}B$ is one of
(296,297,298,299,300, 301,310,314,315).

We assume that the numerical class of $f_{\ast}B$ is (296). Let
$p^{\ast}B_{1,0}=4\sum_{i=1}^{m}C_{i}$ where $C_{i}$ is a smooth curve for
$i=1,\ldots,m$. Since the degree of $p$ is $|G|$, by $(C\cdot F)=1$, we get
that $|G|=4m(C_{1}\cdot p^{\ast}f^{\ast}F)$, and hence $|G|$ is a multiple of
$4m$. Since $f_{\ast}B_{1,0}=C$, $(B_{1,0},B_{1,0})\leq(C\cdot C)=-5$. By
$\frac{|G|}{16}(B_{1,0}\cdot B_{1,0})=-2m+2\sum_{1\leq i<j\leq m}(C_{i}\cdot
C_{j})$, we get that $m=\frac{|G|}{4}$. Since the numerical class of
$f_{\ast}B$ is (296), there must be positive integers $a_{1},a_{2},a_{3},b$
such that

$1+\frac{b-1}{b}=\frac{3}{4}a_{1}+\frac{1}{2}a_{2}+\frac{3}{4}a_{3},$

and either $a_{1}$ or $a_{2}$ is 0. The integers satisfying the above
conditions is only $(a_{1},a_{2},a_{3},b)=(1,0,1,2)$, and hence $f(e_{i})\in
f_{\ast}B_{1,5}\cap f_{\ast}B_{1,6}$ for each $i=1,\ldots,l$. Since
$(f_{\ast}B_{1,5}\cdot f_{\ast}B_{1,6})=1$, $f_{\ast}B_{1,5}\cap
f_{\ast}B_{1,6}$ is one point. We put $x:=f_{\ast}B_{1,5}\cap
f_{\ast}B_{1,6}$. Let $q:{\rm
Blow}_{x}{\mathbb{F}}_{5}\rightarrow{\mathbb{F}}_{5}$ be the blow-up of
${\mathbb{F}}_{5}$ at $x$. Then there is a birational morphism
$g:X/G\rightarrow{\rm Blow}_{x}{\mathbb{F}}_{5}$ such that $f=q\circ g$. Let
$C^{\prime}:=g_{\ast}B_{1,0}$. Let $E$ be the exceptional divisor of $q$.
Since $f(e_{i})=x$ for each $i=1,\ldots,l$, $g(e_{i})\in E$ for each
$i=1,\ldots,l$. Since
$g_{\ast}B=4C^{\prime}+2g_{\ast}B_{1,5}+4g_{\ast}B_{1,6}+2E$, if $g$ is not an
isomorphism, then there must be integers $a_{1},a_{2},a_{3},a_{4},b$ such that

$2+\frac{b-1}{b}=\frac{3}{4}a_{1}+\frac{1}{2}a_{2}+\frac{3}{4}a_{3}+\frac{1}{2}a_{4},$

and if $a_{1}$ is not 0, then either $a_{2}=a_{3}=0$. However, there are not
such positive integers. Therefore, $g$ is an isomorphism, i.e. $X/G={\rm
Blow}_{x}{\mathbb{F}}_{5}$, and hence $B=4B_{1,0}+2B_{1,5}+4B_{1,6}+2E$ and
$(B_{1,0}\cdot E)=1$. We put $p^{\ast}E=2\sum_{j=1}^{u}C^{\prime}_{j}$ where
$C^{\prime}_{j}$ is a smooth curve for $j=1,\ldots,u$. Since
$m=\frac{|G|}{den}$,
$\frac{|G|}{2}=|G|(C_{1}\cdot\sum_{j=1}^{u}C^{\prime}_{j})$. This is a
contradiction. Therefore, the numerical class of $B$ is not (296). As for the
case of (296), the numerical class of $B$ is not one of (310,314).

We assume that the numerical class of $f_{\ast}B$ is (297). Then there must be
integers $a_{1},a_{2},a_{3},a_{4},b$ such that

$1+\frac{b-1}{b}=\frac{3}{4}a_{1}+\frac{1}{2}a_{2}+\frac{3}{4}a_{3}+\frac{3}{4}a_{4},$

and if $a_{1}$ is not zero, then $a_{2}=a_{3}=0$. The integers satisfying the
above condition is $(a_{1},a_{2},a_{3},a_{4},b)=(1,0,0,1,2)$ or $(0,0,1,1,2)$.
Therefore, for each $i=1,\ldots,l$, we get that $f(e_{i})\in
f_{\ast}B_{1,0}\cap f_{\ast}B_{0,1}$ or $f(e_{i})\in f_{\ast}B^{2}_{1,5}\cap
f_{\ast}B_{0,1}$. If $f(e_{i})\in f_{\ast}B^{2}_{1,5}\cap f_{\ast}B_{0,1}$ for
all $i=1,\ldots,l$, then $(B_{1,0}\cdot B_{1,0})=-5$ and $(B_{1,0}\cdot
B_{0,1})=1$. However, as for the case of $X/G\cong{\mathbb{F}}_{n}$, we can
see that such things can not happen. Therefore, $f(e_{i})\in
f_{\ast}B_{1,0}\cap f_{\ast}B_{0,1}$ for some $i=1,\ldots,l$. By using the
blow-up of ${\mathbb{F}}_{5}$ at $x:=f_{\ast}B_{1,0}\cap f_{\ast}B_{0,1}$, as
for the case of (296), this is a contradiction. Therefore, the numerical class
of $B$ is not (297). As for the case of (297), the numerical class of $B$ is
not (315).

We assume that the numerical class of $f_{\ast}B$ is (298). Then there must be
integers $a_{1},a_{2},a_{3},b$ such that

$1+\frac{b-1}{b}=\frac{5}{6}a_{1}+\frac{1}{2}a_{2}+\frac{2}{3}a_{3}.$

The integers satisfying the above condition is only
$(a_{1},a_{2},a_{3},b)=(1,0,1,2)$, and hence $f(e_{i})\in f_{\ast}B_{1,5}\cap
f_{\ast}B_{1,6}$ for each $i=1,\ldots,l$. Since $(f_{\ast}B_{1,5}\cdot
f_{\ast}B_{1,6})=1$, $f_{\ast}B_{1,5}\cap f_{\ast}B_{1,6}$ is one point. We
put $x:=f_{\ast}B_{1,5}\cap f_{\ast}B_{1,6}$. Let $q:{\rm
Blow}_{x}{\mathbb{F}}_{5}\rightarrow{\mathbb{F}}_{5}$ be the blow-up of
${\mathbb{F}}_{5}$ at $x$. As for the case of (296), since there are no
integers $a_{1},a_{2},a_{3},a_{4},b$ such that

$2+\frac{b-1}{b}=\frac{3}{4}a_{1}+\frac{1}{2}a_{2}+\frac{3}{4}a_{3}+\frac{1}{2}a_{4},$

we get that $X/G={\rm Blow}_{x}{\mathbb{F}}_{5}$, and hence
$B=6B_{1,0}+2B_{1,6}+3B_{1,6}+2E$, and $(B_{1,0}\cdot E)=1$. We put
$p^{\ast}E=2\sum_{j=1}^{u}C^{\prime}_{j}$ where $C^{\prime}_{j}$ is a smooth
curve for $j=1,\ldots,u$. Since $(E\cdot E)=-1$, we get that
$u=\frac{|G|}{4}+\sum_{1\leq i<j\leq u}(C^{\prime}_{i}\cdot C^{\prime}_{j})$,
and hence $u\geq\frac{|G|}{4}$. Since $(B_{1,0}\cdot E)=1$, $\frac{|G|}{12}$
is a multiple of $u$. This is a contradiction. Therefore, the numerical class
of $B$ is not (298).

We assume that the numerical class of $f_{\ast}B$ is (299). Then there must be
positive integers $a_{1},a_{2},a_{3},a_{4},b$ such taht

$1+\frac{b-1}{b}=\frac{5}{6}a_{1}+\frac{1}{2}a_{2}+\frac{2}{3}a_{3}+\frac{2}{3}a_{4},$

and $a_{1}a_{3}=0$. The integers satisfying the above conditions is
$(a_{1},a_{2},a_{3},a_{4},b)=(1,0,0,1,2)$ or $(0,1,1,1,6)$. Therefore, for
each $i=1,\ldots,l$, we get that $f(e_{i})\in f_{\ast}B_{1,0}\cap
f_{\ast}B_{0,1}$ or $f(e_{i})\in f_{\ast}B_{1,6}\cap f_{\ast}B_{1,5}\cap
f_{\ast}B_{0,1}$. If $f(e_{i})\in f_{\ast}B_{1,6}\cap f_{\ast}B_{1,5}\cap
f_{\ast}B_{0,1}$ for all $i=1,\ldots,l$, then $(B_{1,0}\cdot B_{1,0})=-5$ and
$(B_{1,0}\cdot B_{0,1})=1$. We get that this is not established in the same
way as in the case of $X/G\cong{\mathbb{F}}_{n}$. By using the blow-up of
${\mathbb{F}}_{5}$ at $x:=f_{\ast}B_{1,0}\cap f_{\ast}B_{0,1}$, as for the
case of (298), we get that there is no case where $f(e_{i})\in
f_{\ast}B_{1,0}\cap f_{\ast}B_{0,1}$ for some $i=1,\ldots,l$. Therefore, the
numerical class of $B$ is not (299). As for the case of (299), the numerical
class of $B$ is not one of (300,301). ∎

######  Corollary 4.8.

Let $X$ be a $K3$ surface and $G$ be a finite subgroup of Aut$(X)$ such that
$X/G$ is smooth. If there is a birational morphism
$f:X/G\rightarrow{\mathbb{F}}_{n}$ from the quotient space $X/G$ to a
Hirzebruch surface ${\mathbb{F}}_{n}$ where $n=6,8$, or $12$, then $f$ is an
isomorphism, i.e. $X/G$ is a Hirzebruch surface.

###### Proof.

Let $n\geq 1$ and $C_{-n}\subset{\mathbb{F}}_{n}$ be the unique irreducible
curve such that $(C_{-n}\cdot C_{-n})=-n$. Since for $x\in{\mathbb{F}}_{n}$,
if $x\in C_{-n}$, then ${\rm Blow}_{x}{\mathbb{F}}_{n}={\rm
Blow}_{y}{\mathbb{F}}_{n+1}$ where $y\in{\mathbb{F}}_{n+1}\backslash
C_{-(n+1)}$, and if $x\not\in C_{-n}$, then ${\rm
Blow}_{x}{\mathbb{F}}_{n}={\rm Blow}_{y}{\mathbb{F}}_{n-1}$ where $y\in
C_{-(n-1)}$, by Theorem 4.7, we get this corollary. ∎

######  Theorem 4.9.

Let $X$ be a $K3$ surface and $G$ be a finite abelian subgroup of ${\rm
Aut}(X)$. If $X/G$ is smooth, then $G$ is isomorphic to one of
${\mathcal{A}G}$ as groups.

###### Proof.

Since $X/G$ is smooth, the quotient space $X/G$ is an Enriques surface or a
rational surface. If $X/G$ is Enriques, then
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}$ as a group and
${\mathbb{Z}}/2{\mathbb{Z}}\in{\mathcal{A}G}$. By Section 3, if
$X/G\cong{\mathbb{F}}_{n}$, then $G$ is isomorphic to one of ${\mathcal{A}G}$
as a group. By [References], if $X/G\cong{\mathbb{P}}^{2}$, then $G$ is
isomorphic to one of ${\mathcal{A}G}$ as a group. Therefore, we assume that
$X/G$ is rational, and $X/G\not={\mathbb{P}}^{2}$ or ${\mathbb{F}}_{n}$.

Let $f:X/G\rightarrow{\mathbb{F}}_{n}$ be a birational morphism where $0\leq
n\leq 12$, and $B$ be the branch divisor of $G$. By Theorem 4.7 and Corollary
4.8, we may assume that $0\leq n\leq 4$. By the proof of Theorem 4.1, the
numerical class of $f_{\ast}B$ is one of the list in section 6.

We assume that the numerical class of $f_{\ast}B$ is one of
(4,5,6,10,11,12,14,15,16,19,20,
25,26,27,28,32,33,36,37,38,41,42,46,51,52,57,58,59,60,79,80,81,82,85,87,88,89,91,94,
96,98,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,176,177,178,180,
181,182,183,184,185,186,187,189,190,191,192,195,196,197,199,200,202,203,206,216,
217,241,242,243,244,245,246,249,250,270,271,272,273,274,275,276,277,279, 282)
of
the list in section 6. By Theorem 2.8, $G$ is generated by automorphisms
$g_{1},\ldots,g_{m}$ where $1\leq m\leq 5$ and the order of $g_{i}$ is two for
$i=1,\ldots,m$. Therefore, $G$ is ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus a}$
where $1\leq a\leq 5$ as a group.

We assume that the numerical class of $f_{\ast}B$ is one of
(1,2,3,17,18,22,23,24,39,54,
55,194,198,201,204,205,212,218,219,228,229,284,285,289,290) of the list in
section 6. By Theorem 2.8, $G$ is generated by automorphisms
$g_{1},\ldots,g_{m}$ where $1\leq m\leq 3$ and the order of $g_{i}$ is 3 for
$i=1,\ldots,m$. Therefore, $G$ is ${\mathbb{Z}}/3{\mathbb{Z}}^{\oplus b}$
where $1\leq b\leq 3$ as a group.

We assume that the numerical class of $f_{\ast}B$ is one of
(29,34,40,44,49,50,53,56,62,63,
64,66,67,68,69,71,77,83,84,92,93,102,106,107,108,127,128,133,134,135,137,138,145,
146,147,148,149,151,153,154,163,164,165,166,167,168,169,174,175,179,188,193,211,
214,220,221,223,224,225,226,227,230,236,237,238,239,240,248,251,252,254,256,258,
259,260,265,266,267,268,269,283,286,288,292,293,294,295) of the list in
section 6. By Theorem 2.8, $G$ is generated by automorphisms
$g_{i},\ldots,g_{m}$, $h_{1},\ldots h_{n}$, where $1\leq m\leq 3$, $1\leq
n\leq 2$, the order of $g_{i}$ is 2 for $i=1,\ldots,m$, and the order of
$h_{j}$ is 3 for $j=1,\ldots,n$. Therefore, $G$ is
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus d}\oplus{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus
e}$ where (d,e)=(1,1),(1,2),(1,3),(2,1),(2,2), (3,1),(3,2) as a group.

We assume that the numerical class of $f_{\ast}B$ is one of
(7,8,9,13,21,30,31,35,43,45,47,
48,61,86,90,
97,99,100,103,104,105,109,110,130,131,139,140,141,142,155,156,157,158,
161,162,207,208,209,210, 213,215,222,231,232,233,234,
255,257,261,262,263,264,278,
280,281,287,291) of the list in section 6. By Theorem 2.8, $G$ is generated by
automorphisms $g_{i},\ldots,g_{m}$, $h_{1},\ldots h_{n}$ where the order of
$g_{i}$ is 2 for $i=1,\ldots,m$, the order of $h_{j}$ is 4 for $j=1,\ldots,n$,
and $(n,m)$ is one of (0,1),(0,2), (0,3), (1,1),(1,2), (2,1),(3,1). Therefore,
$G$ is ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
f}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus g}$ where (f,g)=(0,1),(0,2),
(0,3),(1,1), (1,2),(2,1),(3,1) as a group.

We assume that the numerical class of $f_{\ast}B$ is (65) of the list in
section 6. We denote $B$ by
$3B^{1}_{1,0}+6B^{2}_{1,0}+2B_{1,1}+4B^{1}_{0,1}+4B^{2}_{0,1}+\sum_{j=1}^{l}b^{\prime}_{i}B^{\prime}_{i}$
where $f_{\ast}B^{i}_{1,0}=(1,0)$, $f_{\ast}B^{i}_{0,1}=(0,1)$ in
Pic$({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})$, and $B^{\prime}_{j}$ is an
exceptional divisor of $f$ for $j=1,\ldots,l$. By Theorem 2.8,
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ where
$i=$0 or 1. If
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$,
then $G$ is one of ${\mathcal{A}G}$ as a group. We assume that
$G\cong{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$. By Remark
4.5, there are integers $\beta,a_{j}\geq 0$ such that

$1+\frac{\beta-1}{\beta}=\frac{5}{6}a_{1}+\frac{1}{2}a_{2}+\frac{2}{3}a_{3}+\frac{11}{12}a_{4}+\frac{11}{12}a_{5}.$

Since $G\cong{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$,
$\beta$=1,2,3,4,6, or 12. Since
$f_{\ast}B=3(1,0)+6(1,0)+2(1,1)+4(0,1)+4(0,1)$, the support of $f_{\ast}B$ is
simple normal crossing. Since each irreducible component of $f_{\ast}B$ is
smooth, $a_{j}=0$ or $1$ for each $1\leq j\leq 5$. Since
$f_{\ast}B=3(1,0)+6(1,0)+2(1,1)+4(0,1)+4(0,1)$, the non-zero element of
$\\{a_{1},a_{2}\\}$ is just one, and the non-zero element of
$\\{a_{4},a_{5}\\}$ is just one. The integers which satisfy the above
condition are $(\beta,a_{1},a_{2},a_{3})=(12,1,0,1)$ and $(a_{4},a_{5})=(1,0)$
or (0,1). Therefore, $f(e_{i})\not\in f_{\ast}B^{2}_{1,0}$ for $i=1,\ldots,l$.
By the fact that $f_{\ast}B^{2}_{1,0}=(1,0)$ and $f_{\ast}B_{1,1}=(1,1)$ in
Pic$({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})$ and the fact that
$f(e_{i})\not\in f_{\ast}B^{2}_{1,0}$ for $i=1,\ldots,l$, we get that
$B^{2}_{1,0}\cap B_{1,1}$ is not an empty set, and hence
$p^{-1}(B^{2}_{1,0})\cap p^{-1}(B_{1,1})$ is an empty set. Since
$G\cong{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$, the number
of subgroup of $G$ which is generated by a non-symplectic automorphism of
order 2 is one. Since each ramification index of $B^{2}_{1,0}$ and $B_{1,1}$
is divided by 2, by Theorem 2.8, there is a non-symplectic automorphism $g$ of
order 2 such that Fix$(g)\supset f^{-1}B^{2}_{1,0}$ and Fix$(g)\supset
f^{-1}B_{1,1}$. Since $p^{-1}(B^{2}_{1,0})\cap p^{-1}(B_{1,1})\not=\emptyset$,
this is a contradiction. Therefore, if the numerical class of $f_{\ast}B$ is
(65), then $G$ is one of ${\mathcal{A}G}$ as a group.

As for the case of (65), if the numerical class of $f_{\ast}B$ is one of
(95,136,150,159, 235,247,253) of the list in section 6, then $G$ is one of
${\mathcal{A}G}$ as a group.

We assume that the numerical class of $f_{\ast}B$ is (70) of the list in
section 6. We denote $B$ by
$2B^{1}_{1,0}+3B^{2}_{1,0}+6B^{3}_{1,0}+2B^{1}_{0,1}+4B^{2}_{0,1}+4B^{3}_{0,1}+\sum_{j=1}^{l}b^{\prime}_{i}B^{\prime}_{i}$
where $f_{\ast}B^{i}_{1,0}=(1,0)$, $f_{\ast}B^{i}_{0,1}=(0,1)$, and
$B^{\prime}_{j}$ is an exceptional divisor of $f$ for $j=1,\ldots,l$. By
Theorem 2.8, $G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ where
$i=0,1$,or 2. There are some integers $\beta,a_{j}$ such that

$1+\frac{\beta-1}{\beta}=\frac{1}{2}a_{1}+\frac{2}{3}a_{2}+\frac{5}{6}a_{3}+\frac{1}{2}a_{4}+\frac{3}{4}a_{5}+\frac{3}{4}a_{6}.$

Since $G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ where
$i=0,1$, or 2, we get $\beta$=1,2,3,4,6, or 12. Since
$f_{\ast}B=2(1,0)+3(1,0)+6(1,0)+2(0,1)+4(0,1)+4(0,1)$, the support of
$f_{\ast}B$ is simple normal crossing. Since each irreducible component of
$f_{\ast}B$ is smooth, $a_{j}=0$ or $1$ for each $1\leq j\leq 6$, and by
Theorem 4.3 the non-zero element of $\\{a_{1},a_{2},a_{3}\\}$ is just one, and
the non-zero element of $\\{a_{4},a_{5},a_{6}\\}$ is just one. From the above,
$(\beta,a_{1},a_{2},a_{3},b_{1},b_{2},b_{3})=(1,1,0,0,1,0,0)$. Therefore,
$f(e_{j})\in f_{\ast}(B^{1}_{1,0})\cap f_{\ast}(B^{1}_{0,1})$ for
$j=1,\ldots,l$. Since $((1,0)\cdot(1,0))=0$, we get that
$(p^{\ast}B^{i}_{1,0}\cdot p^{\ast}B^{i}_{1,0})=0$ for $i=2,3$. Since $X$ is a
$K3$ surface, the support of $p^{\ast}B^{i}_{1,0}$ is a union of elliptic
curves for $i=2,3$. Since $G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ where
$i=0,1$, or 2, the number of subgroups of $G$ which are generated by a non-
symplectic automorphism of order 3 is one, and hence there is a non-symplectic
automorphism $g$ of order 3 such that Fix$(g)$ has at least two elliptic
curves. By [References,References], this is a contradiction. Therefore, the
numerical class of $f_{\ast}B$ is not (70).

As for the case of (70), the numerical class of $f_{\ast}B$ is not one of
(75,143) of the list in section 6.

If the numerical class of $f_{\ast}B$ is (72) of the list in section 6, then
by Theorem 2.8, $G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus j}$ where $(i,j)$ is one of (0,1),
(0,2), (1,1), (1,2), (2,1), (2,2), (3,1). We assume that
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$. Since $G$ is generated by non-
symplectic automorphism of order $2$ and $4$, $G_{s}:=\\{g\in G:\ g\ {\rm is\
symplectic}\\}\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$. By the classification of finite
symplectic groups ([References,References,References]), we see that there is
no $G_{s}$ where $G_{s}\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$. Therefore,
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus j}$ where $(i,j)$ is one of (0,1),
(0,2), (1,1), (1,2), (2,1), (3,1), and if the numerical class of $f_{\ast}B$
is (72), then $G$ is one of ${\mathcal{A}G}$ as a group.

As for the case of (72), if the numerical class of $f_{\ast}B$ one of
(74,78,111,144) of the list in section 6, then $G$ is one of ${\mathcal{A}G}$
as a group.

We assume that the numerical class of $f_{\ast}B$ is (73) of the list in
section 6. We denote $B$ by
$2B^{1}_{1,0}+4B^{2}_{1,0}+4B^{3}_{1,0}+3B^{1}_{0,1}+3B^{2}_{0,1}+3B^{3}_{0,1}+\sum_{j=1}^{l}b^{\prime}_{i}B^{\prime}_{i}$
By Theorem 2.8, $G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ where
$i=0,1$, or 2. As for the case of (68), there are integers $\beta,a_{j}$ such
that

$1+\frac{\beta-1}{\beta}=\frac{1}{2}a_{1}+\frac{3}{4}a_{2}+\frac{3}{4}a_{3}+\frac{2}{3}a_{4}+\frac{2}{3}a_{5}+\frac{2}{3}a_{6},$

and $a_{j}=0$ or $1$ for each $1\leq j\leq 6$, $\beta=1,2,3,4,6$, or 12, the
non-zero element of $\\{a_{1},a_{2},a_{3}\\}$ is only one, and the non-zero
element of $\\{a_{4},a_{5},a_{6}\\}$ is only one, however, integers which
satisfy the above condition do not exist. Therefore, the numerical class of
$f_{\ast}B$ is not (73).

As for the case of (73), the numerical class of $f_{\ast}B$ is not one of
(101,129,132,
152,160,170,171,172,173) of the list in section 6.

We assume that the numerical class of $f_{\ast}B$ is (76) of the list in
section 6. We denote $B$ by
$2B^{1}_{1,0}+4B^{2}_{1,0}+4B^{3}_{1,0}+2B^{1}_{0,1}+2B^{2}_{0,1}+2B^{3}_{0,1}+2B^{4}_{0,1}+\sum_{i=1}^{n}b^{\prime}_{i}B^{\prime}_{i}$,
where $f_{\ast}B^{i}_{1,0}=(1,0)$, $f_{\ast}B^{i}_{0,1}=(0,1)$ in
Pic$({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})$, and $f_{\ast}B^{\prime}_{i}=0$.
By Theorem 2.8, $G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$, where $i=0,1,2,3,$ or 4. We assume that
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$.
By Theorem 2.8, $G=G^{1}_{1,0}\oplus G^{2}_{1,0}\oplus G^{1}_{0,1}\oplus
G^{2}_{0,1}\oplus G^{3}_{0,1}$. As for the case of $(\ref{67,1})$, we get that
$f(e_{i})\in B^{1}_{1,0}\cap B^{j}_{0,1}$ for each $i=1,\ldots,m$ where
$j=1,2,3,4$. Therefore, we get $(B^{2}_{1,0}\cdot B^{j}_{0,1})=1$. Let $s\in
G^{2}_{1,0}$ be a generator. Since $G=G^{1}_{1,0}\oplus G^{2}_{1,0}\oplus
G^{1}_{0,1}\oplus G^{2}_{0,1}\oplus G^{3}_{0,1}$, by Theorem 2.8, there is a
non-symplectic automorphism $t\in G^{j}_{0,1}$ for some $j=1,2,3$ such that
Fix$(t\circ s)$ does not have a curve. Since $(B^{2}_{1,0}\cdot
B^{j}_{0,1})=1$ and $|G|=2^{3}4$, we get that $|p^{-1}(B^{2}_{1,0})\cap
p^{-1}(B^{j}_{0,1})|=8$. By [References, Proposition 1], the number of
isolated points of Fix$(t\circ s)$ is 4. This is a contradiction. Therefore,
if the numerical class of $f_{\ast}B$ is (76), then
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus i}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$
where $i=0,1,2$, or 3, and hence $G$ is one of ${\mathcal{A}G}$ as a group.

We assume that the numerical class of $f_{\ast}B$ is (150) of the list in
section 6. We denote $B$ by
$3B_{1,0}+2B^{1}_{1,1}+6B^{2}_{1,1}+4B^{1}_{0,1}+12B^{2}_{0,1}+\sum_{j=1}^{l}b^{\prime}_{i}B^{\prime}_{i}$
where $f_{\ast}B^{i}_{s,t}=sC+tF$, in Pic$({\mathbb{F}}_{1})$, and
$B^{\prime}_{j}$ is an exceptional divisor of $f$ for $j=1,\ldots,l$. By
Theorem 2.8, $G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
i}\oplus{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ where
$i=0$ or $1$. Then the number of subgroup of $G$ which is generated by a non-
symplectic automorphism of order 3 is one. By the above, for $e_{i}$, there
are integers $\beta,a_{j}\geq 0$ such that

$1+\frac{\beta-1}{\beta}=\frac{2}{3}a_{1}+\frac{1}{2}a_{2}+\frac{5}{6}a_{3}+\frac{3}{4}a_{4}+\frac{11}{12}a_{5}.$

Since $G\cong{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$,
$\beta$=1,2,3,4,6, or 12. Since
$f_{\ast}B=3(1,0)+6(1,0)+2(1,1)+4(0,1)+4(0,1)$, the support of $f_{\ast}B$ is
simple normal crossing. Since each irreducible component of $f_{\ast}B$ is
smooth, $a_{j}=0$ or $1$ for each $1\leq j\leq 5$. The integers which satisfy
the above condition are $(\beta,a_{1},a_{2},a_{3},a_{4},a_{5})=(4,0,0,1,0,1)$.
Therefore, $f(e_{i})\not\in f_{\ast}B_{1,0}\cap f_{\ast}B^{2}_{0,1}$ for
$i=1,\ldots,l$, and hence $p^{-1}(B_{1,0})\cap p^{-1}(B^{2}_{0,1})$ is not an
empty set. Since $G_{1,0}\cong{\mathbb{Z}}/3{\mathbb{Z}}$,
$G^{2}_{0,1}\cong{\mathbb{Z}}/12{\mathbb{Z}}$, and $p^{-1}(B_{1,0})\cap
p^{-1}(B^{2}_{0,1})$ is not an empty set, we get that the number of subgroup
of $G$ which is generated by a non-symplectic automorphism of order 3 is at
least two. This is a contradiction. Therefore, the numerical class of
$f_{\ast}B$ is not (150).

As for the case of (150), the numerical class of $f_{\ast}B$ is not (159) of
the list in section 6. ∎

##  5\. Abelian groups of Enriques surfaces with smooth quotient

Let $E$ be an Enriques surface and $H$ be a finite abelian subgroup of
Aut$(E)$ such that $E/H$ is smooth. Let $X$ be the $K3$-cover of $E$, and
$G:=\\{s\in{\rm Aut}(X):s\ {\rm is\ a\ lift\ of\ some}\ h\in H\\}$. Then $G$
is a finite abelian group, $G$ has a non-symplectic involution whose fixed
locus is empty, $X/G=E/H$, and the branch divisor of $G$ is that of $H$.

######  Theorem 5.1.

Let $E$ be an Enriques surface and $H$ be a finite subgroup of Aut$(E)$. We
assume that the quotient space $E/H$ is smooth and there is a birational
morphism from $E/H$ to a Hirzebruch surface ${\mathbb{F}}_{n}$, where $0\leq
n$. Then $0\leq n\leq 4$.

###### Proof.

Let $f:E/H\rightarrow{\mathbb{F}}_{n}$ be a birational morphism, and
$B:=\sum_{i=1}^{k}b_{i}B_{i}$ be the branch divisor of the quotient map
$E\rightarrow E/H$. Since the canonical line bundle of an Enriques surface is
numerically trivial, by Theorem 2.7, the numerical class of $f_{\ast}B$ is one
of Section 3. By [References, Proposition 4.5], $G$ does not have a non-
symplectic automorphism whose order is odd. Therefore, $b_{i}$ is even number
for each $i=1,\ldots,k$ by Theorem 2.8. By the list of the numerical class of
Section 3, we get the claim. ∎

######  Theorem 5.2.

For each numerical classes (6,8,9,11,12,13,16,89,90,91,94,96,97,98,
101,203,206,209,210,281) of the list in section 6, there are an Enriques
surface $E$ and a finite abelian subgroup $H$ of Aut$(E)$ such that $E/H$ is a
Hirzebruch surface ${\mathbb{F}}_{n}$, and the numerical class of the branch
divisor $B$ of the quotient map $E\rightarrow E/H$ is
(6,8,9,11,12,13,16,89,90,91,94,96,97,98,101,203,206,209,210,281).

Furthermore, for a pair $(E,H)$ of an Enriques surface $E$ and a finite
abelian subgroup $H$ of Aut$(E)$, if $E/H\cong{\mathbb{F}}_{n}$ and the
numerical class of the branch divisor $B$ of the quotient map $E\rightarrow
E/H$ is (6,8,9,11,12,13,16,89,90,91,94,96,97,98,101,203, 206, 209, 210, 281),
then $H$ is ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$,
${\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$,
${\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$,
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$, ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
3}$, ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$,
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$, ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}$, ${\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$,
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$, ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
3}$, ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$,
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$,
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$,
${\mathbb{Z}}/4{\mathbb{Z}}\oplus{\mathbb{Z}}/8{\mathbb{Z}}$,
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$, ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
3}$, ${\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$,
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$,
${\mathbb{Z}}/4{\mathbb{Z}}\oplus{\mathbb{Z}}/8{\mathbb{Z}}$, in order, as a
group.

###### Proof.

Let $X$ be the $K3$-cover of $E$, $G:=\\{s\in{\rm Aut}(X):s\ {\rm is\ a\ lift\
of\ some}\ h\in H\\}$, and $p:X\rightarrow X/G$ be the quotient map. Then $G$
is a finite abelian group, $X/G\cong{\mathbb{F}}_{n}$, and the branch divisor
of $p$ is $B$. Since $b_{i}$ is power of two for each $i=1,\ldots,k$,
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
s}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus
t}\oplus{\mathbb{Z}}/8{\mathbb{Z}}^{\oplus u}$ where $s,t,u\geq 0$. By Theorem
2.8, and the assumption that $G$ has a non-symplectic automorphism of order 2
such that whose fixed locus is an empty set, we get $s+t+u\geq 3$, and hence
the numerical class of $B$ is one of
(6,8,9,10,11,12,13,15,16,19,20,81,82,87,88,89,90,91,94,96,97,98,100,101,199,
200,203,206,208,209,210,281,282) of the list in section 6.

We assume that the numerical class of $B$ is (6). We denote $B$ by
$2B^{1}_{1,0}+2B^{2}_{1,0}+2B_{2,2}+2B^{1}_{0,1}+2B^{2}_{0,1}$. By Proposition
3.9, $G=G^{1}_{1,0}\oplus G_{2,2}\oplus
G^{1}_{0,1}\cong\mathbb{Z}/2\mathbb{Z}^{\oplus 3}$. Let $s,t,u,\in G$ be
generators of $G^{1}_{1,0}$, $G^{1}_{0,1}$, and $G_{2,2}$ respectively. Then
the non-symplectic automorphisms of $G$ are $s$, $t$, $u$, and $s\circ t\circ
u$.

From here, we will show that Fix$(s\circ t\circ u)$ is an empty set. We assume
that the curves of Fix$(s)$ are only $p^{-1}(B^{1}_{1,0})$. Since $s$ is a
non-symplectic automorphism of order 2, the quotient space $X/\langle
s\rangle$ is a smooth rational surface. The quotient map $q:X/\langle
s\rangle\rightarrow X/G\cong{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ is the
Galois cover such that the branch divisor is
$2B^{2}_{0,1}+2B_{2,2}+2B^{1}_{0,1}+2B^{2}_{0,1}$, and the Galois group is
isomorphic to ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ as a group. By Theorem
3.5, there is the Galois cover $g:Y\rightarrow X/G$ whose branch divisor is
$2B_{2,2}+2B^{1}_{0,1}+2B^{2}_{0,1}$ and Galois group is isomorphic to
${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ as a group. Since Fix$(s)$ is not an
empty set and the order of $s$ is 2, $X/\langle s\rangle$ is a smooth rational
surface. By Theorem 2.4, there is the Galois cover $h:X/\langle
s\rangle\rightarrow Y$ such that $q=g\circ h$. Since the degree of $q$ is 4
and that of $g$ is 4, $h$ is an isomorphism. Since the branch divisor of $q$
is not that of $g$, this is a contradiction. Therefore, Fix$(s)$ is
$p^{-1}(B^{1}_{1,0})\cup p^{-1}(B^{2}_{1,0})$. In the same way, Fix$(t)$ is
$p^{-1}(B^{1}_{0,1})\cup p^{-1}(B^{2}_{0,1})$. Therefore, by Theorem 2.8,
Fix$(s\circ t\circ u)$ is an empty set, and hence $E:=X/\langle s\circ t\circ
u\rangle$ is an Enriques surface. Let $H:=G/\langle s\circ t\circ u\rangle$.
Then $E/H\cong{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$,
$H\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$, and the branch divisor of $H$
is $B$. It is easy to show that for an Enriques surface $E$ and a finite
abelian subgroup $H$ of Aut$(E)$ such that
$E/H\cong{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ if the numerical class of $H$
is (6), then $H\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$.

As for the case of (6), the claim is established for (89).

We assume that the numerical class of $B$ is (8). We denote $B$ by
$4B^{1}_{1,0}+4B^{2}_{1,0}+2B_{1,1}+4B^{1}_{0,1}+4B^{2}_{0,1}$. By Proposition
3.9, $G=G^{1}_{1,0}\oplus G_{1,1}\oplus
G^{1}_{0,1}\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}^{\oplus
2}$. Let $s,t,u,\in G$ be generators of $G^{1}_{1,0}$, $G^{1}_{0,1}$, and
$G_{1,1}$ respectively. By Theorem 2.8, $s$ and $t$ are non-symplectic
automorphism of order 4 and $u$ is a non-symplectic automorphism of order 2.
By Theorem 2.8, $G^{2}_{1,0}$ is generated by $s\circ t^{2x}\circ u^{y}$ where
$x,y=0$ or 2. Since $(s\circ t^{2x}\circ u^{y})^{2}=s^{2}$ for $x,y=0$ or 2,
we get that Fix$(s^{2})$ is $p^{-1}(B^{1}_{1,0})\cup p^{-1}(B^{2}_{1,0})$. As
for the case of (6), we get the claim for (8).

As for the case of (8), the claim is established for (101).

We assume that the numerical class of $B$ is (9). We denote $B$ by
$4B^{1}_{1,0}+4B^{2}_{1,0}+2B_{1,2}+2B^{1}_{0,1}+2B^{2}_{0,1}$. By Proposition
3.9, $G=G^{1}_{1,0}\oplus G_{1,2}\oplus
G^{1}_{0,1}\cong\mathbb{Z}/2\mathbb{Z}^{\oplus
2}\oplus\mathbb{Z}/4\mathbb{Z}$. Let $s,t,u\in G$ be generators of
$G^{1}_{1,0}$, $G^{1}_{0,1}$, and $G_{1,2}$ respectively. As for the case of
(6), Fix$(t)$ is $p^{-1}(B^{1}_{0,1})\cup p^{-1}(B^{2}_{0,1})$. As for the
case of (8), Fix$(s)$ is the support of $p^{-1}(B^{1}_{1,0})\cup
p^{-1}(B^{2}_{1,0})$. As for the case of (6), we get the claim for (101).

We assume that the numerical class of $B$ is (10). We denote $B$ by
$2B^{1}_{1,0}+2B^{2}_{1,0}+2B^{3}_{1,0}+2B_{1,4}$. Let $s_{1},s_{2},t\in G$ be
generators of $G^{1}_{1,0}$, $G^{2}_{1,0}$, and $G_{1,4}$ respectively. By
Proposition 3.9, $G=G^{1}_{1,0}\oplus G^{2}_{1,0}\oplus
G_{1,4}\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$. Then the non-symplectic
involutions of $G$ are $s_{1},s_{2},t$ and $s_{1}\circ s_{2}\circ t$.

We assume that Fix$(s_{1})$ is $p^{-1}(B^{1}_{1,0})\cup p^{-1}(B^{3}_{1,0})$.
Then $X/\langle s_{1}\rangle$ is a smooth rational surface, and the quotient
map $q:X/\langle s_{1}\rangle\rightarrow
X/G\cong{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ is the Galois cover such that
the branch divisor is $2B^{2}_{0,1}+2B_{1,4}$, and the Galois group is
isomorphic to ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ as a group. Since
${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}\backslash B^{2}_{1,0}$ is simply
connected, in the same way of the proof of Theorem 2.8, this is a
contradiction. Therefore, Fix$(s_{i})$ is $p^{-1}(B^{i}_{1,0})$ for $i=1,2$,
and hence Fix$(s_{1}\circ s_{2}\circ t)$ is $p^{-1}(B^{3}_{1,0})$. There are
not an Enriques surface $E$ and a finite abelian subgroup $H$ of Aut$(E)$ such
that $E/H\cong{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ and the numerical class
of the branch divisor of $H$ is (10).

As for the case of (10), we get the claim for (87,100).

We assume that the numerical class of $B$ is (11). We denote $B$ by
$2B^{1}_{1,0}+2B^{2}_{1,0}+2B^{3}_{1,0}+2B_{1,1}+2B^{1}_{0,1}+2B^{2}_{0,1}+2B^{3}_{0,1}$.
By Proposition 3.9, $G=\oplus_{i=1}^{2}G^{i}_{1,0}\oplus
G_{1,1}\oplus_{i=1}^{2}G^{i}_{0,1}$, and hence the number of non-symplectic
automorphisms of order 2 of $G$ is 16. By Theorem 2.8, $G$ has a non-
symplectic automorphism of order 2 whose fixed locus is an empty set.
Furthermore, it is easy to show that for an Enriques surface $E$ and a finite
abelian subgroup $H$ of Aut$(E)$ such that
$E/H\cong{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ if the numerical class of $H$
is (11), then $H\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$.

As for the case of (11), the claim is established for
(12,13,16,91,94,96,97,98,206,
210).

We assume that the numerical class of $B$ is (15). We denote $B$ by
$2B^{1}_{1,0}+2B^{2}_{1,0}+2B^{1}_{1,2}+2B^{2}_{1,2}$. By Proposition 3.11,
$G=G^{1}_{1,0}\oplus G^{1}_{1,2}\oplus G^{2}_{1,2}$. Let $s,t,u\in G$ be
generators of $G^{1}_{1,0}$, $G^{1}_{1,2}$, and $G^{2}_{1,2}$ respectively.
Then the non-symplectic automorphisms of order 2 of $G$ are $s$, $t$, $u$, and
$s\circ t\circ u$. We assume that Fix$(s\circ t\circ u)$ is an empty set.
Since $(B^{i}_{1,0}\cdot B^{j}_{1,2})\not=0$ for $i,j=1,2$, Fix$(s)$ is
$p^{-1}(B^{1}_{1,0})\cup p^{-1}(B^{2}_{1,0})$. Since
$(B^{1}_{1,0}+B^{2}_{1,0}\cdot B^{1}_{1,2})=4$, $X/(G^{1}_{1,0}\oplus
G^{1}_{1,2})$ is smooth. Since $G=G^{1}_{1,0}\oplus G^{1}_{1,2}\oplus
G^{2}_{1,2}$, the branch divisor of the quotient map $X/(G^{1}_{1,0}\oplus
G^{1}_{1,2})\rightarrow X/G\cong{\mathbb{F}}_{2}$ is $2B^{2}_{1,0}$ and its
degree is 2. Since
$\frac{B^{2}_{1,0}}{2}\not\in$Pic$({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})$
and $X/(G^{1}_{1,0}\oplus G^{1}_{1,2})$ is smooth, this is a contradiction .
Therefore, there are not an Enriques surface $E$ and a finite abelian subgroup
$H$ of Aut$(E)$ such that $E/H\cong{\mathbb{F}}_{1}$ and the numerical class
of branch divisor of $H$ is (15).

As for the case of (15), we get that there are not an Enriques surface $E$ and
a finite abelian subgroup $H$ of Aut$(E)$ such that $E/H\cong{\mathbb{F}}_{n}$
and the numerical class of the branch divisor of $H$ is (88).

We assume that the numerical class of $B$ is (19). We denote $B$ by
$2B^{1}_{1,1}+2B^{2}_{1,1}+2B^{3}_{1,1}+2B^{4}_{1,1}$. By Proposition 3.15,
$G=G^{1}_{1,1}\oplus G^{2}_{1,1}\oplus G^{3}_{1,1}$. Let $s_{i}\in
G^{i}_{1,1}$ be a generator of $G^{i}_{1,1}$ for $i=1,2,3,4$. By Theorem 2.8,
Fix$(s_{i})$ is not an empty set for $i=1,2,3,4$. Since
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$, $s_{4}=s_{1}\circ s_{2}\circ
s_{3}$, and hence $G$ does not have a non-symplectic automorphism of order 2
whose fixed locus is an empty set. Therefore, there are not an Enriques
surface $E$ and a finite abelian subgroup $H$ of Aut$(E)$ such that
$E/H\cong{\mathbb{F}}_{1}$ and the numerical class of the branch divisor of
$H$ is (19).

As for the case of (19), we get that there are not an Enriques surface $E$ and
a finite abelian subgroup $H$ of Aut$(E)$ such that $E/H\cong{\mathbb{F}}_{n}$
and the numerical class of the branch divisor of $H$ is (19,20,81,82,200).

We assume that the numerical class of $B$ is (90). We denote $B$ by
$2B_{1,0}+2B_{1,1}+2B_{2,2}+4B^{1}_{0,1}+4B^{2}_{0,1}$. By Corollary 3.12,
$G=G_{1,1}\oplus G_{2,2}\oplus G^{1}_{0,1}$. Let $q:X/\langle
G_{1,0},G_{1,1},G_{2,2}\rangle\rightarrow X/G\cong{\mathbb{F}}_{1}$ be the
quotient map. Then the branch divisor of $q$ is $4B^{1}_{0,1}+4B^{2}_{0,1}$.
By Theorem 2.4, $X/\langle
G_{1,0},G_{1,1},G_{2,2}\rangle\cong{\mathbb{F}}_{4}$, and the branch divisor
of $\langle G_{1,0},G_{1,1},G_{2,2}\rangle$ is
$2B_{1,0}+2q^{\ast}B_{1,1}+2q^{\ast}B_{2,2}$. Let $s,t,u\in G$ be generators
of $G_{1,1}$, $G_{2,2}$, and $G^{1}_{0,1}$ respectively. Then Fix$(s)$ is the
support of $p^{\ast}B_{1,0}$ and that of $p^{\ast}B_{1,1}$. Then as for the
case of (6), we get the claim.

As for the case of (90), the claim is established for (203,209,281).

We assume that the numerical class of $B$ is (199). We denote $B$ by
$2B_{1,0}+2B_{1,4}+2B^{1}_{1,2}+2B^{2}_{1,2}$. By Corollary 3.16,
$G=G_{1,4}\oplus G^{1}_{1,2}\oplus G^{2}_{1,2}$. Let $s,t,u\in G$ be
generators of $G_{1,4}$, $G^{1}_{1,2}$, and $G^{2}_{1,2}$ respectively. Then
the non-symplectic automorphisms of $G$ are $s$, $t$, $u$, and $s\circ t\circ
u$. Since each fixed locus of $s$, $t$, and $u$ is not an empty set, by
Theorem 2.8, if $G$ has a non-symplectic automorphism of order 2 whose fixed
locus is an empty set, then that is $s\circ t\circ u$. We assume that
Fix$(s\circ t\circ u)$ is an empty set. Then we may assume that Fix$(t)$ is
$p^{-1}(B_{1,0})\cup p^{-1}(B^{1}_{1,2})$. Since $(B_{1,0}+B^{1}_{1,2}\cdot
B_{1,4})=6$, we get $|p^{-1}(B_{1,0}\cup B^{1}_{1,2})\cap
p^{-1}(B_{1,4})|=12$. Since $s\circ t$ is a symplectic automorphism of order 2
and $p^{-1}(B_{1,0}\cup B^{1}_{1,2})\cap p^{-1}(B_{1,4})$ is contained in
Fix$(s\circ t)$, this is a contradiction. Therefore, there are not an Enriques
surface $E$ and a finite abelian subgroup $H$ of Aut$(E)$ such that
$E/H\cong{\mathbb{F}}_{1}$ and the numerical class of the branch divisor of
$H$ is (199).

We assume that the numerical class of $B$ is (208). We denote $B$ by
$4B_{1,0}+2B_{1,3}+4B_{1,2}+2B^{1}_{0,1}+2B^{2}_{0,1}$. By Proposition 3.18,
$G=G_{1,3}\oplus G_{1,2}\oplus G^{1}_{0,1}$. Let $s,t,u\in G$ be generators of
$G_{1,3}$, $t\in G_{1,2}$, and $u\in G^{1}_{0,1}$ respectively. Then the non-
symplectic automorphisms of $G$ are $s$, $t^{2}$, $u$, and $s\circ t^{2}\circ
u$. Since each fixed locus of $s$, $t^{2}$, and $u$ is not an empty set by
Theorem 2.8, if $G$ has a non-symplectic automorphism of order 2 whose fixed
locus is an empty set, then that is $s\circ t^{2}\circ u$.

We assume that Fix$(s\circ t^{2}\circ u)$ is an empty set. Then Fix$(t^{2})$
is $p^{-1}(B_{1,0})\cup p^{-1}(B_{1,2})$, and Fix$(u)$ is
$p^{-1}(B^{1}_{1,0})\cup p^{-1}(B^{2}_{1,0})$. Since $(B_{1,3}\cdot
B^{1}_{0,1}+B^{2}_{0,1})=4$, we get that $X/(G_{1,3}\oplus G^{1}_{0,1})$ is
smooth, and the branch divisor of the quotient map $f:X/(G_{1,3}\oplus
G^{1}_{0,1})\rightarrow X/G\cong{\mathbb{F}}_{2}$ is $4B_{1,0}+4B_{1,2}$, and
the Galois group is ${\mathbb{Z}}/4{\mathbb{Z}}$, which is induced by $t$.
Furthermore, since $(B_{1,3}\cdot B_{1,0}+B_{1,2})=4$ and $(B_{1,3}\cdot
B^{1}_{0,1}+B^{2}_{0,1})=4$, $G/\langle s,t^{2},u\rangle$ is smooth, and the
branch divisor of the quotient map $g:X/\langle s,t^{2},u\rangle\rightarrow
X/G\cong{\mathbb{F}}_{2}$ is $2B_{1,0}+2B_{1,2}$, and the Galois group is
isomorphic to ${\mathbb{Z}}/2{\mathbb{Z}}$ as a group. Let $E_{1,0}$ and
$E_{1,2}$ be the support of $g^{\ast}B_{1,0}$ and $g^{\ast}B_{1,2}$
respectively. Then $g^{\ast}B_{1,0}=2E_{1,0}$ and $g^{\ast}B_{1,2}=2E_{1,2}$.
Moreover, by Theorem 3.5, there is the double cover $h:X/(G_{1,3}\oplus
G^{1}_{0,1})\rightarrow X/\langle s,t^{2},u\rangle$ such that $f=g\circ h$ and
the branch divisor is $2E_{1,0}+2E_{1,2}$. Since $X/(G_{1,3}\oplus
G^{1}_{0,1})$ and $X/\langle s,t^{2},u\rangle$ are smooth, we get
$\frac{E_{1,0}+E_{1,2}}{2}\in$Pic$(X/\langle s,t^{2},u\rangle)$. Since
$g^{\ast}B_{1,2}=g^{\ast}B_{1,0}+2g^{\ast}F$ in Pic$(X/\langle
s,t^{2},u\rangle)$,

$2E_{1,2}=2E_{1,0}+2g^{\ast}F\ {\rm in\ Pic}(X/\langle s,t^{2},u\rangle).$

Since $X/\langle s,t^{2},u\rangle$ is a smooth rational surface,
Pic$(X/(G_{1,3}\oplus G^{1}_{0,1}))$ is torsion free. Therefore, we get

$E_{1,2}=E_{1,0}+g^{\ast}F\ {\rm in\ Pic}(X/\langle s,t^{2},u\rangle),$

and hence

$E_{1,2}+E_{1,0}=2E_{1,0}+g^{\ast}F\ {\rm in\ Pic}(X/\langle
s,t^{2},u\rangle).$

Since $\frac{E_{1,2}+E_{1,0}}{2}\in$Pic$(X/\langle s,t^{2},u\rangle)$, we get

$\frac{g^{\ast}F}{2}\in{\rm Pic}(X/\langle s,t^{2},u\rangle).$

Since $(B_{1,0}\cdot F)=1$, the degree of $g$ is two, and
$\frac{g^{\ast}B_{1,0}}{2}$ and $\frac{g^{\ast}F}{2}$ are elements of
Pic$(X/\langle s,t^{2},u\rangle)$, this is a contradiction. Therefore, there
are not an Enriques surface $E$ and a finite abelian subgroup $H$ of Aut$(E)$
such that $E/H\cong{\mathbb{F}}_{1}$ and the numerical class of the branch
divisor of $H$ is (208).

We assume that the numerical class of $B$ is (282). We denote $B$ by
$2B_{1,0}+2B^{1}_{1,4}+2B^{2}_{1,4}+2B^{3}_{1,4}$. By Corollary 3.16,
$G=\oplus_{i=1}^{3}G^{i}_{1,4}$, Let $s_{i}\in G^{i}_{1,4}$ be a generator for
$i=1,2,3$. Then the non-symplectic automorphisms of $G$ are $s_{i}$ and
$s_{1}\circ s_{2}\circ s_{3}$ where $i=1,2,3$. Since each fixed locus of
$s_{i}$ is not an empty set for each $i=1,2,3$ by Theorem 2.8, if $G$ has a
non-symplectic automorphism of order 2 whose fixed locus is an empty set, then
that is $s_{1}\circ s_{2}\circ s_{3}$. We assume that Fix$(s_{1}\circ
s_{2}\circ s_{3})$ is an empty set. Then we may assume that Fix$(s_{1})$ is
$p^{-1}(B_{1,0})\cup p^{-1}(B^{1}_{1,4})$. Since $(B_{1,0}+B^{1}_{1,4}\cdot
B_{1,4})=4$, we get that $X/(G^{1}_{1,4}\oplus G^{2}_{1,4})$ is smooth, and
the branch divisor of the quotient map $X/(G^{1}_{1,4}\oplus
G^{1}_{1,4})\rightarrow X/G\cong{\mathbb{F}}_{4}$ is $2B^{3}_{1,4}$. This is a
contradiction as the degree of the quotient map is 2. Therefore, there are not
an Enriques surface $E$ and a finite abelian subgroup $H$ of Aut$(E)$ such
that $E/H\cong{\mathbb{F}}_{4}$ and the numerical class of the branch divisor
of $H$ is (282). ∎

By Theorem 5.2, we get Theorem 1.9.

######  Theorem 5.3.

Let $E$ be an Enriques surface and $H$ be a finite abelian subgroup of ${\rm
Aut}(E)$. If $E/H$ is smooth, then $H$ is isomorphic to one of
${\mathcal{A}G}(E)$ as a group.

###### Proof.

Let $X$ be the $K3$-cover of $E$, $G:=\\{s\in{\rm Aut}(X):s\ {\rm is\ a\ lift\
of\ some}\ h\in H\\}$, and $p:X\rightarrow X/G$ be the quotient map. Then $G$
is a finite abelian group, $X/G=E/H$, and the branch divisor of $p$ is $B$. We
classified $H$ for the case of $E/H\cong{\mathbb{F}}_{n}$ in Theorem 5.2. From
here, we assume that $E/H$ is smooth and $E/H\not\cong{\mathbb{F}}_{n}$ or
${\mathbb{P}}^{2}$. Since $G$ does not have a non-symplectic automorphism
whose order is odd ([References]), by Theorem 2.8 and 1.5,
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
s}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus
t}\oplus{\mathbb{Z}}/8{\mathbb{Z}}^{\oplus u}$ where $s,t,u\geq 0$. By the
assumption that $G$ has a non-symplectic automorphism of order 2 such that
whose fixed locus is an empty set, and the fact that $G$ is generated by non-
symplectic automorphisms whose fixed locus have a curve, we get $s+t+u\geq 3$.
Therefore, $G$ is one of the following as a group:

$\\{{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus a},\ {\mathbb{Z}}/4{\mathbb{Z}}^{\oplus
3},\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
f}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus g},\
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}\oplus{\mathbb{Z}}/8{\mathbb{Z}}:$
$3\leq a\leq 5,\ (f,g)=(1,2),(2,1),(3,1)\\}.$

If $G$ is one of

$\\{{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus a},\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus f}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus
g}:3\leq a\leq 5,\ (f,g)=(1,2),(2,1),(3,1)\\}$

as a group, then quotient group $G/K$ of $G$ by a subgroup
$K\cong{\mathbb{Z}}/2{\mathbb{Z}}$ is one of

$\\{{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus a},\ {\mathbb{Z}}/4{\mathbb{Z}}^{\oplus
2},\ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
f}\oplus{\mathbb{Z}}/4{\mathbb{Z}}:a=2,3,4\ f=1,2\\}\subset{\mathcal{A}G}(E)$

as a group. Let $f:X/G\rightarrow{\mathbb{F}}_{n}$ be the birational morphism.
We assume that $G\cong{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 3}$. By the
assumption that $G\cong{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 3}$ and Theorem 2.8,
the numerical class of $f_{\ast}B$ is only (142). We denote $B$ by
$2B_{1,0}+4B^{1}_{1,4}+4B^{2}_{1,4}+4B^{1}_{0,1}+4B^{2}_{0,1}+\sum_{i=1}^{n}b^{\prime}_{i}B^{\prime}_{i}$
where $f_{\ast}B_{1,0}=C$, $f_{\ast}B^{i}_{1,4}=C+4F$,
$f_{\ast}B^{i}_{0,1}=F$, and $f_{\ast}B^{\prime}_{i}=0$ in
Pic$({\mathbb{F}}_{4})$. Since $G\cong{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 3}$,
by Theorem 2.8, we get that $G=G^{1}_{1,4}\oplus G^{2}_{1,4}\oplus
G^{1}_{0,1}$. Let $s\in G^{1}_{1,4}$, $t\in G^{2}_{1,4}$, and $u\in
G^{1}_{0,1}$ be generators respectively. The non-symplectic involutions of $G$
are $s^{2}$, $t^{2}$, $u^{2}$, and $s^{2}\circ t^{2}\circ u^{2}$. Since each
fixed locus of $s^{2}$, $t^{2}$, and $u^{2}$ is not an empty set, if $G$ has a
non-symplectic automorphism of order 2 whose fixed locus is an empty set, then
that is $s^{2}\circ t^{2}\circ u^{2}$. If the fixed locus of $s^{2}\circ
t^{2}\circ u^{2}$ is an empty set, then the fixed locus of $s\circ t\circ u$
is an empty set. By [References], this is a contradiction. Therefore, $G$ is
not ${\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 3}$ as a group.

We assume that
$G\cong{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}\oplus{\mathbb{Z}}/8{\mathbb{Z}}$.
By Theorem 2.8, the numerical class of $f_{\ast}B$ is only (101). By the proof
of Theorem 4.9, $f$ is an isomorphism, i.e. $X/G\cong{\mathbb{F}}_{1}$. By
Theorem 5.2, we get the claim. ∎

By Theorem 5.2 and 5.3, we get Theorem 1.10.

##  6\. Non-Abelian case

Let $S$ be a smooth rational surface such which is neither $\mathbb{P}^{2}$
nor an Enriques surface. In Theorem 6.5, we will describe the existence of a
$K3$ surface $X^{\prime}$ and a finite subgroup $G^{\prime}$ of
Aut$(X^{\prime})$ such that $X^{\prime}/G^{\prime}\cong S$ by the result of
$[\ref{bio:5}$ $,\,{\rm Proposition}\,5.1$ $\,{\rm and\ Theorem}\,5.2]$. It
does not assume that $G^{\prime}$ is an abelian group, but it does not
elucidate the group structure of $G^{\prime}$. First, we prepare a little.
Next, we state Theorem 6.5. Let

${\mathcal{A}S}(K3):=\left\\{\begin{aligned} &{\mathbb{Z}}/n{\mathbb{Z}},\
{\mathbb{Z}}/m{\mathbb{Z}}^{\oplus 2},\
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/k{\mathbb{Z}},\
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus l}\\\ &n=2,\ldots,8,\ m=2,3,4,\ k=4,6,\
l=3,4\end{aligned}\right\\}$

######  Theorem 6.1.

$([\ref{bio:12}])$ Let $X$ be a $K3$ surface, and $G$ be a finite abelian
subgroup of Aut$(X)$. If $G$ is symplectic, then $G\in{\mathcal{A}S}(K3)$ as a
group. Conversely, for each an abelian group
$G^{\prime}\in{\mathcal{A}S}(K3)$, there is a $K3$ surface $X^{\prime}$ such
that $G^{\prime}$ acts faithfully on $X^{\prime}$ as symplectic.

######  Theorem 6.2.

$([\ref{bio:5},\,{\rm Proposition}\,5.1])$ Let $X$ be a $K3$ surface, and $G$
be a finite abelian subgroup of Aut$(X)$ such that $G$ acts symplectically on
$X$. Let $Y$ be a $K3$ surface such that $Y$ is a minimal resolution of the
quotient space $X/G$, and $E_{G}$ be the lattices spanned by the curves on $Y$
arising from the resolution of the singularities of $X/G$.

Then $E_{G}$ is one of the following root lattices:

$G$ | ${\mathbb{Z}}/2{\mathbb{Z}}$ | ${\mathbb{Z}}/3{\mathbb{Z}}$ | ${\mathbb{Z}}/4{\mathbb{Z}}$ | ${\mathbb{Z}}/5{\mathbb{Z}}$ | ${\mathbb{Z}}/6{\mathbb{Z}}$ | ${\mathbb{Z}}/7{\mathbb{Z}}$ | ${\mathbb{Z}}/8{\mathbb{Z}}$
---|---|---|---|---|---|---|---
$E_{G}$ | $A^{8}_{1}$ | $A^{6}_{2}$ | $A_{3}^{4}\oplus A_{1}^{2}$ | $A_{4}^{4}$ | $A_{5}^{2}\oplus A_{2}^{2}\oplus A_{1}^{2}$ | $A_{6}^{3}$ | $A_{7}^{2}\oplus A_{3}\oplus A_{1}$

$G$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$ | ${\mathbb{Z}}/2{\mathbb{Z}}\oplus\mathbb{Z}/4\mathbb{Z}$
---|---|---|---|---
$E_{G}$ | $A^{12}_{1}$ | $A^{14}_{1}$ | $A_{1}^{15}$ | $A_{3}^{4}\oplus A_{1}^{4}$

$G$ | ${\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/6\mathbb{Z}$ | ${\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ | ${\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$
---|---|---|---
$E_{G}$ | $A^{3}_{5}\oplus A_{1}^{3}$ | $A^{8}_{2}$ | $A_{3}^{6}$

Let $M_{G}$ be the minimal primitive sublattice of ${\rm NS}(Y)$ which
contains $E_{G}$. Then $M_{G}$ is an overlattice of finite index ${\rm r}_{G}$
of $E_{G}$ and its properties are the followings

$G$ | ${\mathbb{Z}}/2{\mathbb{Z}}$ | ${\mathbb{Z}}/3{\mathbb{Z}}$ | ${\mathbb{Z}}/4{\mathbb{Z}}$ | ${\mathbb{Z}}/5{\mathbb{Z}}$ | ${\mathbb{Z}}/6{\mathbb{Z}}$
---|---|---|---|---|---
${\rm r}_{G}$ | $2$ | $3$ | $4$ | $5$ | $6$
${\rm rank}M_{G}$ | $8$ | $12$ | $14$ | $16$ | $16$
$M_{G}^{\vee}/M_{G}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 6}$ | ${\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 4}$ | $({\mathbb{Z}}/2{\mathbb{Z}}\oplus\mathbb{Z}/4\mathbb{Z})^{\oplus 2}$ | ${\mathbb{Z}}/5{\mathbb{Z}}^{\oplus 2}$ | ${\mathbb{Z}}/6{\mathbb{Z}}^{\oplus 2}$

$G$ | ${\mathbb{Z}}/7{\mathbb{Z}}$ | ${\mathbb{Z}}/8{\mathbb{Z}}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$
---|---|---|---|---|---
${\rm r}_{G}$ | $7$ | $8$ | $2^{2}$ | $2^{3}$ | $2^{4}$
${\rm rank}M_{G}$ | $18$ | $18$ | $12$ | $14$ | $15$
$M_{G}^{\vee}/M_{G}$ | ${\mathbb{Z}}/7{\mathbb{Z}}$ | ${\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 8}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 8}$ | ${\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 7}$

$G$ | ${\mathbb{Z}}/2{\mathbb{Z}}\oplus\mathbb{Z}/4\mathbb{Z}$ | ${\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/6\mathbb{Z}$ | ${\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ | ${\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$
---|---|---|---|---
${\rm r}_{G}$ | $8$ | $12$ | $3^{2}$ | $4^{2}$
${\rm rank}M_{G}$ | $16$ | $18$ | $16$ | $18$
$M_{G}^{\vee}/M_{G}$ | $({\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/6{\mathbb{Z}})^{\oplus 2}$ | ${\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/6{\mathbb{Z}}$ | ${\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 4}$ | ${\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$

######  Theorem 6.3.

$([\ref{bio:5},\,{\rm Theorem}\,5.2])$ For $G\in{\mathcal{A}S}(K3)$ and a $K3$
surface $X$ such that $G$ acts faithfully on $X$ as symplectic, a $K3$ surface
$Y$ is a minimal resolution of the quotient space $X/G$ if and only if $M_{G}$
is primitively embedded in ${\rm NS}(Y)$.

In what follows, we explain Theorem 6.5. Let $X$ be a $K3$ surface, and $G$ be
a finite subgroup of Aut$(X)$ such that $X/G$ is smooth. If $X/G$ is neither
$\mathbb{P}^{2}$ nor an Enriques surface, then $X/G$ is a Hirzebruch surface
or its blow-ups. Let $p:X\rightarrow X/G$ be the quotient morphism, and
$B:=\sum_{i=1}^{k}b_{i}B_{i}$ be the branch divisor of $p:X\rightarrow X/G$.
Since a Hirzebruch surface $\mathbb{F}_{n}$ has the unique fibre structure,
there is the fibre structure $f:X/G\rightarrow\mathbb{P}^{1}$ induced by a
Hirzebruch surface. Then the general fibre of $f\circ
p:X\rightarrow\mathbb{P}^{1}$ is a disjoint union of elliptic curves.

Let $G_{s}$ be the symplectic subgroup of $G$ generated by symplectic
automorphisms of $G$. We set $G_{h}:=\\{g\in G_{s}\,|\,g\ \ {\rm does\ not\
replace\ irreducible\ components\ of\ }(f\circ p)^{-1}(u)\ \ {\rm for\ any\
}u\in\mathbb{P}^{1}.\\}$.

######  Lemma 6.4.

The group $G_{h}$ is an abelia group.

###### Proof.

Since $G_{h}$ is a finite symplectic group, $\bigcup_{g\in G_{h}}{\rm Fix}(g)$
is a finite set. There is a point $u\in\mathbb{P}^{1}$ such that $G_{s}$ acts
freely on $(f\circ p)^{-1}(u)$. Since the general fibre of $f\circ
p:X\rightarrow\mathbb{P}^{1}$ is a disjoint union of elliptic curves, we get
that $G_{h}$ is an abelian group. ∎

Let $X_{s}:=X/G_{s}$ be the quotient space, $p_{1}:X_{s}\rightarrow
X/G=X_{s}/(G/G_{s})$ be the quotient map, and $f_{1}:=f\circ
p_{1}:X_{s}\rightarrow\mathbb{P}^{1}$ be the fibre structure. Then
$p_{1}:X_{s}\rightarrow X/G=X_{s}/(G/G_{s})$ is a cyclic cover whose branch
divisor is $B$. Let $X_{h}:=X/G_{h}$ be the quotient space,
$p_{2}:X_{h}\rightarrow X_{s}=X_{h}/(G_{s}/G_{h})$ be the quotient map.

For $f\circ p_{1}\circ p_{2}:X_{h}\rightarrow\mathbb{P}^{1}$, by the Stein
factorization, there are the fibre structure
$h:X_{h}\rightarrow\mathbb{P}^{1}$ and a finite map
$k:\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$ such that $f\circ p_{1}\circ
p_{2}=k\circ h$, i.e. the following diagram is commutative:

$\textstyle{X_{s}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f\circ
p_{1}}$$\textstyle{\mathbb{P}^{1}}$$\textstyle{X_{h}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{2}}$$\scriptstyle{h}$$\textstyle{\mathbb{P}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces.}$$\scriptstyle{k}$

Then $X_{h}$ is the normalization of the fibre product of $f\circ
p_{1}:X_{s}\rightarrow\mathbb{P}^{1}$ and
$k:\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$. By the definition of $G_{h}$, we
get that $k:\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$ is a Galois cover whose
Galois group is isomorphic to $G_{s}/G_{h}$ as a group. Recall that $G_{h}$ is
an abelian group, and by Theorem 4.6 $p_{1}:X_{s}\rightarrow X/G$ is a cyclic
cover of order $b$ such that the branch divisor is $B$ and the Galois group is
$G/G_{s}$ where $b$ is the least common multiple of $b_{1},\ldots,b_{k}$. By
the above, we have the following theorem.

######  Theorem 6.5.

Let $S$ be a smooth rational surface which is neither $\mathbb{P}^{2}$ nor an
Enriques surface. We fix a fibre structure $f:S\rightarrow\mathbb{P}^{1}$, and
an effective divisor $B:=\sum_{i=1}^{k}b_{i}B_{i}$ on $S$. Let $b$ be the
least common multiple of $b_{1},\ldots,b_{k}$.

Then there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G=S$ and the branch divisor of the quotient morphism $p:X\rightarrow
X/G$ is $B$ if and only if there are a Galois cover
$k:\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$ and a cyclic cover $q:T\rightarrow
S$ of degree $b$ whose the branch divisor is $B$ such that the minimal
singular resolution $X^{\prime}$ of the normalization of the fibre product of
$f\circ q:T\rightarrow\mathbb{P}^{1}$ and
$k:\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$ is a $K3$ surface, and
$M_{G}^{\prime}$ is primitively embedded in ${\rm NS}(X^{\prime})$ for some
$G^{\prime}\in{\mathcal{A}S}(K3)$.

##  7\. the list of a numerical class

Here, we will give the list of a numerical class of an effective divisor
$B=\sum_{i=1}^{k}b_{i}B_{i}$ on ${\mathbb{F}}_{n}$ such that $B_{i}$ is a
smooth curve for each $i=1,\ldots,k$ and
$K_{{\mathbb{F}}_{n}}+\sum_{i=1}^{k}\frac{b_{i}-1}{b_{i}}B_{i}=0$ in
Pic$({\mathbb{F}}_{n})$.

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G={\mathbb{F}}_{0}\cong{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$, then
by Theorem 2.7 the numerical class of $B$ is one of the following:

(1) $\displaystyle 3(3C+3F)\ \ {\mathbb{Z}}/3{\mathbb{Z}}$ (2) $\displaystyle
3C+3C+3(C+3F)\ \ {\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (3) $\displaystyle
3C+3C+3(C+F)+3F+3F\ \ {\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 3}$ (4)
$\displaystyle 2(4C+4F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}$ (5) $\displaystyle
2C+2C+2(2C+4F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ (6) $\displaystyle
2C+2C+2(2C+2F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (7)
$\displaystyle 4C+4C+2(C+4F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (8) $\displaystyle
4C+4C+2(C+F)+4F+4F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$ (9)
$\displaystyle 4C+4C+2(C+2F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (10) $\displaystyle 2C+2C+2C+2(C+4F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (11) $\displaystyle
2C+2C+2C+2(C+F)+2F+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 5}$ (12)
$\displaystyle 2C+2C+2C+2(C+2F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
4}$ (13) $\displaystyle 2C+2C+2C+2(C+F)+4F+4F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (14)
$\displaystyle 2(2C+2F)+2(2C+2F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$
(15) $\displaystyle 2C+2C+2(C+2F)+2(C+2F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (16) $\displaystyle
2C+2C+2(C+F)+2(C+F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$ (17)
$\displaystyle 3(C+F)+3(C+F)+3(C+F)\ \ {\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$
(18) $\displaystyle 3C+3(C+F)+3(C+2F)\ \ {\mathbb{Z}}/3{\mathbb{Z}}^{\oplus
2}$ (19) $\displaystyle 2(C+F)+2(C+F)+2(C+F)+2(C+F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (20) $\displaystyle
2C+2(C+F)+2(C+F)+2(C+2F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (21)
$\displaystyle 2(C+F)+4(2C+2F)$ (22) $\displaystyle 3(C+F)+3(2C+2F)$ (23)
$\displaystyle 3(C+2F)+3(2C+F)$ (24) $\displaystyle 3C+3(2C+3F)$ (25)
$\displaystyle 2C+2(3C+4F)$ (26) $\displaystyle 2(C+F)+2(3C+3F)$ (27)
$\displaystyle 2(C+2F)+2(3C+2F)$ (28) $\displaystyle 2(C+3F)+2(3C+F)$ (29)
$\displaystyle 2(C+F)+3(C+F)+6(C+F)$ (30) $\displaystyle 2(C+F)+4(C+F)+4(C+F)$
(31) $\displaystyle 2C+4(2C+2F)+2F$ (32) $\displaystyle 4C+2(2C+2F)+4F$ (33)
$\displaystyle 2C+2(3C+3F)+2F$ (34) $\displaystyle 3C+6C+2(C+4F)$ (35)
$\displaystyle 4C+2(C+F)+4(C+2F)$ (36) $\displaystyle 2C+2(C+F)+2(2C+3F)$ (37)
$\displaystyle 2C+2(C+2F)+2(2C+2F)$ (38) $\displaystyle 2C+2(C+3F)+2(2C+F)$
(39) $\displaystyle 3C+3(2C+2F)+3F$ (40) $\displaystyle 2C+6C+3(C+3F)$ (41)
$\displaystyle 2(C+F)+2(C+F)+2(2C+2F)$ (42) $\displaystyle
2(C+2F)+2(C+F)+2(2C+F)$ (43) $\displaystyle 2C+4C+4(C+2F)+2F$ (44)
$\displaystyle 3C+6C+2(C+3F)+2F$ (45) $\displaystyle 4C+4C+2(C+3F)+2F$ (46)
$\displaystyle 2C+2C+2(2C+3F)+2F$ (47) $\displaystyle 2C+4(C+F)+4(C+F)+2F$
(48) $\displaystyle 4C+2(C+F)+4(C+F)+4F$ (49) $\displaystyle
2C+3(C+F)+6(C+F)+2F$ (50) $\displaystyle 6C+2(C+F)+3(C+F)+6F$ (51)
$\displaystyle 2C+2(C+F)+2(2C+2F)+2F$ (52) $\displaystyle
2C+2(C+2F)+2(2C+F)+2F$ (53) $\displaystyle 3C+2(C+F)+6(C+F)+3F$ (54)
$\displaystyle 3C+3(C+F)+3(C+F)+3F$ (55) $\displaystyle 3C+3C+3(C+2F)+3F$ (56)
$\displaystyle 2C+6C+3(C+2F)+3F$ (57) $\displaystyle 2C+2C+2(C+F)+2(C+3F)$
(58) $\displaystyle 2C+2(C+F)+2(C+F)+2(C+F)+2F$ (59) $\displaystyle
2C+2C+2C+2(C+3F)+2F$ (60) $\displaystyle 2C+2C+2(C+F)+2(C+2F)+2F$ (61)
$\displaystyle 2C+4C+4(C+F)+2F+4F$ (62) $\displaystyle 2C+3C+6(C+F)+2F+3F$
(63) $\displaystyle 2C+6C+3(C+F)+2F+6F$ (64) $\displaystyle
3C+6C+2(C+F)+3F+6F$ (65) $\displaystyle 3C+6C+2(C+F)+4F+4F$ (66)
$\displaystyle 2C+6C+3(C+F)+3F+3F$ (67) $\displaystyle 3C+6C+2(C+2F)+2F+2F$
(68) $\displaystyle 3C+6C+2(C+F)+2F+2F+2F$ (69) $\displaystyle
2C+3C+6C+2F+3F+6F$ (70) $\displaystyle 2C+3C+6C+2F+4F+4F$ (71) $\displaystyle
2C+3C+6C+3F+3F+3F$ (72) $\displaystyle 2C+4C+4C+2F+4F+4F$ (73) $\displaystyle
2C+4C+4C+3F+3F+3F$ (74) $\displaystyle 3C+3C+3C+3F+3F+3F$ (75) $\displaystyle
2C+3C+6C+2F+2F+2F+2F$ (76) $\displaystyle 2C+4C+4C+2F+2F+2F+2F$ (77)
$\displaystyle 3C+3C+3C+2F+2F+2F+2F$ (78) $\displaystyle
2C+2C+2C+2C+2F+2F+2F+2F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{1}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(79) $\displaystyle 2(4C+6F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}$ (80)
$\displaystyle 2(2C+4F)+2(2C+2F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$
(81) $\displaystyle 2C+2(C+2F)+2(C+2F)+2(C+2F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (82) $\displaystyle
2(C+3F)+2(C+F)+2(C+F)+2(C+F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (83)
$\displaystyle 3(3C+3F)+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}$ (84)
$\displaystyle 3C+3(2C+2F)+6F+6F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (85)
$\displaystyle 2(4C+4F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ (86)
$\displaystyle 2C+2(3C+3F)+4F+4F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (87)
$\displaystyle 2C+2(3C+3F)+2F+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$
(88) $\displaystyle 2C+2(C+F)+2(2C+3F)+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (89) $\displaystyle
2(2C+2F)+2(2C+2F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (90)
$\displaystyle 2C+2(C+F)+2(2C+2F)+4F+4F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (91) $\displaystyle
2C+2(C+F)+2(2C+2F)+2F+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$ (92)
$\displaystyle 3(C+F)+3(C+F)+3(C+F)+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (93)
$\displaystyle 3C+3(C+F)+3(C+F)+6F+6F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}3{\mathbb{Z}}^{\oplus 3}$ (94)
$\displaystyle 2C+2(C+2F)+2(C+F)+2(C+F)+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$ (95) $\displaystyle
6C+2(C+F)+3(C+F)+12F+12F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (96) $\displaystyle
2(C+F)+2(C+F)+2(C+F)+2(C+F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$
(97) $\displaystyle 2C+2(C+F)+2(C+F)+2(C+F)+4F+4F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (98)
$\displaystyle 2C+2(C+F)+2(C+F)+2(C+F)+2F+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 5}$ (99) $\displaystyle 2C+4(2C+2F)+4F+4F\
\ {\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$ (100) $\displaystyle
2C+4(2C+2F)+2F+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (101) $\displaystyle
4C+2(C+F)+4(C+F)+8F+8F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}\oplus{\mathbb{Z}}/8{\mathbb{Z}}$
(102) $\displaystyle 3(2C+2F)+3(C+F)+2F+2F$ (103) $\displaystyle
4(2C+2F)+2(C+3F)$ (104) $\displaystyle 4(2C+2F)+2(C+2F)+2F$ (105)
$\displaystyle 4(2C+2F)+2(C+F)+2F+2F$ (106) $\displaystyle
2(C+3F)+3(C+F)+6(C+F)$ (107) $\displaystyle 2(C+2F)+3(C+F)+6(C+F)+2F$ (108)
$\displaystyle 2(C+F)+3(C+F)+6(C+F)+2F+2F$ (109) $\displaystyle
2(C+3F)+4(C+F)+4(C+F)$ (110) $\displaystyle 2(C+2F)+4(C+F)+4(C+F)+2F$ (111)
$\displaystyle 2(C+F)+4(C+F)+4(C+F)+2F+2F$ (112) $\displaystyle 2(4C+5F)+2F$
(113) $\displaystyle 2(3C+aF)+2(C+(6-a)F),a\geq 3$ (114) $\displaystyle
2(3C+4F)+2(C+F)+2F$ (115) $\displaystyle 2(3C+3F)+2(C+2F)+2F$ (116)
$\displaystyle 2(3C+3F)+2(C+F)+2F+2F$ (117) $\displaystyle 2(2C+3F)+2(2C+3F)$
(118) $\displaystyle 2(2C+3F)+2(2C+2F)+2F$ (119) $\displaystyle
2(2C+4F)+2(C+F)+2(C+F)$ (120) $\displaystyle 2(2C+3F)+2(C+2F)+2(C+F)$ (121)
$\displaystyle 2(2C+3F)+2(C+F)+2(C+F)+2F$ (122) $\displaystyle
2(2C+2F)+2(C+2F)+2(C+2F)$ (123) $\displaystyle 2(2C+2F)+2(C+2F)+2(C+F)+2F$
(124) $\displaystyle 2(2C+2F)+2(C+F)+2(C+F)+2F+2F$ (125) $\displaystyle
2(C+2F)+2(C+2F)+2(C+F)+2(C+F)$ (126) $\displaystyle
2(C+2F)+2(C+F)+2(C+F)+2(C+F)+2F$ (127) $\displaystyle 3C+3(2C+3F)+2F+2F$ (128)
$\displaystyle 3C+3(2C+2F)+2F+2F+3F$ (129) $\displaystyle 3C+3(2C+2F)+4F+12F$
(130) $\displaystyle 2C+4(2C+4F)$ (131) $\displaystyle 2C+4(2C+3F)+4F$ (132)
$\displaystyle 2C+4(2C+2F)+3F+6F$ (133) $\displaystyle 2C+3(C+2F)+6(C+2F)$
(134) $\displaystyle 2C+3(C+2F)+6(C+F)+6F$ (135) $\displaystyle
2C+3(C+F)+6(C+2F)+3F$ (136) $\displaystyle 2C+3(C+F)+6(C+F)+4F+4F$ (137)
$\displaystyle 2C+3(C+F)+6(C+F)+3F+6F$ (138) $\displaystyle
2C+3(C+F)+6(C+F)+2F+2F+2F$ (139) $\displaystyle 2C+4(C+2F)+4(C+2F)$ (140)
$\displaystyle 2C+4(C+F)+4(C+3F)$ (141) $\displaystyle 2C+4(C+F)+4(C+2F)+4F$
(142) $\displaystyle 2C+4(C+F)+4(C+F)+4F+4F$ (143) $\displaystyle
2C+4(C+F)+4(C+F)+3F+6F$ (144) $\displaystyle 2C+4(C+F)+4(C+F)+2F+2F+2F$ (145)
$\displaystyle 3C+2(C+3F)+6(C+F)+3F$ (146) $\displaystyle
3C+2(C+2F)+6(C+F)+2F+3F$ (147) $\displaystyle 3C+2(C+F)+6(C+3F)$ (148)
$\displaystyle 3C+2(C+F)+6(C+2F)+6F$ (149) $\displaystyle
3C+2(C+F)+6(C+F)+6F+6F$ (150) $\displaystyle 3C+2(C+F)+6(C+F)+4F+12F$ (151)
$\displaystyle 3C+2(C+F)+6(C+F)+2F+2F+3F$ (152) $\displaystyle
3C+3(C+F)+3(C+F)+4F+12F$ (153) $\displaystyle 3C+3(C+2F)+3(C+F)+2F+2F$ (154)
$\displaystyle 3C+3(C+F)+3(C+F)+2F+2F+3F$ (155) $\displaystyle
4C+2(C+3F)+4(C+2F)$ (156) $\displaystyle 4C+2(C+3F)+4(C+F)+4F$ (157)
$\displaystyle 4C+2(C+2F)+4(C+2F)+2F$ (158) $\displaystyle
4C+2(C+2F)+4(C+F)+2F+4F$ (159) $\displaystyle 4C+2(C+F)+4(C+F)+6F+12F$ (160)
$\displaystyle 4C+2(C+F)+4(C+F)+5F+20F$ (161) $\displaystyle
4C+2(C+F)+4(C+2F)+2F+2F$ (162) $\displaystyle 4C+2(C+F)+4(C+F)+2F+2F+4F$ (163)
$\displaystyle 6C+2(C+3F)+3(C+F)+6F$ (164) $\displaystyle 6C+2(C+2F)+3(C+3F)$
(165) $\displaystyle 6C+2(C+2F)+3(C+2F)+3F$ (166) $\displaystyle
6C+2(C+2F)+3(C+F)+3F+3F$ (167) $\displaystyle 6C+2(C+2F)+3(C+F)+2F+6F$ (168)
$\displaystyle 6C+2(C+F)+3(C+3F)+2F$ (169) $\displaystyle
6C+2(C+F)+3(C+2F)+2F+3F$ (170) $\displaystyle 6C+2(C+F)+3(C+F)+10F+15F$ (171)
$\displaystyle 6C+2(C+F)+3(C+F)+9F+18F$ (172) $\displaystyle
6C+2(C+F)+3(C+F)+8F+24F$ (173) $\displaystyle 6C+2(C+F)+3(C+F)+7F+42F$ (174)
$\displaystyle 6C+2(C+F)+3(C+F)+2F+3F+3F$ (175) $\displaystyle
6C+2(C+F)+3(C+F)+2F+2F+6F$ (176) $\displaystyle 2C+2(3C+6F)$ (177)
$\displaystyle 2C+2(3C+5F)+2F$ (178) $\displaystyle 2C+2(3C+4F)+2F+2F$ (179)
$\displaystyle 2C+2(3C+3F)+3F+6F$ (180) $\displaystyle 2C+2(C+4F)+2(2C+2F)$
(181) $\displaystyle 2C+2(C+3F)+2(2C+3F)$ (182) $\displaystyle
2C+2(C+2F)+2(2C+4F)$ (183) $\displaystyle 2C+2(C+F)+2(2C+5F)$ (184)
$\displaystyle 2C+2(C+3F)+2(2C+2F)+2F$ (185) $\displaystyle
2C+2(C+2F)+2(2C+3F)+2F$ (186) $\displaystyle 2C+2(C+2F)+2(2C+2F)+2F+2F$ (187)
$\displaystyle 2C+2(C+F)+2(2C+4F)+2F$ (188) $\displaystyle
2C+2(C+F)+2(2C+2F)+3F+6F$ (189) $\displaystyle 2C+2(C+4F)+2(C+F)+2(C+F)$ (190)
$\displaystyle 2C+2(C+3F)+2(C+2F)+2(C+F)$ (191) $\displaystyle
2C+2(C+3F)+2(C+F)+2(C+F)+2F$ (192) $\displaystyle
2C+2(C+2F)+2(C+2F)+2(C+F)+2F$ (193) $\displaystyle
2C+2(C+F)+2(C+F)+2(C+F)+3F+6F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{2}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(194) $\displaystyle 3(3C+6F)\ \ {\mathbb{Z}}/3{\mathbb{Z}}$ (195)
$\displaystyle 2(4C+8F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}$ (196) $\displaystyle
2(2C+4F)+2(2C+4F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ (197)
$\displaystyle 2C+2(C+2F)+2(2C+6F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$
(198) $\displaystyle 3(C+2F)+3(C+2F)+3(C+2F)\ \
{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (199) $\displaystyle
2C+2(C+4F)+2(C+2F)+2(C+2F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (200)
$\displaystyle 2(C+2F)+2(C+2F)+2(C+2F)+2(C+2F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (201) $\displaystyle 3C+3(2C+4F)+3F+3F\
\ {\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (202) $\displaystyle
2C+2(3C+6F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ (203)
$\displaystyle 2C+2(C+2F)+2(2C+4F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
3}$ (204) $\displaystyle 3C+3(C+3F)+3(C+3F)\ \
{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (205) $\displaystyle
3C+3(C+2F)+3(C+2F)+3F+3F\ \ {\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 3}$ (206)
$\displaystyle 2C+2(C+2F)+2(C+2F)+2(C+2F)+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 4}$ (207) $\displaystyle 2C+4(2C+4F)+2F+2F\
\ {\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (208)
$\displaystyle 4C+2(C+3F)+4(C+2F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (209) $\displaystyle
4C+2(C+2F)+4(C+2F)+4F+4F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}^{\oplus 2}$ (210)
$\displaystyle 4C+2(C+2F)+4(C+2F)+2F+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (211)
$\displaystyle 6C+2(C+2F)+3(C+2F)+6F+6F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (212) $\displaystyle
3(C+2F)+3(2C+4F)$ (213) $\displaystyle 2(C+2F)+4(2C+4F)$ (214) $\displaystyle
2(C+2F)+3(C+2F)+6(C+2F)$ (215) $\displaystyle 2(C+2F)+4(C+2F)+4(C+2F)$ (216)
$\displaystyle 2(3C+6F)+2(C+2F)$ (217) $\displaystyle
2(2C+4F)+2(C+2F)+2(C+2F)$ (218) $\displaystyle 3C+3(2C+6F)$ (219)
$\displaystyle 3C+3(2C+5F)+3F$ (220) $\displaystyle 3C+3(2C+4F)+2F+6F$ (221)
$\displaystyle 2C+3(C+2F)+6(C+2F)+2F+2F$ (222) $\displaystyle
2C+4(C+2F)+4(C+2F)+2F+2F$ (223) $\displaystyle 3C+2(C+3F)+6(C+3F)$ (224)
$\displaystyle 3C+2(C+3F)+6(C+2F)+6F$ (225) $\displaystyle
3C+2(C+2F)+6(C+3F)+2F$ (226) $\displaystyle 3C+2(C+2F)+6(C+2F)+3F+3F$ (227)
$\displaystyle 3C+2(C+2F)+6(C+2F)+2F+6F$ (228) $\displaystyle
3C+3(C+2F)+3(C+4F)$ (229) $\displaystyle 3C+3(C+2F)+3(C+3F)+3F$ (230)
$\displaystyle 3C+3(C+2F)+3(C+2F)+2F+6F$ (231) $\displaystyle
4C+2(C+5F)+4(C+2F)$ (232) $\displaystyle 4C+2(C+4F)+4(C+2F)+2F$ (233)
$\displaystyle 4C+2(C+2F)+4(C+4F)$ (234) $\displaystyle 4C+2(C+2F)+4(C+3F)+4F$
(235) $\displaystyle 4C+2(C+2F)+4(C+2F)+3F+6F$ (236) $\displaystyle
6C+2(C+4F)+3(C+3F)$ (237) $\displaystyle 6C+2(C+4F)+3(C+2F)+3F$ (238)
$\displaystyle 6C+2(C+3F)+3(C+3F)+2F$ (239) $\displaystyle
6C+2(C+3F)+3(C+2F)+2F+3F$ (240) $\displaystyle 6C+2(C+2F)+3(C+3F)+2F+2F$ (241)
$\displaystyle 2C+2(3C+8F)$ (242) $\displaystyle 2C+2(3C+7F)+2F$ (243)
$\displaystyle 2C+2(C+4F)+2(2C+4F)$ (244) $\displaystyle 2C+2(C+3F)+2(2C+5F)$
(245) $\displaystyle 2C+2(C+3F)+2(2C+4F)+2F$ (246) $\displaystyle
2C+2(C+2F)+2(2C+5F)+2F$ (247) $\displaystyle 6C+2(C+2F)+3(C+2F)+4F+12F$ (248)
$\displaystyle 6C+2(C+2F)+3(C+2F)+2F+2F+3F$ (249) $\displaystyle
2C+2(C+3F)+2(C+3F)+2(C+2F)$ (250) $\displaystyle
2C+2(C+3F)+2(C+2F)+2(C+2F)+2F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{3}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(251) $\displaystyle 3C+3(2C+6F)+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}$ (252)
$\displaystyle 3C+3(C+3F)+3(C+3F)+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (253)
$\displaystyle 6C+2(C+3F)+3(C+3F)+4F+4F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$
(254) $\displaystyle 6C+2(C+3F)+3(C+3F)+2F+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}\oplus{\mathbb{Z}}/3{\mathbb{Z}}$ (255)
$\displaystyle 2C+4(2C+6F)+2F$ (256) $\displaystyle 2C+3(C+3F)+6(C+3F)+2F$
(257) $\displaystyle 2C+4(C+3F)+4(C+3F)+2F$ (258) $\displaystyle
3C+2(C+5F)+6(C+3F)$ (259) $\displaystyle 3C+2(C+4F)+6(C+3F)+2F$ (260)
$\displaystyle 3C+2(C+3F)+6(C+3F)+2F+2F$ (261) $\displaystyle
4C+2(C+4F)+4(C+4F)$ (262) $\displaystyle 4C+2(C+4F)+4(C+3F)+4F$ (263)
$\displaystyle 4C+2(C+3F)+4(C+4F)+2F$ (264) $\displaystyle
4C+2(C+3F)+4(C+3F)+2F+4F$ (265) $\displaystyle 6C+2(C+6F)+3(C+3F)$ (266)
$\displaystyle 6C+2(C+5F)+3(C+3F)+2F$ (267) $\displaystyle
6C+2(C+4F)+3(C+3F)+2F+2F$ (268) $\displaystyle 6C+2(C+3F)+3(C+4F)+6F$ (269)
$\displaystyle 6C+2(C+3F)+3(C+3F)+3F+6F$ (270) $\displaystyle 2C+2(3C+10F)$
(271) $\displaystyle 2C+2(3C+9F)+2F$ (272) $\displaystyle 2C+2(C+4F)+2(2C+6F)$
(273) $\displaystyle 2C+2(C+3F)+2(2C+7F)$ (274) $\displaystyle
2C+2(C+3F)+2(2C+6F)+2F$ (275) $\displaystyle 2C+2(C+4F)+2(C+3F)+2(C+3F)$ (276)
$\displaystyle 2C+2(C+3F)+2(C+3F)+2(C+3F)+2F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{4}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(277) $\displaystyle 2C+2(3C+12F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}$ (278)
$\displaystyle 2C+4(2C+8F)\ \ {\mathbb{Z}}/4{\mathbb{Z}}$ (279) $\displaystyle
2C+2(C+4F)+2(2C+8F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}$ (280)
$\displaystyle 4C+2(C+6F)+4(C+4F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (281)
$\displaystyle 4C+2(C+4F)+4(C+4F)+2F+2F\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus
2}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (282) $\displaystyle
2C+2(C+4F)+2(C+4F)+2(C+4F)\ \ {\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 3}$ (283)
$\displaystyle 6C+2(C+4F)+3(C+4F)+3F+3F\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$ (284)
$\displaystyle 3C+3(2C+9F)$ (285) $\displaystyle 3C+3(2C+8F)+3F$ (286)
$\displaystyle 2C+3(C+4F)+6(C+4F)$ (287) $\displaystyle 2C+4(C+4F)+4(C+4F)$
(288) $\displaystyle 3C+2(C+4F)+6(C+4F)+3F$ (289) $\displaystyle
3C+3(C+4F)+3(C+5)$ (290) $\displaystyle 3C+3(C+4F)+3(C+4F)+3F$ (291)
$\displaystyle 4C+2(C+5F)+4(C+4F)+2F$ (292) $\displaystyle
6C+2(C+5F)+3(C+4F)+6F$ (293) $\displaystyle 6C+2(C+4F)+3(C+6F)$ (294)
$\displaystyle 6C+2(C+4F)+3(C+5F)+3F$ (295) $\displaystyle
6C+2(C+4F)+3(C+4F)+2F+6F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{5}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(296) $\displaystyle 4C+2(C+5F)+4(C+6F)$ (297) $\displaystyle
4C+2(C+5F)+4(C+5F)+4F$ (298) $\displaystyle 6C+2(C+6F)+3(C+6F)$ (299)
$\displaystyle 6C+2(C+6F)+3(C+5F)+3F$ (300) $\displaystyle
6C+2(C+5F)+3(C+6F)+2F$ (301) $\displaystyle 6C+2(C+5F)+3(C+5F)+2F+3F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{6}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(302) $\displaystyle 3C+3(2C+12F)\ \ {\mathbb{Z}}/3{\mathbb{Z}}$ (303)
$\displaystyle 3C+3(C+6F)+3(C+6F)\ \ {\mathbb{Z}}/3{\mathbb{Z}}^{\oplus 2}$
(304) $\displaystyle 6C+2(C+6F)+3(C+6F)+2F+2F\ \
{\mathbb{Z}}/2{\mathbb{Z}}^{\oplus 2}\oplus{\mathbb{Z}}/3{\mathbb{Z}}$ (305)
$\displaystyle 3C+2(C+6F)+6(C+6F)$ (306) $\displaystyle 4C+2(C+7F)+4(C+6F)$
(307) $\displaystyle 4C+2(C+6F)+4(C+6F)+2F$ (308) $\displaystyle
6C+2(C+8F)+3(C+6F)$ (309) $\displaystyle 6C+2(C+7F)+3(C+6F)+2F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{7}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(310) $\displaystyle 6C+2(C+7F)+3(C+7F)+6F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{8}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(311) $\displaystyle 4C+2(C+8F)+4(C+8F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/4{\mathbb{Z}}$ (312)
$\displaystyle 6C+2(C+8F)+3(C+9F)$ (313) $\displaystyle 6C+2(C+8F)+3(C+8F)+3F$

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{9}$, then by Theorem 2.7 the numerical class of
$B$ is one of the following:

(314) $\displaystyle 6C+2(C+10F)+3(C+9F)$ (315) $\displaystyle
6C+2(C+9F)+3(C+9F)+2F$

By Theorem 2.7 there are not a $K3$ surface $X$ and a finite subgroup $G$ of
Aut$(X)$ such that $X/G\cong{\mathbb{F}}_{l}$ for $l=10,11$.

If there are a $K3$ surface $X$ and a finite subgroup $G$ of Aut$(X)$ such
that $X/G\cong{\mathbb{F}}_{12}$, then by Theorem 2.7 the numerical class of
$B$ is the following:

(316) $\displaystyle 6C+2(C+12F)+3(C+12F)\ \
{\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}$

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