# Cancellation for $(G,n)$-complexes and the Swan finiteness obstruction

Johnny Nicholson  Department of Mathematics, UCL, Gower Street, London, WC1E
6BT, U.K<EMAIL_ADDRESS>
###### Abstract.

In previous work, we related homotopy types of $(G,n)$-complexes when $G$ has
periodic cohomology to projective $\mathbb{Z}G$ modules representing the Swan
finiteness obstruction. We use this to determine when $X\vee S^{n}\simeq Y\vee
S^{n}$ implies $X\simeq Y$ for $(G,n)$-complexes $X$ and $Y$, and give lower
bounds for the number of minimal homotopy types of $(G,n)$-complexes when this
fails. The proof involves constructing projective $\mathbb{Z}G$ modules as
lifts of locally free modules over orders in products of quaternion algebras,
whose existence follows from the Eichler mass formula. In the case $n=2$,
difficulties arise which lead to a new approach to finding a counterexample to
Wall’s D2 problem.

###### 2010 Mathematics Subject Classification:

Primary 55P15; Secondary 11R52, 20C05, 57M20

## Introduction

For a finitely presented group $G$ and $n\geq 2$, let $HT(G,n)$ be the set of
homotopy types of $(G,n)$-complexes, i.e. finite $n$-dimensional CW-complexes
$X$ for which $\pi_{1}(X)\cong G$ and the universal cover $\widetilde{X}$ is
$(n-1)$-connected. This is a graded tree with edges between each $X$ and
$X\vee S^{n}$ and with grading coming from
$\boldsymbol{\chi}(X)=(-1)^{n}\chi(X)$. The first basic question is whether or
not $HT(G,n)$ has cancellation, i.e. has the property that $X\vee S^{n}\simeq
Y\vee S^{n}$ implies that $X\simeq Y$.

This question is completely inaccessible in general. However, a solution in
the case where $G$ is finite abelian, which includes non-cancellation
examples, follows from work of Browning [3], Dyer-Sieradski [13] and Metzler
[26], and further examples of non-cancellation have appeared elsewhere in the
literature [12, 22]. These examples are of special interest due to their
applications to smooth $4$-manifolds [2, 18, 21], Wall’s D2 problem [20, 31]
and combinatorial group theory [27].

In this article, we will consider the case where $G$ has periodic cohomology.
Let $m_{\mathbb{H}}(G)$ denote the number of one-dimensional quaternionic
representations and recall that $G$ has the D2 property if every
cohomologically $2$-dimensional finite complex $X$ with $\pi_{1}(X)\cong G$ is
homotopic to a $2$-complex [44]. Let $PHT(G,n)$ denote the tree of polarised
homotopy types of $(G,n)$-complexes, i.e. the homotopy types of pairs
$(X,\rho)$ where $X$ is a $(G,n)$-complex and $\rho:\pi_{1}(X)\to G$ is an
isomorphism.

Our main result is that cancellation for $HT(G,n)$ is completely determined by
$m_{\mathbb{H}}(G)$. This generalises [31, Theorem A] which considered
$PHT(G,2)$ when $k=4$.

######  Theorem A.

Let $G$ have $k$-periodic cohomology and let $n$ be such that $k\mid n$ or
$n+2$ and, if $n=2$, suppose $G$ has the D2 property. Then the following are
equivalent:

  1. (i)

$HT(G,n)$ has cancellation

  2. (ii)

$PHT(G,n)$ has cancellation

  3. (iii)

$m_{\mathbb{H}}(G)\leq 2$.

This is in spite of the fact that, in [32, Section 8], we showed that there
exists $G$ with $4$-periodic cohomology such that $HT(G,n)\neq PHT(G,n)$ for
$n>2$ is even.

Recall that, if $G$ is finite and $n$ is even, then $HT(G,n)$ is a fork in
that it has a single vertex at each non-minimal grade $\boldsymbol{\chi}(X)$
and finitely many at the minimal level [32, Corollary 3.7]. Let
$N(G,n)=\\#\\{X\in HT(G,n):\boldsymbol{\chi}(X)\text{ is minimal}\\}$. Our
second result is the following:

######  Theorem B.

Let $G$ have $k$-periodic cohomology and let $n$ be such that $k\mid n$ or
$n+2$ and, if $n=2$, suppose $G$ has the D2 property. If
$m=m_{\mathbb{H}}(G)$, then

$N(G,n)\geq e^{\tfrac{m\log m}{8\log\log m}+O(m\log\log m)}$

In particular, for all $\lambda>0$, we have that $N(G,n)/e^{\lambda
m}\to\infty$ as $m\to\infty$.

We now consider the case $n=2$ in more detail. Recall that every connected
finite $2$-complex $X$ with $\pi_{1}(X)\cong G$ is the presentation complex of
some presentation $\mathcal{P}$ of $G$ and so $N(G,2)$ is the number of
homotopy classes of presentations of $G$ with maximal deficiency. Recall also
that Wall’s D2 problem asks whether every cohomologically 2-dimensional finite
complex is homotopic to a finite 2-complex or, equivalently, whether every
finitely presented group $G$ has the D2 property. This remains open, in
contrast to all other dimensions [37, 38, 44].

It is a consequence of the theorems above that, if Wall’s D2 problem can be
answered in the affirmative, then we would expect non-cancellation examples
for finite 2-complexes over $G$ provided $m_{\mathbb{H}}(G)$ is sufficiently
large. For example, this applies when $G=Q_{28}$ is the quaternion group of
order 28 since $G$ has the D2 property and $m_{\mathbb{H}}(G)=3$ [31, Theorem
7.7].

However, if we are interested in the D2 property itself, we could instead view
the above theorems as a constraint that needs to be satisfied in order for the
D2 property to hold. For example, if the quaternion groups $Q_{4n}$ have the
D2 property for all $n\geq 2$, then $m_{\mathbb{H}}(Q_{4n})=\lfloor
n/2\rfloor$ implies that $N(Q_{4n},2)\gg e^{\lambda n}$ for all $\lambda>0$
and $n$ sufficiently large. In [24, Section 3], a family of presentations
$\mathcal{E}_{n,r}$ for $1\leq r\leq n$ are given which are proposed to
contain all presentations for $Q_{4n}$ up to homotopy. These give at most $n$
presentations and so, for $n$ sufficiently large, either $Q_{4n}$ does not
have the D2 property or $Q_{4n}$ has a presentation not of the form
$\mathcal{E}_{n,r}$. This leads to a new approach to finding a counterexample
to the D2 problem.

Using previous work [32], Theorems A and B can be reduced to pure algebra as
follows. Let $n=ik$ or $ik-2$, let $P_{(G,n)}$ be a projective $\mathbb{Z}G$
module representing the Swan finiteness obstruction $\sigma_{ik}(G)$ and, if
$n=2$, suppose that $G$ has the D2 property. Then, by [32, Theorems A, B],
there are isomorphisms of graded trees

$\Psi:PHT(G,n)\to[P_{(G,n)}],\quad\bar{\Psi}:HT(G,n)\to[P_{(G,n)}]/\operatorname{Aut}(G)$

where $[P_{(G,n)}]$ is the set of projective $\mathbb{Z}G$ modules $P$ with
$P\oplus\mathbb{Z}G^{i}\cong P_{(G,n)}\oplus\mathbb{Z}G^{j}$ for some $i,j\geq
0$ and the action by $\theta\in\operatorname{Aut}(G)$ is given by
$\theta:P\mapsto(I,\psi_{k}(\theta)^{i})\otimes P_{\theta}$ for a map
$\psi_{k}:\operatorname{Aut}(G)\to(\mathbb{Z}/|G|)^{\times}$ which depends
only on $G$ and $k$ [32, Section 6].

The paper will be structured as follows. In Section 1, we begin by
establishing the necessary group-theoretic facts on groups with periodic
cohomology. This includes calculating $m_{\mathbb{H}}(G)$ for each group and
relating its value to the vanishing of $\sigma_{k}(G)$. In Section 2, we will
combine the results in Section 1 with [31, Theorem 4.1] to show that, if
$m_{\mathbb{H}}(G)\leq 2$, then $[P_{(G,n)}]$ has cancellation. Since the
converse also holds, this leads to a complete determination of when
cancellation occurs for a representative of $\sigma_{k}(G)$ and implies (ii)
$\Leftrightarrow$ (iii) in Theorem A.

The work to establish the non-cancellation results will be done in Sections
3-5. In Section 3 we will show that, if $f:G\twoheadrightarrow Q_{4n}$ then
there is an induced map
$f_{*}:\operatorname{Aut}(G)\to\operatorname{Aut}(Q_{4n})$ and a surjection
$[P_{(G,n)}]/\operatorname{Aut}(G)\twoheadrightarrow[\widebar{P_{(G,n)}}]/\operatorname{Im}(f_{*})$.
This allows us to show non-cancellation occurs for $G$ by considering the case
$Q_{4n}$.

In Section 4, we discuss locally free modules over orders in quaternion
algebras and the Eichler mass formula. In Section 5 we study the orders
$\Lambda=\Lambda_{n_{1},\cdots,n_{k}}$ in
$A=\prod_{i=1}^{k}\mathbb{Q}[\zeta_{n_{i}},j]$ which arise as quotients of
$\mathbb{Z}Q_{4n}$. If $k=1$, then $A$ is a quaternion algebra and, using
results in Section 4, we can determine when $\Lambda$ has stably free
cancellation, i.e. every stably free $\Lambda$-module is free. In Theorem
5.13, we use this to determine precisely when $\Lambda$ has stably free
cancellation, completing the classification done by Swan in [41, Section 8].
In Sections 6 and 7, we then apply these results to prove Theorems A and B.

Finally, note that projective $\mathbb{Z}G$ modules of the form $P_{(G,n)}$
have two special properties. Let $C(\mathbb{Z}G)$ denote the projective class
group and $T_{G}$ the Swan subgroup which is generated by the $(I,r)$ where
$(r,|G|)=1$. Then we have the following:

  1. (1)

If $\theta\in\operatorname{Aut}(G)$, then
$[(P_{(G,n)})_{\theta}]=[P_{(G,n)}]\in C(\mathbb{Z}G)/T_{G}$

  2. (2)

$[P_{(G,n)}]$ has cancellation if and only if
$[P_{(G,n)}]/\operatorname{Aut}(G)$ has cancellation.

where (1) follows from the definition of the action of $\operatorname{Aut}(G)$
using $\psi:\operatorname{Aut}(G)\to(\mathbb{Z}/|G|)^{\times}$, and (2)
follows from Theorem A.

In Section 8, we show that each property fails in general for projective
$\mathbb{Z}G$ modules even when $G$ has periodic cohomology. For (1), we show
that, if $G=C_{p}$ for $p\geq 23$ prime, then there exists $[P]\in
C(\mathbb{Z}G)$ and $\theta\in\operatorname{Aut}(G)$ such that
$[P_{\theta}]\neq[P]\in C(\mathbb{Z}G)/T_{G}$. This can be deduced non-
explicitly as an application of Chevalley’s ambiguous class number formula
[5]. For (2) note that, if $G=Q_{28}$ is the quaternion group of order 28,
then $T_{G}=0$ and $\operatorname{Aut}(G)$ acts trivially on $C(\mathbb{Z}G)$.
For the action $P\mapsto P_{\theta}$, we show:

######  Theorem 8.4.

There exists $[P]\in C(\mathbb{Z}G)$ such that $[P]$ has non-cancellation but
$[P]/\operatorname{Aut}(G)$ has cancellation.

This work can be viewed as an attempt to properly amalgamate the techniques
and results obtained by Swan in [41] with the wider literature on applications
of the Swan finiteness obstruction [20, 28, 31]. As such, we will rely heavily
on calculations done in [41], though we will give alternate proofs where
possible.

Acknowledgment. I would like to thank my supervisor F. E. A. Johnson whose
suggestion that I read ‘Swan’s long paper’ [41] was largely the inspiration
for this work. I would also like to thank Pol van Hoften for helpful comments.

##  1\. Groups with periodic cohomology

Recall that a binary polyhedral group is a non-cyclic finite subgroup of
$\mathbb{H}^{\times}$ where $\mathbb{H}$ is the real quaternions. They are the
generalised quaternion groups

$Q_{4n}=\langle x,y\mid x^{n}=y^{2},yxy^{-1}=x^{-1}\rangle$

for $n\geq 2$ and the binary tetrahedral, octahedral and icosahedral groups
$\widetilde{T}$, $\widetilde{O}$, $\widetilde{I}$ which are the preimages of
the dihedral groups $D_{2n}$ and the symmetry groups $T$, $O$, $I$ under the
double cover of Lie groups $f:\mathbb{H}^{\times}\cong S^{3}\to SO(3)$.

We say that a group $G$ has $k$-periodic cohomology for some $k\geq 1$ if its
Tate cohomology groups satisfy
$\hat{H}^{i}(G;\mathbb{Z})=\hat{H}^{i+k}(G;\mathbb{Z})$ for all
$i\in\mathbb{Z}$. For example, it is easy to show that the binary polyhedral
groups have $4$-periodic cohomology. The following can be found in [4, Theorem
11.6].

######  Proposition 1.1.

If $G$ is a finite group, then the following are equivalent:

  1. (i)

$G$ has periodic cohomology

  2. (ii)

$G$ has no subgroup of the form $C_{p}\times C_{p}$ for $p$ prime

  3. (iii)

The Sylow subgroups of $G$ are cyclic or generalised quaternionic $Q_{2^{n}}$.

Let $\operatorname{SL}_{2}(\mathbb{F}_{p})$ be the special linear group of
degree $2$ over $\mathbb{F}_{p}$, let $\operatorname{TL}_{2}(\mathbb{F}_{p})$
be the non-split extension of $C_{2}$ by
$\operatorname{SL}_{2}(\mathbb{F}_{p})$ [45, Proposition 1.2 (iii)] and recall
that $\widetilde{T}\cong\operatorname{SL}_{2}(\mathbb{F}_{3})$,
$\widetilde{O}\cong\operatorname{TL}_{2}(\mathbb{F}_{3})$ and
$\widetilde{I}\cong\operatorname{SL}_{2}(\mathbb{F}_{5})$. Let $O(G)$ be the
unique maximal normal subgroup of odd order. If $G$ has periodic cohomology,
then the type of $G$ is determined by $G/O(G)$ as follows [45, Corollary 2.6].
For reasons that will become apparent later, we will split II and V into two
classes.

Type I IIa IIb III IV Va Vb VI $G/O(G)$ $C_{2^{n}}$ $Q_{8}$ $Q_{2^{n}}$,
$n\geq 4$ $\widetilde{T}$ $\widetilde{O}$ $\widetilde{I}$
$\operatorname{SL}_{2}(\mathbb{F}_{p})$, $p\geq 7$
$\operatorname{TL}_{2}(\mathbb{F}_{p})$, $p\geq 5$

For the rest of this section, we will assume all groups are finite and will
write $f:G\twoheadrightarrow H$ to denote a surjective group homomorphism. We
will also assume basic facts about quaternion groups; for example, $Q_{2^{n}}$
has proper quotients $C_{2}$ and the dihedral groups $D_{2^{m}}$ for $1<m<n$.
We begin with the following observation.

######  Proposition 1.2.

Let $f:G\twoheadrightarrow H$ where $G$ and $H$ have periodic cohomology. If
$|H|>2$, then $G$ and $H$ have the same type.

###### Proof.

Note that $f(O(G))\leq H$ has odd order and so is contained in $O(H)$. In
particular, $f$ induces a quotient $f:G/O(G)\twoheadrightarrow H/O(H)$. Hence
it suffices to show that there are no (proper) quotients among groups in the
family

$\mathscr{F}=\\{C_{2^{n}},Q_{2^{m}},\operatorname{SL}_{2}(\mathbb{F}_{p}),\operatorname{TL}_{2}(\mathbb{F}_{p}):n\geq
2,m\geq 3,p\geq 3\text{ prime}\\}$

unless both are cyclic. Firstly, the quotients of $Q_{2^{n}}$ are $D_{2^{m}}$
for $1<m<n$ and $C_{2}$ which are not in $\mathscr{F}$. It is easy to verify
that the quotients of $\operatorname{SL}_{2}(\mathbb{F}_{3})$ are $C_{3}$,
$A_{4}$ and the quotients of $\operatorname{TL}_{2}(\mathbb{F}_{3})$ are
$C_{2}$, $S_{3}$, $S_{4}$, none of which are in $\mathscr{F}$.

For $p\geq 5$, it is well known [10] that
$\operatorname{SL}_{2}(\mathbb{F}_{p})$ has one (proper) normal subgroup
$C_{2}$ with quotient $\operatorname{PSL}_{2}(\mathbb{F}_{p})$ and similarly
$\operatorname{TL}_{2}(\mathbb{F}_{p})$ has normal subgroups $C_{2}$,
$\operatorname{SL}_{2}(\mathbb{F}_{p})$ with quotients
$\text{PGL}_{2}(\mathbb{F}_{p})$, $C_{2}$. That these groups are not in
$\mathscr{F}$ follows, for example, from [45, Proposition 1.3]. ∎

We will split this section into three parts. Firstly, we will determine the
binary polyhedral quotients of groups $G$ with periodic cohomology. We will
then use this to determine the number of one dimensional quaternionic
representations $m_{\mathbb{H}}(G)$, and finally we compare this with the Swan
finiteness obstruction $\sigma_{k}(G)$.

###  1.1. Binary polyhedral quotients

If $G$ is a finite group, we say that two quotients $f_{1}:G\twoheadrightarrow
H_{1}$, $f_{2}:G\twoheadrightarrow H_{2}$ are equivalent, written $f_{1}\equiv
f_{2}$, if $\operatorname{Ker}(f_{1})=\operatorname{Ker}(f_{2})$ are equal as
sets (and so $H_{1}\cong H_{2}$).

For a prime $p$, let $G_{p}$ be the isomorphism class of the Sylow
$p$-subgroup of $G$. It is useful to note that, if $1\to N\to G\to H\to 1$ is
an extension, then there is an extension of abstract groups $1\to N_{p}\to
G_{p}\to H_{p}\to 1$.

######  Lemma 1.3.

Let $f:G\twoheadrightarrow H$ where $G$ and $H$ have periodic cohomology and
$4\mid|H|$. If $f^{\prime}:G\twoheadrightarrow H^{\prime}$ and
$|H|=|H^{\prime}|$, then $f\equiv f^{\prime}$, i.e. $H\cong H^{\prime}$ and
$\operatorname{Ker}(f)=\operatorname{Ker}(f^{\prime})$.

###### Proof.

Let $H=G/N$, $H^{\prime}=G/N^{\prime}$ and define $\bar{G}=G/(N\cap
N^{\prime})$. Since there are successive quotients
$G\twoheadrightarrow\bar{G}\twoheadrightarrow H$, we have
$G_{p}\twoheadrightarrow\bar{G}_{p}\twoheadrightarrow H_{p}$ for all primes
$p$. If $G_{p}$ is cyclic, then this implies $\bar{G}_{p}$ is cyclic. If not,
then $p=2$ and $G_{2}=Q_{2^{n}}$ which has proper quotients $D_{2^{m}}$ for
$2\leq m\leq n-1$ and $C_{2}$. Since $H$ has periodic cohomology, $H_{2}$ is
cyclic or generalised quaternionic and so $H_{2}=Q_{2^{n}}$ since
$4\mid|H_{2}|$. Hence $\bar{G}_{2}=Q_{2^{n}}$ since $G_{2}\twoheadrightarrow
H_{2}$ factors through $\bar{G}_{2}$, and so $\bar{G}$ has periodic
cohomology.

Now note that $K=N/(N\cap N^{\prime})$ and $K^{\prime}=N^{\prime}/(N\cap
N^{\prime})$ are disjoint normal subgroups of $\bar{G}$ and so $K\cdot
K^{\prime}=K\times K^{\prime}\leq\bar{G}$ by the recognition criteria for
direct products. Hence $K\times K^{\prime}\leq\bar{G}$ and, since $\bar{G}$
has periodic cohomology, Proposition 1.1 (ii) implies that $|K|$ and
$|K^{\prime}|$ are coprime. Since $|N|=|N^{\prime}|$, this implies that
$|K|=|K^{\prime}|=1$ and so $|N|=|N\cap N^{\prime}|=|N^{\prime}|$ and
$N=N^{\prime}$ as required. ∎

Now let $\mathcal{B}(G)$ denote the set of equivalence classes of quotients
$f:G\twoheadrightarrow H$ where $H$ is a binary polyhedral group. Since
$4\mid|H|$, applying Lemma 1.3 again gives:

######  Corollary 1.4.

Let $G$ have periodic cohomology and let $f_{1},f_{2}\in\mathcal{B}(G)$. Then
$f_{1}\equiv f_{2}$ if and only if
$\operatorname{Im}(f_{1})\cong\operatorname{Im}(f_{2})$.

In particular, this shows that $\mathcal{B}(G)$ is in one-to-one
correspondence with the isomorphism classes of binary polyhedral groups $H$
which are quotients of $G$. For brevity, we will often write
$H\in\mathcal{B}(G)$ when there exists $f:G\twoheadrightarrow H$ with
$f\in\mathcal{B}(G)$.

In order to determine $\mathcal{B}(G)$, it suffices to determine the set of
maximal binary polyhedral quotients
$\operatorname{\mathcal{B}_{\text{max}}}(G)$, i.e. the subset containing those
$f\in\mathcal{B}(G)$ such that $f$ does not factor through any other
$g\in\mathcal{B}(G)$. The rest of this section will be devoted to proving the
following:

######  Theorem 1.5.

If $G$ has periodic cohomology, then the type and the number of maximal binary
polyhedral quotients $\\#\operatorname{\mathcal{B}_{\text{max}}}(G)$ are
related as follows.

Type | I | IIa | IIb | III | IV | Va | Vb | VI
---|---|---|---|---|---|---|---|---
$\\#\operatorname{\mathcal{B}_{\text{max}}}(G)$ | 0,1 | 1 | 1,2,3 | 1 | 1 | 1 | 0 | 0

#### Type I

Recall that $G$ has type I if and only if its Sylow subgroups are cyclic, and
$G$ has a presentation

$G=\langle u,v\mid u^{m}=v^{n}=1,vuv^{-1}=u^{r}\rangle$

for some $r\in\mathbb{Z}/m$ where $r^{n}\equiv 1$ mod $m$ [45, Lemma 3.1] and
$(n,m)=1$. We will write $C_{m}\rtimes_{(r)}C_{n}$ to denote this
presentation, where $C_{n}=\langle u\rangle$ and $C_{m}=\langle v\rangle$. By
[20, p165], we can assume that $m$ is odd.

If $G$ has a binary polyhedral quotient $H$, then Proposition 1.2 implies that
$H=Q_{4a}$ for $a>1$ odd and $4\mid n$ since $m$ is odd.

######  Lemma 1.6.

Let $G=C_{m}\rtimes_{(r)}C_{4n}$. Then $G$ has a quotient $Q_{4a}$ if and only
if $a\mid m$ and $r\equiv-1$ mod $a$.

###### Proof.

Recall that $Q_{4a}=C_{a}\rtimes_{(-1)}C_{4}$. If $a\mid m$ and and
$r\equiv-1$ mod $a$, then $\langle u^{a},v^{4}\rangle\leq G$ is normal since
$r^{4}\equiv 1$ mod $a$ implies $uv^{4}u^{-1}=u^{1-r^{4}}v^{4}\in\langle
u^{a},v^{4}\rangle$. This implies that $G/\langle u^{a},v^{4}\rangle\cong
C_{a}\rtimes_{(r)}C_{4}=Q_{4a}$ since $r\equiv-1$ mod $a$.

Conversely, if $f:G\twoheadrightarrow Q_{4a}$, then $Q_{4a}\cong\langle
f(u)\rangle\rtimes_{(r)}\langle f(v)\rangle$ and $|\langle f(u)\rangle|\mid
m$, $|\langle f(u)\rangle|\mid 4n$. Since $Q_{4a}$ contains a maximal normal
cyclic subgroup $C_{2a}$, and $m$ is odd, we must have $\langle
f(u)\rangle\leq C_{a}$. So $a\mid m$, which implies that $(a,4n)=1$ and
$\langle f(u)\rangle\leq C_{4}$ for some $C_{4}\leq Q_{4a}$. Hence $\langle
f(u)\rangle=C_{a}$ and $\langle f(v)\rangle=C_{4}$ since they generate
$Q_{4a}$. As $C_{a}\leq Q_{4a}$ is unique and $C_{4}\leq Q_{4a}$ is unique up
to conjugation, we can write $Q_{4a}\cong\langle
f(u)\rangle\rtimes_{(-1)}\langle f(v)\rangle$, i.e. $r\equiv-1$ mod $a$. ∎

Now suppose $G$ has two maximal binary polyhedral quotients
$f_{a}:G\twoheadrightarrow Q_{4a}$, $f_{b}:G\twoheadrightarrow Q_{4b}$ for
some $a,b>1$ odd, and we can assume $a$ is maximal. Then Lemma 1.6 implies
that $a,b\mid m$ and $r\equiv-1$ mod $a$ and $r\equiv-1$ mod $b$. If
$d=\text{lcm}(a,b)$, then $d\mid m$ and $r\equiv-1$ mod $d$ and so there is a
quotient $f_{d}:G\twoheadrightarrow Q_{4d}$ by Lemma 1.6. By Corollary 1.4 (or
the proof of Lemma 1.6), $f_{a}$ and $f_{b}$ factor through $f_{d}$ which
implies that $a=b=d$ as $f_{a}$ and $f_{b}$ are maximal. By Corollary 1.4
again, this implies that $f_{a}$ and $f_{b}$ are equivalent. In particular,
this shows that $\\#\operatorname{\mathcal{B}_{\text{max}}}(G)\leq 1$.

#### Type II

Recall that, if $G$ has type II, then $O(G)\leq G$ has cyclic Sylow subgroups
and so there exists $n\geq 3$ and $t$, $s$ odd coprime such that

$G\cong(C_{t}\rtimes_{(r)}C_{s})\rtimes_{(a,b)}Q_{2^{n}}.$

Furthermore, if $C_{t}=\langle u\rangle$, $C_{s}=\langle v\rangle$ and
$Q_{2^{n}}$ is as above, then $Q_{2^{n}}$ acts via

$\varphi_{x}:u\mapsto u^{a},v\mapsto v,\quad\varphi_{y}:u\mapsto
u^{b},v\mapsto v$

for some $a,b\in\mathbb{Z}/t$ with $a^{2}\equiv b^{2}\equiv 1$ mod $t$ [45,
Theorem 3.6]. In what follows we will write $G=C_{t}\rtimes_{(a,b)}Q_{2^{n}}$
when $s=1$.

If $G$ has a binary polyhedral quotient $H$, then the proof of Proposition 1.2
implies that $H/O(H)=Q_{2^{n}}$ and so $H=Q_{2^{n}m}$ for some $m$ odd.

######  Lemma 1.7.

Let $G=(C_{t}\rtimes_{(r)}C_{s})\rtimes_{(a,b)}Q_{2^{n}}$. Then $G$ has a
quotient $Q_{2^{n}m}$ if and only if $m\mid t$, $r\equiv 1$ mod $m$ and
$Q_{2^{n}m}\cong C_{m}\rtimes_{(a,b)}Q_{2^{n}}$.

###### Proof.

If $m\mid t$ and $r\equiv 1$ mod $m$, then $\langle u^{m},v\rangle\leq G$ is
normal since $uvu^{-1}=u^{1-r}v\in\langle u^{m},v\rangle$. This implies that
$G/\langle u^{m},v\rangle\cong C_{t}\rtimes_{(a,b)}Q_{2^{n}}$ which has
quotient $C_{m}\rtimes_{(a,b)}Q_{2^{n}}$ since $m\mid t$. If $Q_{2^{n}m}\cong
C_{m}\rtimes_{(a,b)}Q_{2^{n}}$, then $G$ has quotient $Q_{2^{n}m}$.

Conversely, suppose $f:G\twoheadrightarrow Q_{2^{n}m}$. Let
$h:G\twoheadrightarrow G/\langle u,v\rangle\cong Q_{2^{n}}$ and note that, if
$g:Q_{2^{n}m}\twoheadrightarrow Q_{2^{n}}$, then $\operatorname{Ker}(g\circ
f)=\operatorname{Ker}(h)=\langle u,v\rangle$ by Corollary 1.4 and so
$\operatorname{Ker}(f)\leq\langle u,v\rangle$. By composing $g$ with an
element of $\operatorname{Aut}(Q_{2^{n}})$, we can assume $g\circ f=h$ and so
$Q_{2^{n}m}\cong\operatorname{Ker}(g)\rtimes\langle f(x),f(y)\rangle$. Since
$f(v)\in\operatorname{Ker}(g)$ has a trivial action by $\langle
f(x),f(y)\rangle\cong Q_{2^{n}}$, this implies $f(v)=1$, i.e.
$v\in\operatorname{Ker}(f)$. This implies $\operatorname{Ker}(f)=\langle
u^{\ell},v\rangle$ for some $\ell\mid t$ and we need $\ell=m$ since
$\operatorname{Ker}(f)\leq G$ has index $2^{n}m$. Hence $m\mid t$ and, by
normality, $uvu^{-1}=u^{1-r}v\in\langle u^{m},v\rangle$ and so $r\equiv 1$ mod
$m$. Finally, we have $Q_{2^{n}m}\cong G/\langle u^{m},v\rangle\cong
C_{m}\rtimes_{(a,b)}Q_{2^{n}}$. ∎

######  Lemma 1.8.

If $m\geq 1$, then $Q_{2^{n}m}\cong C_{m}\rtimes_{(a,b)}Q_{2^{n}}$ if and only
if

$(a,b)=\begin{cases}(1,-1),(-1,1),(-1,-1),&\text{if $n=3$}\\\ (1,-1),&\text{if
$n\geq 4$.}\end{cases}$

###### Proof.

It follows easily from the standard presentation that $Q_{2^{n}m}\cong
C_{m}\rtimes_{(1,-1)}Q_{2^{n}}$. If $f:Q_{2^{n}m}\twoheadrightarrow
Q_{2^{n}}$, then $\operatorname{Ker}(f)=C_{m}$ is unique by Corollary 1.4.
Hence $Q_{2^{n}m}\cong C_{m}\rtimes_{(a,b)}Q_{2^{n}}$ if and only if there
exists $\theta\in\operatorname{Aut}(Q_{2^{n}})$ such that
$\varphi_{(a,b)}=\varphi_{(1,-1)}\circ\theta$ where
$\varphi_{(i,j)}:Q_{2^{n}}\to\operatorname{Aut}(C_{m})\cong(\mathbb{Z}/m)^{\times}$
has $\varphi_{(i,j)}(x)=i$, $\varphi_{(i,j)}(y)=j$. This implies that
$\operatorname{Im}(\varphi_{(a,b)})\leq\operatorname{Im}(\varphi_{(1,-1)})=\langle
1,-1\rangle=\\{\pm 1\\}$ and so $a,b\in\\{\pm 1\\}$. If $(a,b)=(1,1)$, then
$Q_{2^{n}m}\cong C_{m}\times Q_{2^{n}}$ which is a contradiction unless $m=1$,
in which case $(1,1)=(1,-1)$. In particular,
$(a,b)\in\\{(1,-1),(-1,1),(-1,-1)\\}$.

If $n=3$, then we $\theta_{1}:x\mapsto y,y\mapsto x$ has
$\varphi_{(1,-1)}\circ\theta_{1}=\varphi_{(1,-1)}$ and $\theta_{2}:x\mapsto
y,y\mapsto xy$ has $\varphi_{(1,-1)}\circ\theta_{2}=\varphi_{(-1,-1)}$. Hence
all $(a,b)$ are possible. If $n\geq 4$, then

$\operatorname{Aut}(Q_{2^{n}})=\\{\theta_{i,j}:x\mapsto x^{i},y\mapsto
x^{j}y\mid i\in(\mathbb{Z}/{2^{n-1}})^{\times},j\in\mathbb{Z}/{2^{n-1}}\\}$

and $\varphi_{(1,-1)}\circ\theta_{i,j}=\varphi_{(1,-1)}$ for all $i,j$ and so
only $(a,b)=(1,-1)$ is possible. ∎

Now suppose $G$ has type IIb, i.e. $G/O(G)=Q_{2^{n}}$ for some $n\geq 4$. By
combining Lemmas 1.7 and 1.8, we get that $G$ has a quotient $Q_{2^{n}m}$ if
and only if $m\mid t$, $r\equiv 1$ mod $m$ and $(a,b)=(1,-1)$ mod $m$. If $G$
has two distinct maximal binary polyhedral quotients
$f_{i}:G\twoheadrightarrow Q_{2^{n}m_{i}}$ for $i=1,2$, then $m_{1},m_{2}\mid
t$, $r\equiv 1$ mod $m_{1},m_{2}$ and $(a,b)\equiv(1,-1)$ mod $m_{1},m_{2}$.
If $m=\text{lcm}(m_{1},m_{2})$, then this implies that $m\mid t$, $r\equiv 1$
mod $m$ and $(a,b)\equiv(1,-1)$ mod $m$ and so $f:G\twoheadrightarrow
Q_{2^{n}m}$. By Corollary 1.4, $m>m_{1},m_{2}$ and $f_{1}$ and $f_{2}$ must
factor through $f$ which is a contradiction. Hence
$\\#\operatorname{\mathcal{B}_{\text{max}}}(G)=1$.

A similar argument works in the case where $G$ has type IIa, i.e.
$G/O(G)=Q_{8}$. If $G$ has four distinct maximal binary polyhedral quotients
$f_{i}:G\twoheadrightarrow Q_{8m_{i}}$ for $i=1,2,3,4$, then Lemmas 1.7 and
1.8 imply there exists $i,j$ for which $(a,b)\equiv(1,-1)$, $(-1,1)$ and
$(-1,-1)$ mod $m_{i},m_{j}$. By a similar argument to the above, this implies
that $f_{i}$, $f_{j}$ factors through $f:G\twoheadrightarrow Q_{8m}$ where
$m=\text{lcm}(m_{i},m_{j})$ which is a contradiction since $m_{i}\neq m_{j}$
and $f_{i}$, $f_{j}$ are maximal. Hence
$1\leq\\#\operatorname{\mathcal{B}_{\text{max}}}(G)\leq 3$.

Furthermore, if $G$ has quotients $Q_{8m_{i}}$ and $Q_{8m_{j}}$, then this
implies that $(a,b)$ mod $m_{i}$ and $(a,b)$ mod $m_{j}$ are distinct which is
a contradiction unless $(m_{i},m_{j})=1$.

#### Types III, IV, Va

If $G$ has type III, IV or Va, then $G/O(G)=\widetilde{T}$, $\widetilde{O}$ or
$\widetilde{I}$. If $f:G\twoheadrightarrow H$ is another binary polyhedral
quotient $H$, then Proposition 1.2 implies that $H\cong G/O(G)$. By Corollary
1.4, $f$ is equivalent to the quotient $G\twoheadleftarrow G/O(G)$. Hence
$\\#\operatorname{\mathcal{B}_{\text{max}}}(G),\\#\mathcal{B}(G)=1$.

#### Types Vb, VI

Suppose $G$ has type Vb or VI. Since no binary polyhedral groups have type Vb
or VI, Proposition 1.2 implies that $G$ has no binary polyhedral quotients.
Hence $\\#\operatorname{\mathcal{B}_{\text{max}}}(G),\\#\mathcal{B}(G)=0$.
This completes the proof of Theorem 1.5.

###  1.2. Quaternionic representations

Recall that $m_{\mathbb{H}}(G)$ denotes the number of copies of $\mathbb{H}$
in the Wedderburn decomposition of $\mathbb{R}G$ for a finite group $G$, i.e.
the number of one-dimensional quaternionic representations, and that $G$
satisfies the Eichler condition if $m_{\mathbb{H}}(G)=0$. The following is
[30, Proposition 3.3].

######  Proposition 1.9.

Let $f:G\twoheadrightarrow H$ be a quotient. Then
$m_{\mathbb{H}}(G)=m_{\mathbb{H}}(H)$ if and only if every
$g\in\mathcal{B}(G)$ factors through $f$, i.e. if
$f^{*}:\mathcal{B}(H)\to\mathcal{B}(G)$ is a bijection.

For example, this shows that $G$ satisfies the Eichler condition if and only
if $G$ has no quotient which is a binary polyhedral group. It also follows
that, if $G$ has a unique maximal binary polyhedral quotient $H$, then
$m_{\mathbb{H}}(G)=m_{\mathbb{H}}(H)$.

We now show how to use this to deduce the following from Theorem 1.5.

######  Theorem 1.10.

If $G$ has periodic cohomology, then type and $m_{\mathbb{H}}(G)$ are related
as follows.

Type | I | IIa | IIb | III | IV | Va | Vb | VI
---|---|---|---|---|---|---|---|---
$m_{\mathbb{H}}(G)$ | $\geq 0$ |  $\geq 1$ odd |  $\geq 2$ even | 1 | 2 | 2 | 0 | 0

#### Type Vb, VI

If $G$ has type Vb or VI, then Theorem 1.5 implies that $G$ has no binary
polyhedral quotients and so $m_{\mathbb{H}}(G)=0$ by Proposition 1.9.

#### Type IIb, III, IV, Va

If $G$ has type IIb, III, IV or Va, then Theorem 1.5 implies that
$\\#\operatorname{\mathcal{B}_{\text{max}}}(G)=1$, i.e. $G$ has a unique
maximal binary polyhedral quotient $H$. By Proposition 1.9, we must have that
$m_{\mathbb{H}}(G)=m_{\mathbb{H}}(H)$. By Proposition 1.2, $H$ has the same
type as $G$. Recall that $m_{\mathbb{H}}(Q_{4n})=\lfloor n/2\rfloor$ [20,
Section 12]. If $G$ has type IIa, then $H=Q_{2^{n}m}$ for $n\geq 4$, $m\geq 1$
odd and $m_{\mathbb{H}}(Q_{2^{n}m})=2^{n-3}m\geq 2$ is even. If $G$ has type
III, IV or Va, then $H=\widetilde{T}$, $\widetilde{O}$ or $\widetilde{I}$
respectively which have $m_{\mathbb{H}}(\widetilde{T})=1$,
$m_{\mathbb{H}}(\widetilde{O})=2$ and $m_{\mathbb{H}}(\widetilde{I})=2$.

#### Type IIa

If $G$ has type IIa, then Theorem 1.5 implies that
$\\#\operatorname{\mathcal{B}_{\text{max}}}(G)=1,2,3$. If
$b=\\#\operatorname{\mathcal{B}_{\text{max}}}(G)$, let
$f_{i}:G\twoheadrightarrow Q_{8m_{i}}$ denote the maximal binary polyhedral
quotients for $1\leq i\leq b$. It follows from the proof of Theorem 1.5 that
the $m_{i}$ are coprime and so the maximal quotient factoring through any two
of the $f_{i}$ is the unique quotient $f:G\twoheadrightarrow Q_{8}$. Since
$m_{\mathbb{H}}(Q_{8m_{i}})=m_{i}$ and $m_{\mathbb{H}}(Q_{8})=1$, it can be
shown using real representation theory that

$m_{\mathbb{H}}(G)=\sum_{i=1}^{b}(m_{\mathbb{H}}(Q_{8m_{i}})-1)+m_{\mathbb{H}}(Q_{8})=\begin{cases}m_{1},&\text{if
$b=1$}\\\ (m_{1}+m_{2})-1,&\text{if $b=2$}\\\ (m_{1}+m_{2}+m_{3})-2,&\text{if
$b=3$}\end{cases}$

which is odd since the $m_{i}$ are odd. This completes the proof of Theorem
1.10.

###  1.3. Vanishing of the Swan finiteness obstruction

Recall that a group $G$ has $k$-periodic cohomology if and only if there
exists a $k$-periodic projective resolution over $\mathbb{Z}G$ [32,
Proposition 2.7]. If $G$ has $k$-periodic cohomology, then Swan [40] defined
an obstruction $\sigma_{k}(G)\in C(\mathbb{Z}G)/T_{G}$, where $T_{G}$ is
generated by $(I,r)$ for $r\in(\mathbb{Z}/|G|)^{\times}$, which vanishes if
and only if there exists a $k$-periodic resolution of free $\mathbb{Z}G$
modules. Determining which groups have $\sigma_{k}(G)=0$ remains a difficult
open problem, and has applications to the classification of spherical space
forms [8].

The main result of this section will be the following extension of Theorem
1.10 which shows how $m_{\mathbb{H}}(G)$ and the vanishing of $\sigma_{k}(G)$
are related to the type of $G$.

######  Theorem 1.11.

If $G$ has periodic cohomology, then type, $m_{\mathbb{H}}(G)$ and
$\sigma_{k}(G)$ occur simultaneously if and only if they are related by a
column in the following table:

Type | I | IIa | IIb | III | IV | V a | V b | VI
---|---|---|---|---|---|---|---|---
$m_{\mathbb{H}}(G)$ | $\geq 0$ |  $\geq 1$ odd |  $\geq 2$ even | $1$ | $2$ | $2$ | $0$ | $0$
$\sigma_{k}(G)$ | $0$ |  $0$ or $\neq 0$ |  $0$ or $\neq 0$ | $0$ |  $0$ or $\neq 0$ | $0$ | $0$ |  $0$ or $\neq 0$

In order to prove Theorem 1.11, we will begin by noting the following which is
proven in [8, Theorem 3.19].

######  Lemma 1.12.

If $G$ has $k$-periodic cohomology and type I, III or V, then
$\sigma_{k}(G)=0$.

This gives the restrictions on the vanishing of $\sigma_{k}(G)$ given in the
above table. It now suffices to construct examples which realise the
constraints in each column.

We begin by constructing the examples with $\sigma_{k}(G)=0$. Firstly the
groups $C_{n}$, $Q_{4n}$ and $\operatorname{SL}_{2}(\mathbb{F}_{p})$,
$\operatorname{TL}_{2}(\mathbb{F}_{p})$ for $p\geq 3$ can all be shown to have
vanishing finiteness obstruction [8, Theorem 3.19 (c)]. Now note that
$m_{\mathbb{H}}(Q_{4n})=\lfloor n/2\rfloor$ [20, Section 12]. In particular,
$m_{\mathbb{H}}(Q_{8n+4})=m_{\mathbb{H}}(Q_{8n})=n$ for all $n\geq 1$ and
$Q_{8n+4}$ has type I and $Q_{8n}$ has type II. Finally, the cyclic groups
$G=C_{n}$ have type I and $m_{\mathbb{H}}(C_{n})=0$.

If $n\geq 3$ and $a,b,c\geq 1$ are odd coprime, then define

$Q(2^{n}a;b,c)=C_{abc}\rtimes_{(r,s)}Q_{2^{n}}$

where $(r,s)\equiv(-1,-1)$ mod $a$, $(r,s)\equiv(-1,1)$ mod $b$ and
$(r,s)\equiv(1,-1)$ mod $c$. The following was shown by Milgram [28, Theorem
D] and Davis [9, Corollary 6.2].

######  Theorem 1.13.

Let $p$, $q$ be distinct odd primes.

  1. (i)

Then $\sigma_{4}(Q(8;p,q))\neq 0$ if $p,q\equiv 3$ mod $4$, or $p\equiv 3$ mod
$4$, $q\equiv 5$ mod $8$ and $p^{n}\not\equiv\pm 1$ mod $q$ for all $n$ odd

  2. (ii)

If $n\geq 4$, then $\sigma_{4}(Q(2^{n};p,1))\neq 0$ if $p\not\equiv 1$ mod $8$
and $p\not\equiv\pm 1$ mod $2^{n-1}$.

We will construct further examples of groups with $\sigma_{k}(G)\neq 0$ as
follows. Let $H$ be a group listed in Theorem 1.13 which has
$\sigma_{4}(H)\neq 0$ and suppose $H\leq G$ where $G$ has $4n$-periodic
cohomology for some $n\geq 1$ odd. If $\sigma_{4n}(G)=0$, then
$\sigma_{4n}(H)=0$ by restricting resolutions. However, since $4n\equiv 4$ mod
8, it follows from [46, Corollary 12.6] that $\sigma_{4}(H)=0$. This is a
contradiction and so $\sigma_{4n}(G)\neq 0$.

#### Type IIa

It suffices to prove the following.

######  Lemma 1.14.

Let $n\geq 3$, $m\geq 1$ be odd coprime. Then there exists distinct odd primes
$p$, $q$ and $r,a,b\in\mathbb{Z}$ such that

$G=(C_{mpq}\rtimes_{(r)}C_{n})\rtimes_{(a,b)}Q_{8}$

has $\sigma_{4n}(G)\neq 0$ and $m_{\mathbb{H}}(G)=m$.

###### Proof.

By Dirichlet’s theorem on primes in arithmetic progression, there exists
distinct primes $p$, $q$ such that $p,q\nmid m$, $p,q\equiv 3$ mod 4 and
$p,q\equiv 1$ mod $n$. Let $a,b\in\mathbb{Z}/mpq$ be such that
$(a,b)\equiv(1,-1)$ mod $m$, $(a,b)\equiv(-1,1)$ mod $p$ and
$(a,b)\equiv(1,-1)$ mod $q$. Since $n\mid p-1,q-1$, we can pick
$r_{p}\in(\mathbb{Z}/p)^{\times}$, $r_{q}\in(\mathbb{Z}/q)^{\times}$ of order
$n$. Let $r\in\mathbb{Z}/mpq$ be such that $r\equiv 1$ mod $m$, $r\equiv
r_{p}$ mod $p$, $r\equiv r_{q}$ mod $q$. This has $r^{n}\equiv 1$ mod $mpq$
and so we can define $G=(C_{mpq}\rtimes_{(r)}C_{n})\rtimes_{(a,b)}Q_{8}$.

By [45, Corollary 4.6], $G$ has period $4\cdot\operatorname{ord}_{mpq}(r)=4n$.
If $C_{mpq}=\langle u\rangle$, $C_{n}=\langle v\rangle$ and $Q_{8}=\langle
x,y\rangle$, then $Q(8;p,q)\cong\langle u^{m},x,y\rangle\leq G$. Since
$p,q\equiv 3$ mod 4, Theorem 1.13 (i) implies that $\sigma_{4}(Q(8;p,q))\neq
0$ and so $\sigma_{4n}(G)\neq 0$.

Since $(a,b)\equiv(1,-1)$ mod $m$ and $r\equiv 1$ mod $m$, Lemmas 1.7 and 1.8
imply that $G$ has a quotient $Q_{8m}$. Similarly, if $G$ has a quotient
$Q_{8m_{0}}$ with $(m_{0},m)=1$, then $m_{0}\mid mpq$ and $r\equiv 1$ mod
$m_{0}$. However, $r\nequiv 1$ mod $p,q$ and so $m_{0}=1$. This implies that
$\operatorname{\mathcal{B}_{\text{max}}}(G)=\\{Q_{8m}\\}$ and so
$m_{\mathbb{H}}(G)=m_{\mathbb{H}}(Q_{8m})=m$ by Proposition 1.9. ∎

#### Type IIb

Let $n\geq 1$. By Dirichlet’s theorem on primes in arithmetic progression,
there exists a prime $p$ such that $p\nmid n$, $p\not\equiv 1$ mod $8$ and
$p\not\equiv\pm 1$ mod $2^{k+3}$ where $k=\nu_{2}(n)$ is the highest power of
$2$ dividing $n$. If $G=Q(16n;p,1)$, then Theorem 1.13 (ii) implies that
$\sigma_{4}(G)\neq 0$. It is easy to see that
$\operatorname{\mathcal{B}_{\text{max}}}(G)=\\{Q_{16n}\\}$ and so
$m_{\mathbb{H}}(G)=m_{\mathbb{H}}(Q_{8n})=n$ by Proposition 1.9.

#### Type IV

In order to give our example, first define $\widetilde{T}_{v}$ for $v\geq 1$
as the unique extension of $\widetilde{T}$ by $C_{3^{v-1}}$ which has periodic
cohomology [45, Lemma 3.3]. Also define $\widetilde{O}_{v}$ for $v\geq 1$ to
be the unique extension of $C_{2}$ by $\widetilde{T}_{v}$ which has periodic
cohomology [45, Lemma 3.4]. Note that $\widetilde{O}_{v}$ is $4$-periodic by
[45, Corollary 4.6] and, by Proposition 1.2, it also has type IV since it has
a quotient $\widetilde{O}$.

By [45, Theorem 4.1], $\widetilde{O}_{v}$ has a subgroup of the form
$Q(16;3^{v-1},1)$ and so has a subgroup $Q(16;3,1)$ for all $v\geq 2$. Since
$3\nequiv\pm 1$ mod 8, Theorem 1.13 (ii) implies that
$\sigma_{4}(Q(16;3,1))\neq 0$ and so $\sigma_{4}(\widetilde{O}_{v})\neq 0$ for
all $v\geq 2$.

#### Type VI

Recall that $\operatorname{TL}_{2}(\mathbb{F}_{p})$ is an extension of $C_{2}$
by $\operatorname{SL}_{2}(\mathbb{F}_{p})$. If
$z\in\operatorname{TL}_{2}(\mathbb{F}_{p})$ maps to the generator of $C_{2}$.
Let $p\geq 5$ and $q\nmid p(p^{2}-1)$ be odd primes and define

$G=C_{q}\rtimes_{\varphi}\operatorname{TL}_{2}(\mathbb{F}_{p})$

where, if $C_{q}=\langle u\rangle$, then $\varphi_{z}:u\mapsto u^{-1}$ and
$\varphi_{x}:u\mapsto u$ for all
$x\in\operatorname{SL}_{2}(\mathbb{F}_{p})\leq\operatorname{TL}_{2}(\mathbb{F}_{p})$.
Since $q$ is coprime to $|\operatorname{TL}_{2}(\mathbb{F}_{p})|=2p(p^{2}-1)$,
[45, Theorem 3.6] implies $G$ has type VI. It follows from [45, Theorem 4.2]
that $G$ has subgroups $Q(4(p+1);q,1)$ and $Q(4(p-1);q,1)$ and so $G$ has a
subgroup $Q(2^{n};q,1)$ where $n=2+\max\\{\nu_{2}(p+1),\nu_{2}(p-1)\\}\geq 3$.
Note also that, by [45, Corollary 4.6], $G$ has period $2(p-1)$.

If $p\equiv 3$ mod $4$, then $2(p-1)\equiv 4$ mod 8. Using Dirichlet’s theorem
of primes in arithmetic progression, we can now pick a prime $q$ such that
$q\nmid p(p^{2}-1)$, $q\nequiv 1$ mod 8 and $q\nequiv 1$ mod $2^{n-1}$. By
Theorem 1.13, we have $\sigma_{4}(Q(2^{n};q,1))\neq 0$. Since
$Q(2^{n};q,1)\leq G$, this implies that $\sigma_{2(p-1)}(G)\neq 0$.

##  2\. Cancellation for the Swan finiteness obstruction

The aim of this section will be to use the results in Section 1 to prove the
following cancellation result, where $G$ has $k$-periodic cohomology and
$P_{(G,n)}$ is a projective $\mathbb{Z}G$ module for which
$[P_{(G,n)}]=\sigma_{k}(G)\in C(\mathbb{Z}G)/T_{G}$.

######  Theorem 2.1.

$[P_{(G,n)}]$ has cancellation if and only if $m_{\mathbb{H}}(G)\leq 2$.

That $m_{\mathbb{H}}(G)\geq 3$ implies $[P_{(G,n)}]$ has non-cancellation is
dealt with in Section 6. For a direct proof, we can combine Lemma 3.4 with
[41, Theorem A] (see also Lemma 6.2). The aim of this section will be to prove
that $m_{\mathbb{H}}(G)\leq 2$ implies $[P_{(G,n)}]$ has cancellation.

Recall that, if $G$ is finite,
$I=\operatorname{Ker}(\varepsilon:\mathbb{Z}G\to\mathbb{Z})$ is the
augmentation ideal and $(r,|G|)=1$, then $(I,r)\subseteq\mathbb{Z}G$ is a
projective ideal. Since $(I,r)\cong(I,s)$ if $r\equiv s$ mod $|G|$, we often
write $r\in(\mathbb{Z}/|G|)^{\times}$, As noted in [32, Section 2], if $P$ is
a projective $\mathbb{Z}G$ module and $r\in(\mathbb{Z}/|G|)^{\times}$, then we
can define $(I,r)\otimes P$ to be a (left) $\mathbb{Z}G$ module since $(I,r)$
is a two-sided ideal. We can use this action to show the following.

######  Proposition 2.2.

Let $G$ be a finite group let $P,Q$ be projective $\mathbb{Z}G$ modules such
that $[P]-[Q]\in T_{G}$. Then there is an isomorphism of graded trees

$[P]\cong[Q]$

given sending $P_{0}\mapsto(I,r)\otimes P_{0}$ for some
$r\in(\mathbb{Z}/|G|)^{\times}$.

###### Proof.

Let $r\in(\mathbb{Z}/|G|)^{\times}$ be such that $[Q]=[P]+[(I,r)]\in
C(\mathbb{Z}G)$. If $P_{0}\in[P]$, then [32, Lemma 2.4] implies that
$[(I,r)\otimes P_{0}]=[(I,r)]+[P_{0}]=[Q]$ and so $P_{0}\mapsto(I,r)\otimes
P_{0}$ defines a map $[P]\to[Q]$ which preserves the rank of the projective
modules. To see that it is bijective note that, if
$r,s\in(\mathbb{Z}/|G|)^{\times}$, then $(I,r)\otimes(I,s)\cong(I,rs)$ by [32,
Remark 2.3] and so the map has inverse $P_{0}\mapsto(I,r^{-1})\otimes P_{0}$.
∎

In particular, if $P$ and $Q$ are both representatives of the finiteness
obstruction $\sigma_{k}(G)$, then $[P]\cong[Q]$. Hence, it suffices to prove
Theorem 2.1 for any representative $P_{(G,n)}$ of the finiteness obstruction.

If $m_{\mathbb{H}}(G)=0$, i.e. $G$ satisfies the Eichler condition, then
$[P_{(G,n)}]$ automatically has cancellation by the Swan-Jacobinski theorem
[42, Theorem 9.3].

######  Theorem 2.3.

If $G$ satisfies the Eichler condition, then $\mathbb{Z}G$ has projective
cancellation, i.e. $[P]$ has cancellation for all projective $\mathbb{Z}G$
modules $P$.

If $\sigma_{k}(G)=0$, then $\mathbb{Z}G$ is a representative of the finiteness
obstruction and we know already from [31, Theorem 6.3] that $[\mathbb{Z}G]$
has cancellation if and only if $m_{\mathbb{H}}(G)\leq 2$. It therefore
suffices to restrict our attention to those groups with $\sigma_{k}(G)\neq 0$
and $m_{\mathbb{H}}(G)\neq 0$ and so, by Theorem 1.11, it remains to prove
Theorem 2.1 in the case where $G$ has type II with $m_{\mathbb{H}}(G)=1,2$ or
type IV and $m_{\mathbb{H}}(G)=2$.

In order to deal with these cases, we will appeal to the following which is a
consequence of [31, Theorem 4.1].

######  Lemma 2.4.

Let $f:G\twoheadrightarrow H$ where $G$ has periodic cohomology and $H$ is a
binary polyhedral group such that $m_{\mathbb{H}}(G)=m_{\mathbb{H}}(H)\leq 2$.
If $P\in P(\mathbb{Z}G)$ is such that $f_{\\#}([P])\in T_{H}$, then $[P]$ has
cancellation.

###### Proof.

Let $r_{H}\in(\mathbb{Z}/|H|)^{\times}$ be such that
$f_{\\#}([P])=[(I,r_{H})]$. By Lemma 3.2, the induced map $T_{G}\to T_{H}$ is
surjective and so there exists $r\in(\mathbb{Z}/|G|)^{\times}$ such that
$f_{\\#}((I,r))\cong(I,r_{H})$, i.e. $r\equiv r_{H}$ mod $|H|$. By [32, Lemma
2.4], we have that

$f_{\\#}([(I,r^{-1})\otimes
P])=f_{\\#}([(I,r^{-1})]+[P])=[(I,r_{H}^{-1})]+[(I,r_{H})]=0\in
C(\mathbb{Z}H).$

By [41, Theorem I], we know that $[\mathbb{Z}H]$ has cancellation and so
$f_{\\#}((I,r^{-1})\otimes P)\cong\mathbb{Z}H$. Since the map
$\mathbb{Z}H^{\times}\to K_{1}(\mathbb{Z}H)$ is surjective [23, Theorems
7.15-7.18], the conditions of [31, Theorem 4.1] are met and so
$[(I,r^{-1})\otimes P]$ has cancellation. By Proposition 2.2, this implies
that $[P]$ has cancellation. ∎

#### Type II

If $G$ has type II and $m_{\mathbb{H}}(G)=1,2$, then
$\mathcal{B}(G)=\\{Q_{8}\\}$ or $\\{Q_{16}\\}$. Hence, for $k=3$ or $4$, we
have $f:G\twoheadrightarrow Q_{2^{k}}$ with
$m_{\mathbb{H}}(G)=m_{\mathbb{H}}(Q_{2^{k}})$. By [41, Theorems III,IV], we
have that $C(\mathbb{Z}Q_{8})=T_{Q_{8}}$ and $C(\mathbb{Z}Q_{16})=T_{Q_{16}}$
and so $f_{\\#}([P_{(G,n)}])\in T_{Q_{2^{k}}}$ automatically. Hence
$[P_{(G,n)}]$ has cancellation, by Lemma 2.4.

#### Type IV

If $G$ has type IV, then $m_{\mathbb{H}}(G)=m_{\mathbb{H}}(\widetilde{O})=2$
and there exists a quotient $f:G\twoheadrightarrow\widetilde{O}$. Recall that
$Q_{12}\leq\widetilde{O}$ and that this is unique up to conjugacy [41, Lemma
14.3]. We will need the following lemma, the proof of which is contained in
the proof of [31, Theorem 6.9].

######  Lemma 2.5.

If $P$ is a projective $\mathbb{Z}\widetilde{O}$ module, then $[P]\in
T_{\widetilde{O}}$ if and only if
$[\operatorname{Res}_{Q_{12}}^{\widetilde{O}}(P)]=0\in C(\mathbb{Z}Q_{12})$.

Let $N=\operatorname{Ker}(f)$ and let $H=f^{-1}(Q_{12})$ which is a subgroup
of $G$ for which $N\unlhd H\leq G$ and $Q_{12}=H/N$. By [31, Lemma 6.6], we
have a diagram

${C(\mathbb{Z}G)}$${C(\mathbb{Z}\widetilde{O})}$${C(\mathbb{Z}H)}$${C(\mathbb{Z}Q_{12})}$$\scriptstyle{f_{\\#}}$$\scriptstyle{\text{Res}^{G}_{H}}$$\scriptstyle{\text{Res}^{\widetilde{O}}_{Q_{12}}}$$\scriptstyle{(f\mid_{H})_{\\#}}$

and, by commutativity, we get that

$\operatorname{Res}_{Q_{12}}^{\widetilde{O}}(f_{\\#}(P_{(G,n)}))=(f\mid_{H})_{\\#}(\operatorname{Res}_{H}^{G}(P_{(G,n)})).$

It follows from general properties of finiteness obstructions [8, Remark 2.18]
that, if $[P_{(G,n)}]=\sigma_{k}(G)\in C(\mathbb{Z}G)/T_{G}$, then
$[\operatorname{Res}_{H}^{G}(P_{(G,n)})]=\sigma_{k}(H)\in
C(\mathbb{Z}H)/T_{H}$. It is easy to see that $H$ has type I and so
$\sigma_{k}(G)=0$ by Theorem 1.11. This implies that
$\operatorname{Res}_{H}^{G}(P_{(G,n)})\in T_{H}$ and so
$(f\mid_{H})_{\\#}(\operatorname{Res}_{H}^{G}(P_{(G,n)}))\in T_{Q_{12}}$.
However, [41, Theorem IV] implies that $T_{Q_{12}}=0$ and so
$[\operatorname{Res}_{Q_{12}}^{\widetilde{O}}(f_{\\#}(P_{(G,n)}))]=0$ and
$f_{\\#}(P_{(G,n)})\in T_{\widetilde{O}}$ by Lemma 2.5. Hence Lemma 2.4
applies and $[P_{(G,n)}]$ has cancellation.

##  3\. Induced representations and the action of $\operatorname{Aut}(G)$

If $G$ is a finite group and $P\in P(\mathbb{Z}G)$, we say that
$\operatorname{Aut}(G)$ acts on $[P]$ if there exists a group homomorphism
$\psi:\operatorname{Aut}(G)\to(\mathbb{Z}/|G|)^{\times}$ for which
$[P]-[P_{\theta}]=[(I,\psi(\theta))]\in C(\mathbb{Z}G)$ where $I$ is the
augmentation ideal. The action is then given by sending
$P_{0}\mapsto(P_{0})_{\theta}\otimes(I,\psi(\theta))$ for $P_{0}\in[P]$. For
example, $\operatorname{Aut}(G)$ acts on the stably free class $[\mathbb{Z}G]$
for all finite groups $G$ via the trivial map $\psi(\theta)=1$ for all
$\theta\in\operatorname{Aut}(G)$. More generally, if $G$ has periodic
cohomology, then $\operatorname{Aut}(G)$ acts on $[P_{(G,n)}]$ by the action
defined in the introduction (see also [32, Section 6]).

Recall that, if $R$ and $S$ are rings and $f:R\to S$ is a ring homomorphism,
then $S$ is an $(S,R)$-bimodule, with right-multiplication by $r\in R$ given
by $x\cdot r=xf(r)$ for any $x\in S$. If $M$ is an $R$-module, we can define
the extension of scalars of $M$ by $f$ as the tensor product

$f_{\\#}(M)=S\otimes_{R}M$

which is defined since $S$ as a right $R$-module and $M$ as a left $R$-module.
We will view this as a left $S$-module where left-multiplication by $s\in S$
is given by $s\cdot(x\otimes m)=(sx)\otimes m$ for any $x\in S$ and $m\in M$.

For example, a group homomorphism $f:G\to H$ can be viewed as a ring
homomorphism $f:\mathbb{Z}G\to\mathbb{Z}H$ by sending $\sum_{g\in
G}x_{g}g\mapsto\sum_{g\in G}x_{g}f(g)$ where $x_{g}\in\mathbb{Z}$. If $M$ is a
$\mathbb{Z}G$ module then $f_{\\#}(M)=\mathbb{Z}H\otimes_{\mathbb{Z}G}M$
coincides with the induced module.

Recall that a subgroup $N\leq G$ is characteristic if every
$\varphi\in\operatorname{Aut}(G)$ has $\varphi(N)=N$ and we say that a
quotient $f:G\to H$ is characteristic if $\operatorname{Ker}(f)\leq G$ is a
characteristic subgroup.

######  Lemma 3.1.

Let $f:G\twoheadrightarrow H$ be a characteristic quotient and let
$\theta\in\operatorname{Aut}(G)$. If $M$ is a $\mathbb{Z}G$ module, then

$f_{\\#}(M_{\theta})\cong f_{\\#}(M)_{\bar{\theta}}.$

###### Proof.

Since $f$ is characteristic, there is a map
$\bar{\cdot}:\operatorname{Aut}(G)\to\operatorname{Aut}(H)$ such that
$f\circ\theta=\bar{\theta}\circ f$ for all $\theta\in\operatorname{Aut}(G)$.
The result now follows from [32, Corollary 7.4]. ∎

######  Lemma 3.2.

Let $f:G\twoheadrightarrow H$ and $r\in(\mathbb{Z}/|G|)^{\times}$, then

$f_{\\#}((I_{G},r))\cong(I_{H},r).$

###### Proof.

By [8, Remark 2.3], we have that $(I_{G},r)\cong(\Sigma_{G},r^{-1})$ where
$\Sigma_{G}=\sum_{g\in G}g$ denotes the group norm. Hence it suffices to prove
instead that $f_{\\#}((\Sigma_{G},r))\cong(\Sigma_{H},r)$. Let $d=|G|/|H|$
and, since $(d,r)=1$, there exists $a,b\in\mathbb{Z}$ such that $ad+br=1$. Now
define

$\psi:(\Sigma_{H},r)\to
f_{\\#}((\Sigma_{G},r))=\mathbb{Z}H\otimes_{\mathbb{Z}G}(\Sigma_{G},r)$

by sending $r\mapsto 1\otimes r$ and $\Sigma_{H}\mapsto
a(1\otimes\Sigma_{G})+b\Sigma_{H}(1\otimes r)$, and it is straightforward to
check that $\psi$ is a $\mathbb{Z}H$ module isomorphism. ∎

Our main result gives a possible way to show how $[P]/\operatorname{Aut}(G)$
has non-cancellation when $f:G\twoheadrightarrow H$ is a characteristic
quotient and
$f_{*}=\bar{\cdot}:\operatorname{Aut}(G)\to\operatorname{Aut}(H)$.

######  Theorem 3.3.

Let $f:G\twoheadrightarrow H$ is a characteristic quotient and $P\in
P(\mathbb{Z}G)$. If $\operatorname{Aut}(G)$ acts on $[P]$ and
$\bar{P}=f_{\\#}(P)$, then there is a surjection of graded trees

$f_{\\#}:[P]/\operatorname{Aut}(G)\twoheadrightarrow[\bar{P}]/\operatorname{Im}(f_{*})$

where the action of $\operatorname{Im}(f_{*})\subseteq\operatorname{Aut}(H)$
on $[\bar{P}]$ is induced by $f$.

###### Proof.

If $\otimes$ denotes the tensor product of $\mathbb{Z}G$ modules, then it is
easy to see that $f_{\\#}(M\otimes N)\cong f_{\\#}(M)\otimes N$ for all $M$,
$N$. Combining with Lemmas 3.1 and 3.2 gives

$\displaystyle f_{\\#}((I_{G},r)\otimes P_{\theta})$
$\displaystyle\cong(I_{H},r)\otimes_{\mathbb{Z}H}P_{\theta}\cong((I_{H},r)\otimes\mathbb{Z}H)\otimes_{\mathbb{Z}H}P_{\theta}$
$\displaystyle\cong(I_{H},r)\otimes(\mathbb{Z}H\otimes_{\mathbb{Z}H}P_{\theta})\cong(I_{H},r)\otimes\bar{P}_{\bar{\theta}}$

where $\bar{\cdot}:\operatorname{Aut}(G)\to\operatorname{Aut}(H)$. Since
$[(I_{G},r)\otimes P_{\theta})]=[P]$, this implies that
$[(I_{H},r)\otimes\bar{P}_{\bar{\theta}}]=[\bar{P}]$ and so
$\operatorname{Aut}(H)$ acts on $[\bar{P}]$ by sending $P_{0}$ to
$(I_{H},\varphi(\theta))\otimes P_{0}$.

Now note that the map $f_{\\#}:[P]\to[\bar{P}]$ is well-defined and surjective
by [41, Theorem A10]. By the argument above, $f_{\\#}$ respects these actions
and so induces a map
$f_{\\#}:[P]/\operatorname{Aut}(G)\to[\bar{P}]/\operatorname{Aut}(H)$ which is
necessarily surjective. ∎

We will now show how this can be applied in the case where $G$ has periodic
cohomology. First note the following:

######  Lemma 3.4.

If $G$ has periodic cohomology and $m_{\mathbb{H}}(G)\geq 3$, then $G$ has a
quotient $Q_{4n}$ for some $n\geq\max\\{\frac{2}{3}m_{\mathbb{H}}(G),6\\}$.

###### Proof.

Since $m_{\mathbb{H}}(G)\geq 3$, the possible types are I and II by Theorem
1.11. If $H$ is the binary polyhedral quotient of maximal order, then
Proposition 1.2 implies $H=Q_{4n}$ for some $n\geq 2$. If
$b=\\#\operatorname{\mathcal{B}_{\text{max}}}(G)$, then Theorem 1.5 implies
$b=1,2,3$.

Let $m=m_{\mathbb{H}}(G)$. If $b=1$, then Proposition 1.9 implies
$m=m_{\mathbb{H}}(Q_{4n})=\lfloor\frac{n}{2}\rfloor\leq\frac{n}{2}$ and so
$n\geq 2m$. If $b=2,3$, then $G$ has type IIa. If $Q_{8m_{i}}$ are the maximal
binary polyhedral quotients for $1\leq i\leq b$, then the proof of Theorem 1.5
implies that $m=(m_{1}+m_{2})-1$ if $b=2$ and $m=(m_{1}+m_{2}+m_{3})-2$ if
$b=3$. Suppose $m_{1}>m_{2}>m_{3}$ so that $m_{1}\geq m_{2}+2$ and $m_{2}\geq
m_{3}+2$ since the $m_{i}$ are odd coprime. Since $n=2m_{1}$, this implies
that $m\leq n-3$ if $b=2$ and $m\leq\frac{3}{2}n-8$ if $b=3$. Hence $n\geq
m+3$ and $n\geq\frac{2}{3}(m+8)$ in the two cases respectively. The bound now
follows since $\min\\{2m,m+3,\frac{2}{3}(m+8)\\}\geq\max\\{6,\frac{2}{3}m\\}$
for all $m\geq 3$. ∎

In fact, the quotient $f:G\twoheadrightarrow Q_{4n}$ is always characteristic
due to the following.

######  Proposition 3.5.

If $f:G\twoheadrightarrow H$ where $G$ has periodic cohomology and $H$ is a
binary polyhedral group, then $f$ is characteristic.

###### Proof.

Let $\varphi\in\operatorname{Aut}(G)$ and consider
$N=\varphi(\operatorname{Ker}(f))\leq G$. Then $N$ is a normal subgroup with
$|N|=|\operatorname{Ker}(f)|$. Since $H$ is a binary polyhedral group, it has
$4\mid|H|$ and so Lemma 1.3 implies that $N=\operatorname{Ker}(f)$. ∎

By combining this with Theorem 3.3, we have:

######  Corollary 3.6.

Let $G$ have $k$-periodic cohomology and $m_{\mathbb{H}}(G)\geq 3$. Then there
is a surjection of graded trees

$f_{\\#}:[P_{(G,k)}]/\operatorname{Aut}(G)\twoheadrightarrow[\widebar{P_{(G,k)}}]/\operatorname{Im}(f_{*})$

where $f:G\twoheadrightarrow Q_{4n}$ for some
$n\geq\max\\{\frac{2}{3}m_{\mathbb{H}}(G),6\\}$ and where the action of
$\operatorname{Im}(f_{*})\subseteq\operatorname{Aut}(Q_{4n})$ on
$[\widebar{P_{(G,k)}}]$ is induced by $f$.

Note that the two bounds for $n$ have distinct uses. If $m_{\mathbb{H}}(G)$ is
small, as is the case when dealing with cancellation in Theorem A, then the
bound $n\geq 6$ will be most useful. If $m_{\mathbb{H}}(G)$ is large, as in
the asymptotic estimates in Theorem B, then we will use the bound
$n\geq\frac{2}{3}m_{\mathbb{H}}(G)$.

##  4\. The Eichler mass formula

Let $K$ be a number field with ring of integers $\mathcal{O}_{K}$ and let
$\Lambda$ be an $\mathcal{O}_{K}$-order in a finite-dimensional semi-simple
$K$-algebra $A$. It is a standard fact (see, for example, [42, Lemma 2.1])
that, if $M$ is a finitely generated $\Lambda$-module, then $M$ is projective
if and only if $M$ is locally projective, i.e. for all $p$ prime,
$M_{p}=M\otimes\mathbb{Z}_{p}$ is projective over
$\Lambda_{p}=\Lambda\otimes\mathbb{Z}_{p}$ where $\mathbb{Z}_{p}$ is the
$p$-adic integers.

In the case where $K=\mathbb{Q}$, $\Lambda=\mathbb{Z}G$ and $A=\mathbb{Q}G$
for $G$ a finite group, then $M$ projective implies that $M_{p}$ is a free
$\mathbb{Z}_{p}G$ module for all $p$ prime [39, Theorem 2.21, 4.2]. In
particular, in this case, $M$ is projective if and only if $M$ is locally
free.

Define the locally free class group $C(\Lambda)$ to be the equivalence classes
of locally free modules up to the relation $P\simeq Q$ if
$P\oplus\Lambda^{i}\cong Q\oplus\Lambda^{j}$ for some $i,j\geq 0$. By abuse of
notation, we write $[P]$ to denote both the class $[P]\in C(\Lambda)$ and,
where convenient, the set of isomorphism classes of projective modules $P_{0}$
where $[P_{0}]=[P]$.

Define the class set $\operatorname{Cls}\Lambda$ as the set of isomorphism
classes of rank one locally free $\Lambda$-modules, which is finite by the
Jordan-Zassenhaus theorem [6, Section 24]. Equivalently, this is the set of
locally principal fractional $\Lambda$-ideals, under the relation $I\sim J$ if
there exists $\alpha\in A^{\times}$ such that $I=\alpha J$ (see [36]). This
comes with the stable class map

$[\,\cdot\,]_{\Lambda}:\operatorname{Cls}\Lambda\to C(\Lambda)$

which sends $P\mapsto[P]$ and is surjective since every locally free
$\Lambda$-module $P$ is of the form $P_{0}\oplus\Lambda^{i}$ where
$P_{0}\in\operatorname{Cls}\Lambda$ and $i\geq 0$ [15]. Define
$\operatorname{Cls}^{[P]}(\Lambda)$ to be $[\,\cdot\,]_{\Lambda}^{-1}([P])$ ,
i.e. the rank one locally free modules in $[P]$, and let
$\operatorname{SF}(\Lambda)$ be $\operatorname{Cls}^{[\Lambda]}(\Lambda)$,
i.e. the set of rank one stably free modules.

We say that $\Lambda$ has locally free cancellation if $P\oplus\Lambda\cong
Q\oplus\Lambda$ implies $P\cong Q$ for all locally free $\Lambda$-modules $P$
and $Q$. It follows from the discussion above that $\Lambda$ has locally free
cancellation if and only if $[\,\cdot\,]_{\Lambda}$ is bijective, i.e.
$\\#\operatorname{Cls}\Lambda=\\#C(\Lambda)$.

Similarly we say that $\Lambda$ has stably free cancellation when
$P\oplus\Lambda^{i}\cong\Lambda^{j}$ implies that $P\cong\Lambda^{j-i}$, or
equivalently, if $\\#\operatorname{SF}(\Lambda)=1$.

If $X\subseteq\operatorname{Cls}\Lambda$, then we can define the mass of $X$
to be

$\operatorname{mass}(X)=\sum_{I\in
X}\frac{1}{[O_{L}(I)^{\times}:\mathcal{O}_{K}^{\times}]}.$

Recall that a quaternion algebra $A$ over $K$ is totally definite if $A$ is
ramified over all archimedean places $\nu$, i.e. $A\otimes K_{\nu}$ is a
division algebra over $K_{\nu}$. Note that every complex place $\nu$ splits
since the only quaternion algebra over $\mathbb{C}$ is $M_{2}(\mathbb{C})$. In
particular, if $A$ is totally definite, then $K$ must be a totally real field.

Let $\zeta_{K}(s)$ be the Dedekind zeta function, let
$h_{K}=|C(\mathcal{O}_{K})|$ be the class number of $K$ and let $\Delta_{K}$
be the discriminant of $K$. The following was proven in [14].

######  Theorem 4.1 (Eichler mass formula).

Let $A$ be a totally definite quaternion algebra over $K$ and let $\Lambda$ be
a maximal $\mathcal{O}_{K}$-order in $A$. If $n=[K:\mathbb{Q}]$, then

$\operatorname{mass}(\operatorname{Cls}\Lambda)=\frac{2\zeta_{K}(2)}{(2\pi)^{2n}}\cdot|\Delta_{K}|^{3/2}\cdot
h_{K}\cdot\prod_{\mathfrak{p}\mid\operatorname{disc}(A)}(N_{K/\mathbb{Q}}(\mathfrak{p})-1).$

The following was first shown by Vignéras in [43], though a simplified proof
can be found in [36, Theorem 5.11].

######  Theorem 4.2.

Let $A$ be a totally definite quaternion algebra over $K$ and let $\Lambda$ be
a maximal $\mathcal{O}_{K}$-order in $A$. If $P$, $Q$ are locally free
$\Lambda$-modules, then

$\operatorname{mass}(\operatorname{Cls}^{[P]}(\Lambda))=\operatorname{mass}(\operatorname{Cls}^{[Q]}(\Lambda)).$

In particular, this implies that
$\operatorname{mass}(\operatorname{Cls}^{[P]}(\Lambda))=\frac{\operatorname{mass}(\operatorname{Cls}\Lambda)}{|C(\Lambda)|}$,
where $C(\Lambda)$ denotes the class group of locally free $\Lambda$-modules.

It was shown by Eichler that

$|C(\Lambda)|=h_{K}\cdot[(\mathcal{O}_{K}^{\times})^{+}:(\mathcal{O}_{K}^{\times})^{2}]$

where $(\mathcal{O}_{K}^{\times})^{+}$ denotes the group of totally positive
units, i.e. those units $u\in\mathcal{O}_{K}^{\times}$ for which $\sigma(u)>0$
for all embeddings $\sigma:K\hookrightarrow\mathbb{R}$. The following can be
shown using the results above as well as lower bounds on
$|\Delta_{K}|^{1/[K:\mathbb{Q}]}$ in terms of $[K:\mathbb{Q}]$.

######  Theorem 4.3.

Let $A$ be a totally definite quaternion algebra over $K$ and let $\Lambda$ be
a maximal $\mathcal{O}_{K}$-order in $A$. If $\Lambda$ has stably free
cancellation, then $[K:\mathbb{Q}]\leq 6$.

######  Remark 4.4.

This was proven by Hallouin-Maire [17, Theorem 1], though it is worth noting
that part of their result was incorrect as stated (see [35]).

In the notation of [41, Section 3], define the Eichler constant

$\operatorname{ei}_{K}=\frac{2\zeta_{K}(2)|\Delta_{K}|^{3/2}}{(2\pi)^{2d}}=\frac{(-1)^{d}\zeta_{K}(-1)}{2^{d-1}}\in\mathbb{Q}$

where $d=[K:\mathbb{Q}]$ and where the second equality comes from the
functional equation for $\zeta_{K}(s)$. This is rational since
$\zeta_{K}(-1)\in\mathbb{Q}$. Another constraint on the fields $K$ over which
stably free cancellation can occur is as follows.

######  Proposition 4.5.

Let $A$ be a totally definite quaternion algebra over $K$ and let $\Lambda$ be
a maximal $\mathcal{O}_{K}$-order in $A$. If $\Lambda$ has stably free
cancellation, then the numerator of $\zeta_{K}(-1)$ (or, equivalently,
$\operatorname{ei}_{K}$) is a power of $2$.

###### Proof.

If $\Lambda$ has stably free cancellation, then
$\operatorname{mass}(\operatorname{SF}(\Lambda))=[\Lambda^{\times}:\mathcal{O}_{K}^{\times}]^{-1}$
since $\Lambda$ a maximal order implies $O_{L}(\Lambda)=\Lambda$, i.e. the
numerator is $1$. By Theorems 4.1 and 4.2, we also have that

$\operatorname{mass}(\operatorname{SF}(\Lambda))=\frac{\operatorname{ei}_{K}}{[(\mathcal{O}_{K}^{\times})^{+}:(\mathcal{O}_{K}^{\times})^{2}]}\cdot\prod_{\mathfrak{p}\mid\operatorname{disc}(A)}(N_{K/\mathbb{Q}}(\mathfrak{p})-1).$

Note that $[(\mathcal{O}_{K}^{\times})^{+}:(\mathcal{O}_{K}^{\times})^{2}]$ is
a power of $2$ since
$(\mathcal{O}_{K}^{\times})^{2}\subseteq(\mathcal{O}_{K}^{\times})^{+}\subseteq\mathcal{O}_{K}^{\times}$
and
$[(\mathcal{O}_{K}^{\times})^{\times}:(\mathcal{O}_{K}^{\times})^{2}]=2^{d}$
by Dirichlet’s unit theorem. Since
$N_{K/\mathbb{Q}}(\mathfrak{p})\in\mathbb{Z}$, this implies that the numerator
of $\operatorname{ei}_{K}$, or equivalently $\zeta_{K}(-1)$, is a power of
$2$. ∎

##  5\. Orders in quaternionic components of $\mathbb{Q}G$

Recall that, for a finite group $G$, the rational group ring $\mathbb{Q}G$ is
semisimple and so admits a decomposition into simple $\mathbb{Q}$-algebras.
For the quaternion groups of order $4n\geq 8$, we have

$\mathbb{Q}Q_{4n}\cong\prod_{d\mid
n}M_{2}(\mathbb{Q}(\zeta_{d}+\zeta_{d}^{-1}))\times\prod_{d\nmid n,\,d\mid
2n}\mathbb{Q}[\zeta_{d},j]$

where $\zeta_{d}=e^{2\pi i/d}\in\mathbb{C}$ is a primitive $d$th root of
unity, and $\mathbb{Q}[\zeta_{d},j]\subseteq\mathbb{H}$ sits inside the real
quaternions. This is stated on [41, p75] though a more detailed proof can be
found in [20, p48-51]. In order to apply the results of Section 4, it will be
helpful to note that

$\mathbb{Q}[\zeta_{n},j]\cong\left(\frac{(\zeta_{n}-\zeta_{n}^{-1})^{2},-1}{\mathbb{Q}(\zeta_{n}+\zeta_{n}^{-1})}\right)$

i.e. that $\mathbb{Q}[\zeta_{n},j]$ is a quaternion algebra with centre
$K=\mathbb{Q}(\zeta_{n}+\zeta_{n}^{-1})$, the totally real subfield of
$\mathbb{Q}(\zeta_{n})$. It is straightforward to check that
$\mathbb{Q}[\zeta_{n},j]$ isfSw83 totally definite for $n\geq 3$ (see, for
example, [41, Lemma 4.3]).

If $n_{i}$ are distinct positive integers such that $n_{i}\nmid n$ and
$n_{i}\mid 2n$ for $1\leq i\leq k$, then define $\Lambda_{n_{1},\cdots,n_{k}}$
to be the image of $\mathbb{Z}G$ under the projection of $\mathbb{Q}Q_{4n}$
onto $A_{n_{1},\cdots,n_{k}}=\prod_{i=1}^{k}\mathbb{Q}[\zeta_{n_{i}},j]$. For
example, $\mathbb{Z}Q_{4n}$ projects onto
$\Lambda_{2n}=\mathbb{Z}[\zeta_{2n},j]$ for all $n\geq 2$.

Fix the standard presentation $Q_{4n}=\langle x,y\mid
x^{n}=y^{2},yxy^{-1}=x^{-1}\rangle$ and recall that, for $n\geq 3$, we have
$\operatorname{Aut}(Q_{4n})=\\{\theta_{a,b}:a\in(\mathbb{Z}/{2n})^{\times},b\in\mathbb{Z}/{2n}\\}$
where $\theta_{a,b}(x)=x^{a}$ and $\theta_{a,b}(y)=x^{b}y$.

######  Lemma 5.1.

Let $4n\geq 12$. If $n_{i}$ are distinct positive integers such that
$n_{i}\nmid n$ and $n_{i}\mid 2n$ for $1\leq i\leq k$, then the map
$f:\mathbb{Z}Q_{4n}\twoheadrightarrow\Lambda_{n_{1},\cdots,n_{k}}$ induces a
map

$f_{*}:\operatorname{Aut}(Q_{4n})\to\operatorname{Aut}(\Lambda_{n_{1},\cdots,n_{k}}).$

###### Proof.

It follows from [20, p48-51] that the map
$f:\mathbb{Q}Q_{4n}\twoheadrightarrow A_{n_{1},\cdots,n_{k}}$ is given by
$x\mapsto(\zeta_{n_{1}},\cdots,\zeta_{n_{k}})$, $y\mapsto(j,\cdots,j)$. If
$\bar{\zeta}=(\zeta_{n_{1}},\cdots,\zeta_{n_{k}})$ and $\bar{j}=(j,\cdots,j)$,
then $\Lambda_{n_{1},\cdots,n_{k}}=\langle\bar{\zeta},\bar{j}\rangle$ as a
$\mathbb{Z}$-order in $A_{n_{1},\cdots,n_{k}}$. For
$a\in(\mathbb{Z}/{2n})^{\times}$ and $b\in\mathbb{Z}/{2n}$, define

$\bar{\theta}_{a,b}:\Lambda_{n_{1},\cdots,n_{k}}\to\Lambda_{n_{1},\cdots,n_{k}}$
$\bar{\zeta}\mapsto\bar{\zeta}^{a}=(\zeta_{n_{1}}^{a},\cdots,\zeta_{n_{k}}^{a}),\quad\bar{j}\mapsto\bar{\zeta}^{b}\bar{j}=(\zeta_{n_{1}}^{b}j,\cdots,\zeta_{n_{k}}^{b}j)$

which we can extend to be a ring homomorphism. Hence
$\bar{\theta}_{a,b}\in\operatorname{Aut}(\Lambda_{n_{1},\cdots,n_{k}})$ and it
is easy to see that $f\circ\theta_{a,b}=\bar{\theta}_{a,b}\circ f$. This
implies that there is an induced map
$f_{*}:\operatorname{Aut}(Q_{4n})\to\operatorname{Aut}(\Lambda_{n_{1},\cdots,n_{k}})$
where $f_{*}(\theta_{a,b})=\bar{\theta}_{a,b}$. ∎

######  Lemma 5.2.

Let $G$ be a finite group, let $r\in(\mathbb{Z}/|G|)^{\times}$ and suppose
$\psi\in\mathbb{Z}G$ is such that $\mathbb{Z}G/(\psi)$ is torsion-free and
$(\varepsilon(\psi),r)=1$. If
$f:\mathbb{Z}G\twoheadrightarrow\mathbb{Z}G/(\psi)$, then

$f_{\\#}((I,r))\cong\mathbb{Z}G/(\psi).$

###### Proof.

First consider the exact sequence of $\mathbb{Z}G$ modules

$0\to(I,r)\xrightarrow[]{i}\mathbb{Z}G\xrightarrow[]{\varepsilon}\mathbb{Z}/r\mathbb{Z}\to
0.$

If $\Lambda=\mathbb{Z}G/(\psi)$, then we can apply
$\Lambda\otimes_{\mathbb{Z}G}-$ to the above sequence to get

$0\to\Lambda\otimes_{\mathbb{Z}G}(I,r)\xrightarrow[]{1\otimes
i}\Lambda\otimes_{\mathbb{Z}G}\mathbb{Z}G\xrightarrow[]{1\otimes\varepsilon}\Lambda\otimes_{\mathbb{Z}G}(\mathbb{Z}/r\mathbb{Z})\to
0$

which is exact since
$\operatorname{Tor}^{1}_{\mathbb{Z}}(\mathbb{Z}/r\mathbb{Z},\Lambda)=0$ since
$\Lambda$ is torsion-free. Since
$f_{\\#}((I,r))\cong\Lambda\otimes_{\mathbb{Z}G}(I,r)$ and
$\Lambda\cong\Lambda\otimes_{\mathbb{Z}G}\mathbb{Z}G$, it suffices to show
that $1\otimes i$ is an isomorphism, i.e. that
$\Lambda\otimes_{\mathbb{Z}G}(\mathbb{Z}/r\mathbb{Z})=0$. Now
$\mathbb{Z}/r\mathbb{Z}\cong\mathbb{Z}\otimes_{\mathbb{Z}G}(\mathbb{Z}/r\mathbb{Z})$
and the associativity of tensor product implies that

$\Lambda\otimes_{\mathbb{Z}G}(\mathbb{Z}/r\mathbb{Z})\cong(\Lambda\otimes_{\mathbb{Z}G}\mathbb{Z})\otimes_{\mathbb{Z}G}(\mathbb{Z}/r\mathbb{Z})\cong(\mathbb{Z}/\varepsilon(\psi))\otimes_{\mathbb{Z}G}(\mathbb{Z}/r\mathbb{Z})=0$

since $r$ and $\varepsilon(\psi)$ are coprime. ∎

The following is an extension of Theorem 3.3 to this setting.

######  Proposition 5.3.

Let $f:\mathbb{Z}Q_{4n}\twoheadrightarrow\Lambda_{n_{1},\cdots,n_{k}}$ where
$n\geq 3$ and $n_{1},\cdots,n_{k}$ are distinct positive integers such that
$n_{i}\nmid n$ and $n_{i}\mid 2n$. If $\operatorname{Aut}(Q_{4n})$ acts on
$[P]$ and $\bar{P}=f_{\\#}(P)$, then there is a surjection of graded trees

$f_{\\#}:[P]/\operatorname{Aut}(Q_{4n})\twoheadrightarrow[\bar{P}]/\operatorname{Im}(f_{*})$

where the action of
$\operatorname{Im}(f_{*})\subseteq\operatorname{Aut}(\Lambda_{n_{1},\cdots,n_{k}})$
on $[\bar{P}]$ is induced by $f$.

###### Proof.

Let $n_{1},\cdots,n_{t}$ be the set of all $n_{i}$ such that $n_{i}\nmid n$
and $n_{i}\mid 2n$, where $k\leq t$. It follows from [20, p48-51] that
$\mathbb{Q}Q_{4n}/(y^{2}+1)\cong A_{n_{1},\cdots,n_{t}}$. In particular, $f$
factors through the map

$g:\mathbb{Z}Q_{4n}\twoheadrightarrow\mathbb{Z}Q_{4n}/(y^{2}+1).$

Now $\varepsilon(y^{2}+1)=2$ and $(2,r)=1$, and so
$g_{\\#}((I,r))\cong\mathbb{Z}Q_{4n}/(y^{2}+1)$. Hence, by Lemma 5.2, we have
that $f_{\\#}((I,r))\cong\Lambda_{n_{1},\cdots,n_{k}}$. The result now follows
using [32, Corollary 7.4] and a similar argument to the proof of Theorem 3.3.
∎

For the rest of this section, we will consider the cancellation problem for
orders of the form $\Lambda_{n_{1},\cdots,n_{k}}$. We begin by considering the
case $k=1$.

###  5.1. Cancellation for quaternionic orders

First note that $\Lambda_{2n}=\mathbb{Z}[\zeta_{2n},j]$ is a
$\mathcal{O}_{K}$-order in the quaternion algebra $\mathbb{Q}[\zeta_{2n},j]$
with centre $K=\mathbb{Q}(\zeta_{2n}+\zeta_{2n}^{-1})$. We can therefore apply
the results in Section 4 to get:

######  Lemma 5.4.

Let $2n=16,22$ or $2n\geq 26$ with $2n\neq 30,42$. Then $\Lambda_{2n}$ does
not have stably free cancellation.

###### Proof.

If $\Lambda_{2n}$ has stably free cancellation, then so does $\Gamma_{2n}$
where $\Lambda_{2n}\subseteq\Gamma_{2n}\subseteq\mathbb{Q}[\zeta_{2n},j]$ is a
maximal order by [41, Theorem A10]. Since
$[\mathbb{Q}(\zeta_{2n}+\zeta_{2n}^{-1}):\mathbb{Q}]=\frac{1}{2}\varphi(2n)$,
it follows from Theorem 4.3 that $\varphi(2n)\leq 12$, i.e. if $2n\leq 30$ or
$2n=36,42$. It remains to consider the cases $2n=16,22,26,28,36$. If
$\operatorname{ei}_{n}=\operatorname{ei}_{\mathbb{Q}(\zeta_{n}+\zeta_{n}^{-1})}$,
then Proposition 4.5 implies that the numerator of $\operatorname{ei}_{2n}$ is
a power of $2$ for $2n=16,22,26,28,36$. However, by [41, Table II], we have
that

$\operatorname{ei}_{16}=\frac{5}{48},\quad\operatorname{ei}_{22}=\frac{5}{132},\quad\operatorname{ei}_{26}=\frac{19}{156},\quad\operatorname{ei}_{28}=\frac{13}{21},\quad\operatorname{ei}_{36}=\frac{31}{36}$

which is a contradiction. ∎

######  Remark 5.5.

Theorem 4.3 and Proposition 4.5 do not characterise which orders have stably
free cancellation, even among the $\Lambda_{2n}$. For example, $\Lambda_{42}$
has $[K:\mathbb{Q}]=6$, $\operatorname{ei}_{K}=\frac{1}{6}$ [41, Table II] but
does not have stably free cancellation [41, p88].

It is possible to show that $\Lambda_{2n}$ has stably free cancellation in all
other cases. It can be shown using [36, Table 2] that $\Lambda_{2n}$ has
cancellation in all classes for all remaining cases other than $2n=20,24$. The
cases $2n=20,24$ can then be dealt with either using a MAGMA program, or by
explicitly computing $\operatorname{mass}(\operatorname{SF}(\Lambda_{2n}))$
and showing it is equal to $[\Lambda^{\times}:\mathcal{O}_{K}^{\times}]^{-1}$.

We also note the following bounds which we will use in the proof of Theorem B.

######  Proposition 5.6.

Let $K=\mathbb{Q}(\zeta_{2n}+\zeta_{2n}^{-1})$. If
$P\in\operatorname{Cls}(\Lambda_{2n})$, then

$\\#\operatorname{Cls}^{[P]}(\Lambda_{2n})\geq\frac{2|\Delta_{K}|^{3/2}}{2^{t_{2n}}(2\pi)^{\varphi(2n)}}\geq
e^{\frac{3}{8}\varphi(n)\log n+O(n\log\log n)}$

where
$t_{2n}=\operatorname{ord}_{2}([(\mathcal{O}_{K}^{\times})^{+}:(\mathcal{O}_{K}^{\times})^{2}])$.

###### Proof.

To get the first inequality, note that
$\\#\operatorname{Cls}^{[P]}(\Lambda_{2n})\geq\operatorname{mass}(\operatorname{Cls}^{[P]}(\Lambda_{2n}))$.
We can then apply Theorems 4.1 and 4.2 and note that
$N_{K/\mathbb{Q}}(\mathfrak{p})\geq 2$ for all
$\mathfrak{p}\mid\operatorname{disc}(\mathbb{Q}[\zeta_{2n},j])$ to get that:

$\operatorname{mass}(\operatorname{Cls}^{[P]}(\Lambda_{2n}))\geq\frac{2\zeta_{K}(2)|\Delta_{K}|^{3/2}}{(2\pi)^{\varphi(2n)}\cdot[(\mathcal{O}_{K}^{\times})^{+}:(\mathcal{O}_{K}^{\times})^{2}]}\geq\frac{2|\Delta_{K}|^{3/2}}{2^{t_{2n}}(2\pi)^{\varphi(2n)}}$

where we note that $\zeta_{K}(2)\geq 1$ by the Euler product formula.

Recall that $(\mathcal{O}_{K}^{\times})^{+}\subseteq\mathcal{O}_{K}^{\times}$
and so $t_{2n}\leq[K:\mathbb{Q}]=\varphi(2n)$. In order to compute
$\Delta_{K}=\Delta_{K/\mathbb{Q}}$, we will use that

$\Delta_{\mathbb{Q}(\zeta_{2n})/\mathbb{Q}}=N_{K/\mathbb{Q}}(\Delta_{\mathbb{Q}(\zeta_{2n})/K})\Delta_{K/\mathbb{Q}}^{[\mathbb{Q}(\zeta_{2n}):K]}.$

Since $\Delta_{\mathbb{Q}(\zeta_{2n})/K}=(\zeta_{2n}-\zeta_{2n}^{-1})^{2}$ and
$\Delta_{\mathbb{Q}(\zeta_{2n})/\mathbb{Q}}=\left(2n\prod_{p\mid
2n}p^{\frac{-1}{p-1}}\right)^{\varphi(2n)}$, we get that
$\Delta_{K}=c_{2n}^{-1}\left(2n\prod_{p\mid
2n}p^{\frac{-1}{p-1}}\right)^{\frac{1}{2}\varphi(2n)}$ where
$c_{2n}=N_{K/\mathbb{Q}}(\zeta_{2n}-\zeta_{2n}^{-1})$ which has $c_{2n}=2$ if
$n=2^{k}$, $c_{2n}=\sqrt{p}$ if $n=p^{k}$ for $p$ an odd prime and $c_{2n}=1$
otherwise. If $B$ denotes the bound above, we have that:

$\displaystyle\textstyle\log B$
$\displaystyle\geq\textstyle\frac{3}{4}\varphi(2n)(\log 2n-\sum_{p\mid
2n}\frac{\log p}{p-1})-\frac{3}{2}\log c_{n}-\varphi(2n)\log 4\pi-\log 2$
$\displaystyle\geq\textstyle\frac{3}{4}\varphi(2n)\log 2n+O(n\log\log n)$

since $c_{2n}\leq n$ and by using that $\sum_{p\mid 2n}\frac{\log
p}{p-1}=O(\log\log n)$ by [41, Lemma 2.7]. This makes sense since
$\varphi(n)\geq O(n/\log\log n)$ by standard results and so

$\varphi(2n)\log 2n\geq O(n\log n/\log\log n)\geq O(n\log\log n)$

since $\log n/(\log\log n)^{2}\to\infty$ as $n\to\infty$. We note finally that
$\varphi(2n)\geq\frac{1}{2}\varphi(n)$ and $\log 2n\geq\log n$ to get the
desired inequality. ∎

###  5.2. Cancellation for higher quaternionic orders

We will now consider the case $k\geq 2$. Firstly note that, for positive
integers $n_{1},\cdots,n_{k}$, there exists $n\geq 1$ such that $n_{i}\nmid
n$, $n_{i}\mid 2n$ if and only if $\nu_{2}(n_{i})=r$ for all $i$ and some
$r\geq 1$, i.e. for all $i$, $n_{i}=2^{r}m_{i}$ for some $m_{i}$ odd. From the
definition, it is possible to show that

$\Lambda_{n_{1},\cdots,n_{k}}\cong\mathbb{Z}Q_{4n}/(\Phi_{n_{1}}(x)\,\Phi_{n_{2}}(x)\,\cdots\,\Phi_{n_{k}}(x))$

where $\Phi_{n}(x)$ denote the $n$the cyclotomic polynomial. It follows that,
if $S\subseteq\\{n_{1},\cdots,n_{k}\\}$, then there is a quotient
$\Lambda_{n_{1},\cdots,n_{k}}\twoheadrightarrow\Lambda_{S}$. By [41, Theorem
A10] this implies that, if $\Lambda_{n_{1},\cdots,n_{k}}$ has stably free
cancellation, then $\Lambda_{S}$ has stably free cancellation for all
$S\subseteq\\{n_{1},\cdots,n_{k}\\}$.

In particular, by Lemma 5.4, we have that $\Lambda_{n_{1},\cdots,n_{k}}$ does
not have stably free cancellation except possibly if, for all $i$, we have
$n_{i}\leq 14$ or $n_{i}=18,20,24,30$. By the discussion above, this is
equivalent to

$\\{n_{1},\cdots,n_{k}\\}\subseteq\begin{cases}\\{2,6,10,14,18,30\\},&\text{if
$r=1$}\\\ \\{4,12,20\\},&\text{if $r=2$}\\\ \\{8,24\\}&\text{if
$r=3$}\end{cases}$

where we note that the cases with $r\geq 4$ do not arise.

For a subset $S\subseteq\mathbb{Z}_{\geq 1}$, define a graph $\mathcal{G}(S)$
with vertices the elements of $S$ and where $a,b\in S$ are connected by an
edge if $a/b=p^{r}$ for some $p$ prime and $r\in\mathbb{Z}$.

######  Lemma 5.7.

Let $r\geq 1$ and let $n_{1},\cdots,n_{k}$ be such that $\nu_{2}(n_{i})=r$ for
all $i$. If $\mathcal{G}(n_{1},\cdots,n_{k})$ is not connected, then there
exists a splitting $\\{n_{1},\cdots,n_{k}\\}=S_{1}\sqcup S_{2}$ with $S_{1}$,
$S_{2}$ non-empty such that
$\Lambda_{n_{1},\cdots,n_{k}}\cong\Lambda_{S_{1}}\times\Lambda_{S_{2}}$.

###### Proof.

If $\mathcal{G}(n_{1},\cdots,n_{k})$ is not connected then, by definition,
there exists a splitting $\\{n_{1},\cdots,n_{k}\\}=S_{1}\sqcup S_{2}$ with
$S_{1}$, $S_{2}$ non-empty and such that there are no edges between any $a\in
S_{1}$ and $b\in S_{2}$. Now consider
$\Lambda_{n_{1},\cdots,n_{k}}=\mathbb{Z}Q_{4n}/(\Phi_{n_{1}}\,\cdots\,\Phi_{n_{k}})$
where, for example, $n=\frac{1}{2}\text{lcm}(n_{1},\cdots,n_{k})$.

If $a\in S_{1}$ and $b\in S_{2}$, then $a/b\neq p^{r}$ for all
$r\in\mathbb{Z}$ and so $(\Phi_{a},\Phi_{b})=1$ by results of Diederichsen
[11] (see also [41, Theorem B1]). In particular, this implies that

$\textstyle(\prod_{n_{i}\in S_{1}}\Phi_{n_{i}},\prod_{n_{i}\in
S_{2}}\Phi_{n_{i}})=1$

and so
$\mathbb{Z}Q_{4n}/(\Phi_{n_{1}}\,\cdots\,\Phi_{n_{k}})\cong\mathbb{Z}Q_{4n}/(\prod_{n_{i}\in
S_{1}}\Phi_{n_{i}})\times\mathbb{Z}Q_{4n}/(\prod_{n_{i}\in
S_{2}}\Phi_{n_{i}})$ which completes the proof since
$\Lambda_{S_{j}}\cong\mathbb{Z}Q_{4n}/\prod_{n_{i}\in S_{j}}\Phi_{n_{i}}$ for
$j=1,2$. ∎

Note that, if
$\Lambda_{n_{1},\cdots,n_{k}}\cong\Lambda_{S_{1}}\times\Lambda_{S_{2}}$, then
$\Lambda_{n_{1},\cdots,n_{k}}$ has stably free cancellation if and only if
$\Lambda_{S_{1}}$ and $\Lambda_{S_{2}}$ have stably free cancellation. It
therefore suffices to consider the case where
$\mathcal{G}(n_{1},\cdots,n_{k})$ is connected.

The following cases were calculated by Swan in [41, Lemma 8.9-8.12, 8.13] and
[41, Corollary 8.17, 10.14].

######  Lemma 5.8.

Let $\Lambda=\Lambda_{2,14}$, $\Lambda_{6,18}$, $\Lambda_{6,30}$,
$\Lambda_{4,12}$, $\Lambda_{4,20}$ or $\Lambda_{8,24}$. Then $\Lambda$ does
not have stably free cancellation.

Let $(**)$ be the set of $\\{n_{1},\cdots,n_{k}\\}$ which satisfy $(*)$ and
also have no subset of the form given in Lemma 5.8. The following is
straightforward:

######  Lemma 5.9.

Let $r\geq 1$ and let $S=\\{n_{1},\cdots,n_{k}\\}$ where $\nu_{2}(n_{i})=r$
for all $i$. Suppose $S$ satisfies $(**)$ and $\mathcal{G}(S)$ is connected.
If $k\geq 2$, then

$S=\begin{cases}\\{2,6\\},\\{2,10\\},\\{2,18\\},\\{10,30\\},&\text{if
$k=2$}\\\ \\{2,6,10\\},\\{2,10,18\\},\\{2,10,30\\},&\text{if $k=3$}\\\
\\{2,10,18,30\\},&\text{if $k=4$}\end{cases}$

and no cases with $r\geq 2$ or $k\geq 5$ arise.

It suffices to determine when stably free cancellation occurs for the orders
$\Lambda_{S}$ where $S$ is given in the above lemma.

######  Lemma 5.10.

If $\Lambda=\Lambda_{2,6}$, $\Lambda_{2,10}$ or $\Lambda_{2,18}$, then
$\Lambda$ has stably free cancellation.

###### Proof.

The case $\Lambda_{2,2p}$ for $p$ an odd prime is dealt with in [41, Lemma
10.13], and it follows from this that $\Lambda_{2,6}$, $\Lambda_{2,10}$ have
cancellation. The case $\Lambda_{2,18}$ is similar, though with added
complications. First recall that
$\Lambda_{2,18}=\mathbb{Z}Q_{36}/(\Phi_{2}\,\Phi_{18})$. If $I=(\Phi_{2})$,
$J=(\Phi_{18})$ are ideals in $\mathbb{Z}Q_{36}$, then $I\cap
J=(\Phi_{2}\,\Phi_{18})$ and $I+J=(\Phi_{2},\Phi_{18})=(\Phi_{2},3)$ by [41,
Theorem B1]. By [7, Example 42.3], we get the following Milnor square

${\Lambda_{2,18}}$${\mathbb{Z}[\zeta_{18},j]}$${\mathbb{Z}[j]}$${\mathbb{F}_{3}[j]}$

since $\Lambda_{18}=\mathbb{Z}[\zeta_{18},j]$ and $\Lambda_{2}=\mathbb{Z}[j]$.
If $\operatorname{Ker}$ denotes the fibre over $(\Lambda_{2},\Lambda_{18})$,
then

$\displaystyle\operatorname{Ker}(\operatorname{Cls}(\Lambda_{2,18})\to\operatorname{Cls}(\Lambda_{2})\times\operatorname{Cls}(\Lambda_{18}))$
$\displaystyle\cong\mathbb{Z}[j]^{\times}\backslash\,\mathbb{F}_{3}[j]^{\times}/\mathbb{Z}[\zeta_{18},j]^{\times}=\\{[1],[1+j]\\}$
$\displaystyle\operatorname{Ker}(C(\Lambda_{2,18})\to C(\Lambda_{2})\times
C(\Lambda_{18}))$
$\displaystyle\cong\frac{K_{1}(\mathbb{F}_{3}[j])}{K_{1}(\mathbb{Z}[j])\times
K_{1}(\mathbb{Z}[\zeta_{18},j])}\cong\mathbb{F}_{3}^{\times}/(\mathbb{F}_{3}^{\times})^{2}$

where the first identifications are standard [42] and the second comes from
[23, Lemma 7.5,7.6] since $h_{9}$ is odd. Since both sets have the same size,
and $\mathbb{Z}[j]$, $\mathbb{Z}[\zeta_{18},j]$ have cancellation, it follows
that $\Lambda_{2,18}$ has stably free cancellation. ∎

######  Lemma 5.11.

Let $n,m\geq 2$ even such that $n/m\neq p^{r}$ for all $p$ prime and
$r\in\mathbb{Z}$. If $\Lambda_{2,n}$, $\Lambda_{2,m}$ have stably free
cancellation, then so does $\Lambda_{2,n,m}$.

###### Proof.

Let $I=(\Phi_{2}\,\Phi_{n})$ and $J=(\Phi_{2},\,\Phi_{m})$ so that $I\cap
J=(\Phi_{2}\,\Phi_{n}\,\Phi_{m})$ and
$I+J=(\Phi_{2}\,\Phi_{n},\Phi_{2}\,\Phi_{m})=(\Phi_{2})$ since
$(\Phi_{n},\Phi_{m})=1$ by [41, Theorem B1]. By [7, Example 42.3], we get the
following Milnor square

${\Lambda_{2,n,m}}$${\Lambda_{2,m}}$${\Lambda_{2,n}}$${\mathbb{Z}[j]}$

since $\Lambda_{2}=\mathbb{Z}[j]$. Since $\mathbb{Z}[j]^{\times}=\\{\pm 1,\pm
j\\}$, we have that
$\Lambda_{2,n}^{\times}\twoheadrightarrow\mathbb{Z}[j]\times$ since the units
$\pm 1,\pm j\in\Lambda_{2,n}$ lift from $\Lambda_{2}$, $\Lambda_{n}$. If
$\operatorname{Ker}$ is the fibre over $(\Lambda_{2,n},\Lambda_{2,m})$, then

$\operatorname{Ker}(\operatorname{Cls}(\Lambda_{2,n,m})\to\operatorname{Cls}(\Lambda_{2,n})\times\operatorname{Cls}(\Lambda_{2,m}))\cong\Lambda_{2,n}^{\times}\,\backslash\mathbb{Z}[j]^{\times}/\Lambda_{2,m}^{\times}=1$

from which the result follows easily. ∎

In particular, since $\Lambda_{2,n}$ has stably free cancellation for
$n=6,10,18$ by Lemma 5.10, this implies that $\Lambda_{2,6,10}$,
$\Lambda_{2,10,18}$ have stably free cancellation.

######  Lemma 5.12.

$\Lambda_{10,30}$ does not have stably free cancellation.

###### Proof.

As above, consider the Milnor square corresponding to the ideals
$I=(\Phi_{10})$ and $J=(\Phi_{30})$ which has $I+J=(\Phi_{10},3)$ by [41,
Theorem B1]:

${\Lambda_{10,30}}$${\mathbb{Z}[\zeta_{30},j]}$${\mathbb{Z}[\zeta_{10},j]}$${\mathbb{F}_{3}[\zeta_{10},j]}$

By [41, p76], $\Lambda_{10}$ is unramified at 3 and so
$\mathbb{F}_{3}[\zeta_{10},j]\cong
M_{2}(\mathbb{F}_{3}[\zeta_{10}+\zeta_{10}^{-1}])$. Since
$\zeta_{10}+\zeta_{10}^{-1}=\frac{1}{2}(1+\sqrt{5})$ and
$\mathbb{F}_{3}(\sqrt{5})\cong\mathbb{F}_{9}$, we get
$K_{1}(\mathbb{F}_{3}[\zeta_{10},j])\cong
K_{1}(M_{2}(\mathbb{F}_{9}))\cong\mathbb{F}_{9}^{\times}$ by Morita
equivalence. Consider the composition

$f:\mathbb{Z}[\zeta_{30},j]^{\times}\to K_{1}(\mathbb{Z}[\zeta_{30},j])\to
K_{1}(\mathbb{F}_{3}[\zeta_{10},j])\cong\mathbb{F}_{9}^{\times}.$

Since 30 is composite, $1+\zeta_{30}\in\mathbb{Z}[\zeta_{30}]^{\times}$ and,
by [23, Lemma 7.5,7.6], we have that
$K_{1}(\mathbb{Z}[\zeta_{30},j])\cong(\mathbb{Z}[\zeta_{30}+\zeta_{30}^{-1}]^{\times})^{+}$.
By the argument given in the proof on [41, Lemma 8.9], we get that

$1+\zeta_{30}\mapsto 2+(\zeta_{30}+\zeta_{30}^{-1})\mapsto
2+(\zeta_{10}+\zeta_{10}^{-1})=2+\frac{1}{2}(1+\sqrt{5})=1-\sqrt{5}\in\mathbb{F}_{9}^{\times}.$

Since $1-\sqrt{5}$ is a generator of
$\mathbb{F}_{9}\cong\mathbb{F}_{3}[\sqrt{5}]$, this implies that $f$ is
surjective. In particular, this implies that

$\operatorname{Ker}(C(\Lambda_{10,30})\to C(\Lambda_{10})\times
C(\Lambda_{30}))\cong\frac{K_{1}(\mathbb{F}_{3}[\zeta_{10},j])}{K_{1}(\mathbb{Z}[\zeta_{10},j])\times
K_{1}(\mathbb{Z}[\zeta_{30},j])}=1.$

It therefore suffices to show that
$\mathbb{Z}[\zeta_{10},j]^{\times}\,\backslash\mathbb{F}_{3}[\zeta_{10},j]^{\times}/\mathbb{Z}[\zeta_{30},j]^{\times}\neq
1$. By [23, Lemma 7.5,7.6], we have that
$\mathbb{Z}[\zeta_{2n},j]^{\times}=\langle\mathbb{Z}[\zeta_{2n}]^{\times},j\rangle$
for all $n$ and so

$\operatorname{Im}(\mathbb{Z}[\zeta_{10},j]^{\times}\times\mathbb{Z}[\zeta_{30},j]^{\times}\to\mathbb{F}_{3}[\zeta_{10},j]^{\times})\cap\mathbb{F}_{3}[j]=\\{\pm
1,\pm j\\}.$

It follows that $[1]\neq[1+j]$, which completes the proof. ∎

Combining all these results together leads to the following, which determines
precisely when stably free cancellation occurs for the
$\Lambda_{n_{1}.\cdots,n_{k}}$.

######  Theorem 5.13.

Let $n\geq 1$ and let $n_{1},\cdots,n_{k}$ be such that $n_{i}\nmid n$ and
$n_{i}\mid 2n$. Then a complete list of the $\Lambda_{n_{1},\cdots,n_{k}}$
which have stably free cancellation is as follows:

  1. (i)

$\Lambda_{2}$, $\Lambda_{4}$, $\Lambda_{6}$, $\Lambda_{8}$, $\Lambda_{10}$,
$\Lambda_{12}$, $\Lambda_{14}$, $\Lambda_{18}$, $\Lambda_{20}$,
$\Lambda_{24}$, $\Lambda_{30}$

  2. (ii)

$\Lambda_{2,6}$, $\Lambda_{2,10}$, $\Lambda_{2,18}$, $\Lambda_{2,30}$,
$\Lambda_{6,10}$, $\Lambda_{6,14}$, $\Lambda_{10,14}$, $\Lambda_{10,18}$,
$\Lambda_{14,18}$, $\Lambda_{14,30}$, $\Lambda_{18,30}$

  3. (iii)

$\Lambda_{2,6,10}$, $\Lambda_{2,10,18}$, $\Lambda_{2,18,30}$,
$\Lambda_{6,10,14}$, $\Lambda_{10,14,18}$, $\Lambda_{14,18,30}$.

###### Proof.

Consider the not-necessarily-connected subsets of the form $(*)$. If $k=1$,
then use Lemma 5.4. By the work above, if $k\geq 2$, then eliminate all
subsets which contain $\\{2,14\\}$, $\\{4,12\\}$, $\\{4,20\\}$, $\\{6,18\\}$,
$\\{6,30\\}$, $\\{8,24\\}$, $\\{10,30\\}$. ∎

It is not much more difficult to construct a list of those orders
$\Lambda_{n_{1},\cdots,n_{k}}$which have cancellation in all classes, as well
as those orders which have non-cancellation in all locally free classes.
However, we will omit this here for brevity.

##  6\. Proof of Theorem A

Let $G$ have $k$-periodic cohomology, let $n=ik$ or $ik-2$ for some $i\geq 1$
and let $P_{(G,n)}\in P(\mathbb{Z}G)$ be a representative for the Swan
finiteness obstruction $\sigma_{ik}(G)\in C(\mathbb{Z}G)/T_{G}$. We will now
proceed to prove the following theorem which, by [32, Theorem B], is
equivalent to Theorem A.

######  Theorem 6.1.

$[P_{(G,n)}]/\operatorname{Aut}(G)$ has cancellation if and only if
$m_{\mathbb{H}}(G)\leq 2$.

If $m_{\mathbb{H}}(G)\leq 2$, then Theorem 2.1 implies that $[P_{(G,n)}]$ has
cancellation and so $[P_{(G,n)}]/\operatorname{Aut}(G)$ has cancellation also.
It therefore remains to show that $m_{\mathbb{H}}(G)\geq 3$ implies
$[P_{(G,n)}]/\operatorname{Aut}(G)$ has non-cancellation.

Let $G$ have periodic cohomology and $m_{\mathbb{H}}(G)\geq 3$. By Lemma 3.4,
there is a quotient $f:G\twoheadrightarrow Q_{4m}$ for some $m\geq 6$. Suppose
$g:\mathbb{Z}Q_{4m}\twoheadrightarrow\Lambda_{n_{1},\cdots,n_{k}}$ and
$n_{1},\cdots,n_{k}$ are distinct positive integers such that $n_{i}\nmid m$
and $n_{i}\mid 2m$. Then, by combining Corollary 3.6 and Proposition 5.3,
there is a surjection of graded trees

$f_{\\#}:[P_{(G,n)}]/\operatorname{Aut}(G)\twoheadrightarrow[\widebar{P_{(G,n)}}]/\operatorname{Im}((g\circ
f)_{*})$

where $\widebar{P_{(G,k)}}=(g\circ f)_{\\#}(P_{(G,n)})$, $(g\circ
f)_{*}:\operatorname{Aut}(G)\to\operatorname{Aut}(\Lambda_{n_{1},\cdots,n_{k}})$
and the action of $\operatorname{Im}((g\circ
f)_{*})\subseteq\operatorname{Aut}(\Lambda_{n_{1},\cdots,n_{k}})$ on
$[\bar{P}]$ is induced by $g\circ f$.

The following is immediate from Theorem 5.13. Note that this shows that
$\mathbb{Z}Q_{4m}$ does not have stably free cancellation for $m\geq 6$ as was
shown in [41, Theorem I].

######  Lemma 6.2.

If $m\geq 6$, then $Q_{4m}$ has a quotient of the form
$\Lambda=\Lambda_{n_{1},\cdots,n_{k}}$ with $n_{i}\nmid m$, $n_{i}\mid 2m$
which does not have stably free cancellation. In particular,

  1. (i)

If $m\neq 6,7,9,10,12,15$, then we can take $\Lambda=\Lambda_{2m}$

  2. (ii)

If $m=6,7,9,10,12,15$, then we can take $\Lambda=\Lambda_{4,12}$,
$\Lambda_{2,14}$, $\Lambda_{6,18}$, $\Lambda_{4,20}$, $\Lambda_{8,24}$,
$\Lambda_{6,30}$ and $\Lambda_{6,42}$ respectively.

Recall that, if $\Lambda\subseteq A$ is a $\mathbb{Z}$-order in a finite-
dimensional semisimple separable $\mathbb{Q}$-algebra $A$ and
$\Lambda\subseteq\Gamma\subseteq A$ is a maximal order, then the defect group
is defined as $D(\Lambda)=\operatorname{Ker}(i_{*}:C(\Lambda)\to C(\Gamma))$
where $i:\Lambda\hookrightarrow\Gamma$, and note that $i_{*}$ is surjective by
[41, Theorem A10]. By [41, Theorem A24], this is independent of the choice of
$\Gamma$ and, if $f:\Lambda_{1}\to\Lambda_{2}$ is a map of
$\mathbb{Z}$-orders, then $f$ induces a map $f_{*}:D(\Lambda_{1})\to
D(\Lambda_{2})$.

It was shown by Milgram [28, Theorem 2.B.1] that $[P_{(G,n)}]\in
D(\mathbb{Z}G)$ and so $[\widebar{P_{(G,n)}}]\in
D(\Lambda_{n_{1},\cdots,n_{k}})$ by the remarks above. In particular, if
$D(\Lambda_{n_{1},\cdots,n_{k}})=0$, then we have that
$[\widebar{P_{(G,n)}}]=0\in C(\Lambda_{n_{1},\cdots,n_{k}})$ automatically.

######  Lemma 6.3.

If $\Lambda=\Lambda_{n_{1},\cdots,n_{k}}\subseteq A_{n_{1},\cdots,n_{k}}$ is
of the following forms, then $D(\Lambda)=0$.

  1. (i)

$\Lambda=\Lambda_{2n}$ for $n\geq 1$

  2. (ii)

$\Lambda=\Lambda_{4,12}$, $\Lambda_{4,20}$, $\Lambda_{8,24}$,
$\Lambda_{6,30}$, $\Lambda_{6,42}$.

###### Proof.

By [41, Corollary 8.3], we have that $D(\Lambda_{2n})=0$. The other orders are
of the form $\Lambda_{d,2n}$ for some $d\neq 2n$ such that $d\nmid n$ and
$d\mid 2n$. Suppose that $d$ and $n$ are one of the give pairs above. It
follows from the proofs of [41, Lemma 8.9-8.12, 8.14] that the projection
$f:\Lambda_{d,2n}\to\Lambda_{2n}$ induces an isomorphism
$f_{*}:C(\Lambda_{d,2n})\cong C(\Lambda_{2n})$. Since $D(\Lambda_{2n})=0$,
this implies that $C(\Lambda_{d,2n})\cong C(\Gamma)$ for any maximal order
$\Lambda_{2n}\subseteq\Gamma\subseteq A_{2n}$ by [41, Theorem A24].

Let $\Lambda_{d,2n}\subseteq\Gamma_{d,2n}\subseteq A_{d,2n}$ be a maximal
order. Since $A_{d,2n}\cong A_{d}\times A_{n}$, we must have that
$\Gamma_{2n,d}\cong\Gamma_{d}\times\Gamma_{2n}$ where $\Gamma_{d}\subseteq
A_{d}$ and $\Gamma_{2n}\subseteq A_{2n}$ are maximal orders [20, p33]. Since
$\Gamma\twoheadrightarrow\Gamma_{2n}$ induces a surjection
$C(\Gamma)\twoheadrightarrow C(\Gamma_{2n})$ [41, Theorem A10], we have that
$|C(\Gamma_{2n})|=|C(\Lambda_{d,2n})|\geq|C(\Gamma_{d,2n})|\geq|C(\Gamma_{2n})|$
by considering the case $\Gamma=\Gamma_{2n}$. This implies that
$D(\Lambda_{d,2n})=0$. ∎

######  Remark 6.4.

This is false for $\Lambda_{2,14}$ and $\Lambda_{6,18}$ and, in fact, all such
orders in the cases $Q_{28}$, $Q_{36}$ which do not have stably free
cancellation. In particular, note that

$\mathbb{Q}Q_{28}\cong\mathbb{Q}D_{14}\times
A_{2,14},\quad\mathbb{Q}Q_{36}\cong\mathbb{Q}D_{18}\times A_{2,6,18}$

and so only options are $\Lambda_{2}$, $\Lambda_{6}$, $\Lambda_{14}$,
$\Lambda_{18}$, $\Lambda_{2,6}$, $\Lambda_{2,18}$ which all have stably free
cancellation, or $\Lambda_{2,14}$, $\Lambda_{6,18}$, $\Lambda_{2,6,18}$ which
all do not have stably free cancellation by Theorem 5.13 but have
$D(\Lambda)\neq 0$ from results in [41].

Suppose $G$ has a quotient $Q_{4m}$ for some $m\geq 6$ and $4m\neq 28,36$.
Then Lemmas 6.2 and 6.3 imply that there exists
$\Lambda_{G}=\Lambda_{n_{1},\cdots,n_{k}}$ such that $D(\Lambda_{G})=0$ and
$[\Lambda_{G}]$ has non-cancellation. Since $[P_{(G,n)}]\in D(\mathbb{Z}G)$,
this implies that $[\widebar{P_{(G,n)}}]=0\in C(\Lambda_{G})$. If $4m=28,36$,
then Lemma 6.2 implies that there exists
$\Lambda_{G}=\Lambda_{n_{1},\cdots,n_{k}}$ such that $[\Lambda_{G}]$ has non-
cancellation. Since $Q_{28}$ and $Q_{36}$ have type I, Proposition 1.2 implies
that $G$ has type I and so $\sigma_{k}(G)=0$ by Theorem 1.11. Hence
$[P_{(G,n)}]\in T_{G}$ and so similarly we have that
$[\widebar{P_{(G,n)}}]=0\in C(\Lambda_{G})$ by Lemma 5.2.

Hence, in both cases, there is a surjection of graded trees

$[P_{(G,n)}]/\operatorname{Aut}(G)\twoheadrightarrow[\Lambda_{G}]/\operatorname{Aut}(\Lambda_{G}).$

By, for example [32, Lemma 7.1], the action of
$\operatorname{Aut}(\Lambda_{G})$ on $[\Lambda_{G}]$ fixes the free module.
Since $[\Lambda_{G}]$ has non-cancellation, this implies that
$[\Lambda_{G}]/\operatorname{Aut}(\Lambda_{G})$ has non-cancellation. This
completes the proof of Theorem 2.1.

##  7\. Proof of Theorem B

Recall that, if $G$ is finite and $n$ is even, then $HT(G,n)$ is a fork in
that it has a single vertex at each non-minimal height and finitely many at
the minimal level [32, Corollary 3.7]. Let $N(G,n)=\\#\\{X\in
HT(G,n):\chi(X)\text{ is minimal}\\}$.

We will now prove the following, which is a restatement of Theorem B.

######  Theorem 7.1.

Let $G$ have $k$-periodic cohomology and let $n$ be such that $k\mid n$ or
$n+2$ and, if $n=2$, suppose $G$ has the D2 property. If
$m=m_{\mathbb{H}}(G)$, then

$N(G,n)\geq e^{\tfrac{m\log m}{8\log\log m}+O(m\log\log m)}$

In particular, for all $\lambda>0$, we have that $N(G,n)/e^{\lambda
m}\to\infty$ as $m\to\infty$.

###### Proof.

By [32, Theorem B], there is a bijection

$HT(G,n)\cong[P_{(G,n)}]/\operatorname{Aut}(G)$

for some $P_{(G,n)}\in P(\mathbb{Z}G)$ such that $[P_{(G,n)}]=\sigma_{n}(G)\in
C(\mathbb{Z}G)/T_{G}$. Now suppose $m=m_{\mathbb{H}}(G)\geq 3$. By Lemma 3.4,
there is a quotient $f:G\twoheadrightarrow Q_{4m_{0}}$ for some $m_{0}\geq
2m/3$. By Corollary 3.6, this implies that

$N(G,n)\geq\\#\operatorname{Cls}^{[P_{(G,n)}]}(\mathbb{Z}G)/\operatorname{Aut}(G)\geq\\#\operatorname{Cls}^{[P]}(\mathbb{Z}Q_{4m_{0}})/\operatorname{Im}(f_{*})$

where $P=f_{\\#}(P_{(G,n)})$. Since there is a quotient of $\mathbb{Z}$-orders
$g:Q_{4m_{0}}\twoheadrightarrow\Lambda_{2m_{0}}$, it follows from [41, Theorem
A10] that

$\\#\operatorname{Cls}^{[P]}(\mathbb{Z}Q_{4m_{0}})\geq\\#\operatorname{Cls}^{[\bar{P}]}(\Lambda)$

where $\bar{P}=g_{\\#}(P)$. Since $\bar{P}=(g\circ f)_{\\#}(P_{(G,n)})$, we
have that $[\bar{P}]=(g\circ f)_{\\#}([P_{(G,n)}])\in C(\Lambda_{2m_{0}})$. It
follows from [28, Theorem 2.B.1] that $[P_{(G,n)}]\in D(\mathbb{Z}G)$ and so
$[\bar{P}]\in D(\Lambda_{2m_{0}})$. By [41, Corollary 8.3], we have that
$D(\Lambda_{2m_{0}})=0$ and so $[\bar{P}]=[\Lambda_{2m_{0}}]$.

Since $|\operatorname{Aut}(Q_{4m_{0}})|=2m_{0}\varphi(2m_{0})$, Proposition
5.6 implies that

$N(G,n)\geq\frac{1}{2m_{0}\varphi(2m_{0})}e^{\frac{3}{8}\varphi(m_{0})\log
m_{0}+O(m_{0}\log\log m_{0})}$

and we can omit $1/2m_{0}\varphi(2m_{0})$ since it is sub-exponential. By [19,
Theorem 328], we have that $\varphi(n)\geq n/2\log\log n$ for $n$ sufficiently
large. Since $m_{0}\geq 2m/3$, this implies that

$\log
N(G,n)\geq\frac{3}{16}\cdot\frac{(2m/3)\log(2m/3)}{\log\log(2m/3)}+O(m\log\log
m)\geq\frac{m\log m}{8\log\log m}+O(m\log\log m)$

since $\log x/\log\log x$ is increasing for $x$ sufficiently large. ∎

We note also the following improvement in the case where $G$ is a quaternion
group, the proof of which is contained in the argument above.

######  Proposition 7.2.

Let $n$ be even and, if $n=2$, assume $Q_{4m}$ has the D2 property. Then we
have

$N(Q_{4m},n)\geq e^{\frac{3}{8}\varphi(m)\log m+O(m\log\log m)}.$

In particular, if $p$ is prime and $m=m_{\mathbb{H}}(Q_{4p})=\lfloor
p/2\rfloor$, then

$N(Q_{4p},n)\geq e^{\frac{3}{8}p\log p+O(p\log\log p)}\geq e^{\frac{3}{4}m\log
m+O(m\log\log m)}.$

##  8\. Examples

Let $G$ have $k$-periodic cohomology, let $n=ik$ or $ik-2$ and let
$P_{(G,n)}\in P(\mathbb{Z}G)$ be such that $[P_{(G,n)}]=\sigma_{ik}(G)\in
C(\mathbb{Z}G)/T_{G}$. The classes $[P_{(G,n)}]\in C(\mathbb{Z}G)$ have two
special properties. Firstly, we have:

  1. (1)

If $\theta\in\operatorname{Aut}(G)$, then
$[(P_{(G,n)})_{\theta}]=[P_{(G,n)}]\in C(\mathbb{Z}G)/T_{G}$.

This was necessary in order to define our action of $\operatorname{Aut}(G)$ on
the class $[P_{(G,n)}]$ since $\theta\in\operatorname{Aut}(G)$ sends
$P\mapsto(I,\psi(\theta))\otimes P_{\theta}$ for some
$\psi:\operatorname{Aut}(G)\to(\mathbb{Z}/|G|)^{\times}$. This is well-defined
since

$[(I,\psi(\theta))\otimes
P_{\theta}]=[(I,\psi(\theta))]+[P_{\theta}]=[P_{(G,n)}],$

i.e. $[P_{(G,n)}]-[(P_{(G,n)})_{\theta}]\in T_{G}$. The second property
follows from Theorem A:

  1. (2)

$[P_{(G,n)}]$ has cancellation if and only if
$[P_{(G,n)}]/\operatorname{Aut}(G)$ has cancellation.

The aim of this section will be to give examples to show that both properties
are false for general projective modules over groups with periodic cohomology.

###  8.1. Cyclic group of order $p$

Here we will consider the case $G=C_{p}$ for a prime $p$. Recall that
$T_{C_{p}}=0$ [40, Corollary 6.1] and so we need only find $[P]\in
C(\mathbb{Z}C_{p})$ and $\theta\in\operatorname{Aut}(C_{p})$ for which
$[P]\neq[P_{\theta}]\in C(\mathbb{Z}C_{p})$.

First define $i:\mathbb{Z}C_{p}\twoheadrightarrow\mathbb{Z}[\zeta_{p}]$ by
mapping a generator $x\in C_{p}$ to $\zeta_{p}=e^{2\pi i/p}\in\mathbb{C}$. It
follows from a theorem of Rim [34, Theorem 6.24] that the map

$i_{*}:C(\mathbb{Z}C_{p})\to C(\mathbb{Z}[\zeta_{p}])$

is an isomorphism. Let
$\tilde{\cdot}:\operatorname{Aut}(C_{p})\to\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$
be the group isomorphism sending $\theta_{i}:x\mapsto x^{i}$ to
$\tilde{\theta}_{i}:\zeta_{p}\mapsto\zeta_{p}^{i}$ for all
$i\in(\mathbb{Z}/p)^{\times}$. The following is easy to check:

######  Lemma 8.1.

If $\theta\in\operatorname{Aut}(C_{p})$ and $[P]\in C(\mathbb{Z}C_{p})$, then
$i_{*}(\theta_{*}([P]))=\tilde{\theta}_{*}(i_{*}([P])$, i.e. there is a
commutative diagram

${C(\mathbb{Z}C_{p})}$${C(\mathbb{Z}[\zeta_{p}])}$${C(\mathbb{Z}C_{p})}$${C(\mathbb{Z}[\zeta_{p}])}$$\scriptstyle{\theta_{*}}$$\scriptstyle{i_{*}}$$\scriptstyle{\tilde{\theta}_{*}}$$\scriptstyle{i_{*}}$

Let
$G=\operatorname{Aut}(C_{p})\cong\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$
and let $C(\mathbb{Z}C_{p})^{G}$ and $C(\mathbb{Z}[\zeta_{p}])^{G}$ denotes
the fixed points under the actions of $\operatorname{Aut}(C_{p})$ and
$\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ respectively. By Lemma
8.1, we have that the map

$i_{*}:C(\mathbb{Z}C_{p})^{G}\to C(\mathbb{Z}[\zeta_{p}])^{G}$

is an isomorphism, which could be viewed as an extension of Rim’s theorem.

######  Proposition 8.2.

$C(\mathbb{Z}C_{p})^{G}=C(\mathbb{Z}[\zeta_{p}])^{G}=0$.

###### Proof.

It follows from the Chevalley’s ambiguous class number formula [5] (see also
[16, Remark 6.2.3]) that

$|C(\mathbb{Z}[\zeta_{p}])^{G}|=\frac{\operatorname{Ram}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})}{[\mathbb{Q}(\zeta_{p}):\mathbb{Q}]\cdot[\mathbb{Z}^{\times}:\mathbb{Z}^{\times}\cap
N_{\mathbb{Q}(\zeta_{p})/\mathbb{Q}}(\mathbb{Q}(\zeta_{p})^{\times})]}$

where $\operatorname{Ram}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ is the product of
the ramification indices at the finite and infinite places of
$\mathbb{Q}(\zeta_{p})$. It is easy to check that
$\operatorname{Ram}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})=2(p-1)$. To compute the
denominator, note that $\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})$ is the fixed
field of $\mathbb{Q}(\zeta_{p})$ under the conjugation action. If
$\alpha\in\mathbb{Q}(\zeta_{p})$, then the transitivity property of norms
$N_{\mathbb{Q}(\zeta_{p})/\mathbb{Q}}=N_{\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})/\mathbb{Q}}\circ
N_{\mathbb{Q}(\zeta_{p})/\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})}$ implies that

$N_{\mathbb{Q}(\zeta_{p})/\mathbb{Q}}(\alpha)=N(\alpha\cdot\bar{\alpha})=N(\alpha)\cdot
N(\bar{\alpha})=|N(\alpha)|^{2}\geq 0$

where $N=N_{\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})/\mathbb{Q}}$. This implies
that there does not exist $\alpha\in\mathbb{Q}(\zeta_{p})^{\times}$ such that
$N_{\mathbb{Q}(\zeta_{p})/\mathbb{Q}}(\alpha)=-1$ and so
$[\mathbb{Z}^{\times}:\mathbb{Z}^{\times}\cap
N_{\mathbb{Q}(\zeta_{p})/\mathbb{Q}}(\mathbb{Q}(\zeta_{p})^{\times})]=2$.
Hence $|C(\mathbb{Z}[\zeta_{p}])^{G}|=1$ and so
$C(\mathbb{Z}C_{p})^{G}=C(\mathbb{Z}[\zeta_{p}])^{G}=0$ by the discussion
above. ∎

Since $G$ acts trivially on the class $0\in C(\mathbb{Z}C_{p})$, Proposition
8.2 implies that $G$ acts non-trivially on every class $[P]\neq 0\in
C(\mathbb{Z}C_{p})$. This shows that $\operatorname{Aut}(C_{p})$ acts non-
trivially on $C(\mathbb{Z}C_{p})$ if and only if $C(\mathbb{Z}C_{p})\cong
C(\mathbb{Z}[\zeta_{p}])$ is non-trivial. This is the case if and only if
$p\leq 23$ by [25]. Hence we have:

######  Theorem 8.3.

If $G=C_{p}$ for $p\geq 23$ prime, then there exists $[P]\in C(\mathbb{Z}G)$
and $\theta\in\operatorname{Aut}(G)$ such that $[P_{\theta}]\neq[P]\in
C(\mathbb{Z}G)/T_{G}$.

More explicitly, let $\operatorname{Aut}(C_{23})=\\{\theta_{i}:x\mapsto
x^{i}\mid i\in(\mathbb{Z}/{23})^{\times}\\}$ and recall that
$C(\mathbb{Z}C_{23})\cong\mathbb{Z}/3$ [29, p30]. We know from above that the
action of $\operatorname{Aut}(C_{23})$ on $C(\mathbb{Z}C_{23})$ is non-
trivial. Since there is a unique map

$\operatorname{Aut}(C_{23})\cong\mathbb{Z}/22\to\operatorname{Aut}(\mathbb{Z}/3)\cong\mathbb{Z}/2$

with non-trivial image, the induces map
$(\theta_{i})_{*}:C(\mathbb{Z}C_{23})\to C(\mathbb{Z}C_{23})$ has
$(\theta_{i})_{*}(1)=2$ is $i$ is odd and $(\theta_{i})_{*}(1)=1$ is $i$ is
even. Hence $|C(\mathbb{Z}C_{23})/\operatorname{Aut}(C_{23})|=2$. This could
also be checked directly by calculating the non-principal ideals in
$C(\mathbb{Z}[\zeta_{23}])$.

###  8.2. Quaternion group of order 28

Note that $C(\mathbb{Z}Q_{28})\cong\mathbb{Z}/2$ [41, Theorem III] and so,
since $\operatorname{Aut}(Q_{28})$ fixes $0$, it must act trivially on
$C(\mathbb{Z}Q_{28})$, i.e. for all $[P]\in C(\mathbb{Z}Q_{28})$ and
$\theta\in\operatorname{Aut}(Q_{28})$, we have $[P]=[P_{\theta}]$. Hence we
can define an action of $\operatorname{Aut}(Q_{28})$ on each class $[P]$ by
letting $\theta\in\operatorname{Aut}(Q_{28})$ send $P_{0}\in[P]$ to
$(P_{0})_{\theta}\in[P]$. This is the only action of the required form since
$T_{Q_{28}}=0$ [41, Theorem VI]. In contrast to our second property, we show:

######  Theorem 8.4.

There exists $[P]\in C(\mathbb{Z}Q_{28})$ such that $[P]$ has non-cancellation
but $[P]/\operatorname{Aut}(G)$ has cancellation.

Our approach will be to use the action of $\operatorname{Aut}(G)$ on Milnor
squares computed in [32, Lemma 7.6], and our computations will be similar to
the case $Q_{24}$ which was worked out in [32, Section 8].

Fix the standard presentation $Q_{28}=\langle x,y\mid
x^{7}=y^{2},yxy^{-1}=x^{-1}\rangle$. Note that $(x^{7}+1)(x^{7}-1)=x^{14}-1=0$
implies that the ideals $I=(x^{7}-1)$ and $J=(x^{7}+1)$ have that $I\cap
J=(0)$ and $I+J=(2,x^{7}-1)$. Let $\psi=x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1$ and
note also that $x^{7}+1=\psi\cdot(x+1)$ implies that the ideals
$I^{\prime}=(x+1)$ and $J^{\prime}=(\psi)$ have $I^{\prime}\cap
J^{\prime}=(x^{7}+1)$ and $I^{\prime}+J^{\prime}=(7,x+1)$.

By [7, Example 42.3], we have the following two Milnor squares where
$\Lambda=\mathbb{Z}Q_{28}/(x^{7}+1)$ and
$\mathbb{Z}D_{14}\cong\mathbb{Z}Q_{28}/(x^{7}-1)$ with $D_{14}$ the dihedral
group of order $14$.

${\mathbb{Z}Q_{28}}$${\Lambda}$${\mathbb{Z}D_{14}}$${\mathbb{F}_{2}D_{14}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{1}}$$\scriptstyle{g_{2}}$$\scriptstyle{g_{1}}$
${\Lambda}$${\mathbb{Z}Q_{28}/(\psi)}$${\mathbb{Z}Q_{28}/(x+1)}$${\mathbb{F}_{7}Q_{28}/(x+1)}$$\scriptstyle{i_{2}}$$\scriptstyle{i_{1}}$$\scriptstyle{j_{2}}$$\scriptstyle{j_{1}}$

######  Lemma 8.5.

$(f_{2})_{*}:\operatorname{Cls}(\mathbb{Z}Q_{28})\to\operatorname{Cls}(\Lambda)$
is a bijection.

###### Proof.

First note that $C(\mathbb{Z}D_{14})\cong
C(\mathbb{Z}[\zeta_{7}+\zeta_{7}^{-1}])$ [33, p328] and
$C(\mathbb{Z}[\zeta_{7}+\zeta_{7}^{-1}])\leq C(\mathbb{Z}[\zeta_{7}])=1$ [25].
Since $\mathbb{Z}D_{14}$ satisfies the Eichler condition, we have cancellation
by Theorem 2.3 and so
$\operatorname{Cls}(\mathbb{Z}D_{14})=\\{\mathbb{Z}D_{14}\\}$. By [41, Theorem
A10], there is a surjection

$((f_{1})_{*},(f_{2})_{*}):\operatorname{Cls}(\mathbb{Z}Q_{28})\twoheadrightarrow\operatorname{Cls}(\mathbb{Z}D_{14})\times\operatorname{Cls}(\Lambda)\cong\operatorname{Cls}(\Lambda)$

and, by the discussion above, the fibres are in bijection with

$\mathbb{Z}D_{14}^{\times}\backslash\mathbb{F}_{2}D_{14}^{\times}/\operatorname{Aut}(P)$

for $P\in\operatorname{Cls}(\Lambda)$. By combining [41, Theorem A18, Lemma
10.1, Lemma 10.11], we have

$\mathbb{Z}D_{14}^{\times}\backslash\mathbb{F}_{2}D_{14}^{\times}\cong\frac{K_{1}(\mathbb{F}_{2}D_{14})}{K_{1}(\mathbb{Z}D_{14})}\cong\frac{\mathbb{F}_{2}[\zeta_{7}+\zeta_{7}^{-1}]^{\times}}{\mathbb{Z}[\zeta_{7}+\zeta_{7}^{-1}]^{\times}}=1$

and so
$\mathbb{Z}D_{14}^{\times}\backslash\mathbb{F}_{2}D_{14}^{\times}/\operatorname{Aut}(P)=1$
and $(f_{2})_{*}$ is bijective. ∎

Recall that, for the standard presentation $Q_{28}=\langle x,y\mid
x^{7}=y^{2},yxy^{-1}=x^{-1}\rangle$, we have
$\operatorname{Aut}(Q_{28})=\\{\theta_{a,b}:a\in(\mathbb{Z}/{14})^{\times},b\in\mathbb{Z}/{14}\\}$
where $\theta_{a,b}(x)=x^{a}$ and $\theta_{a,b}(y)=x^{b}y$ as noted in Section
5. The following is easy to check:

######  Lemma 8.6.

Let $a\in(\mathbb{Z}/{14})^{\times}$, $b\in\mathbb{Z}/{14}$. Then there exists
$\tilde{\theta}_{a,b}\in\operatorname{Aut}(\Lambda)$ such that
$f_{2}\circ\theta_{a,b}=\tilde{\theta}_{a,b}\circ f_{2}$.

Now note that $\mathbb{Z}Q_{28}/(\psi)\cong\mathbb{Z}[\zeta_{14},j]$ and
$\mathbb{Z}Q_{28}/(x+1)\cong\mathbb{Z}[j]$. This allows us to rewrite the
second square as follows:

${\Lambda}$${\mathbb{Z}[\zeta_{14},j]}$${\mathbb{Z}[j]}$${\mathbb{F}_{7}[j]}$$\scriptstyle{i_{2}}$$\scriptstyle{i_{1}}$$\scriptstyle{j_{2}}$$\scriptstyle{j_{1}}$
${x,y}$${\zeta_{14},j}$${-1,j}$${-1,j}$

By [41, Theorem A18], we now have a commutative diagram with exact rows:

${\mathbb{Z}[j]^{\times}\backslash\mathbb{F}_{7}[j]^{\times}/\mathbb{Z}[\zeta_{14},j]^{\times}}$${\operatorname{Cls}(\Lambda)}$${\operatorname{Cls}(\mathbb{Z}[j])\times\operatorname{Cls}(\mathbb{Z}[\zeta_{14},j])}$${\frac{K_{1}(\mathbb{F}_{7}[j])}{K_{1}(\mathbb{Z}[j])\times
K_{1}(\mathbb{Z}[\zeta_{14},j])}}$${C(\Lambda)}$${C(\mathbb{Z}[j])\times
C(\mathbb{Z}[\zeta_{14},j])}$

where the left vertical map is induced by the map of units
$\mathbb{F}_{7}[j]^{\times}\to K_{1}(\mathbb{F}_{7}[j])$.

Note that $\operatorname{Cls}(\mathbb{Z}[j])=\\{\mathbb{Z}[j]\\}$ since
$\mathbb{Z}[j]$ is a PID and
$\operatorname{Cls}(\mathbb{Z}[\zeta_{14},j])=\\{\mathbb{Z}[\zeta_{14},j]\\}$
by [41, p85]. By exactness, this implies that the two inclusions are
bijections.

If $\mathcal{O}$ is a $\mathbb{Z}$-order in a finite-dimensional semisimple
$\mathbb{Q}$-algebra with centre $R$, and $\nu:K_{1}(\mathcal{O})\to
R^{\times}$ is the reduced norm, then we define
$SK_{1}(\mathcal{O})=\operatorname{Ker}(\nu)$ [23, p325].

######  Lemma 8.7.

$\frac{K_{1}(\mathbb{F}_{7}[j])}{K_{1}(\mathbb{Z}[j])\times
K_{1}(\mathbb{Z}[\zeta_{14},j])}\cong\mathbb{F}_{7}^{\times}/(\mathbb{F}_{7}^{\times})^{2}$
where the map
$K_{1}(\mathbb{F}_{7}[j])\cong\mathbb{F}_{7}[j]^{\times}\to\mathbb{F}_{7}^{\times}$
is induced by the norm $N(a+bj)=a^{2}+b^{2}$.

###### Proof.

Note that $K_{1}(\mathbb{Z}[j])\cong\mathbb{Z}[j]^{\times}=\langle j\rangle$
since $\mathbb{Z}[j]$ is a Euclidean domain and
$K_{1}(\mathbb{F}_{7}[j])\cong\mathbb{F}_{7}[j]^{\times}$ since
$\mathbb{F}_{7}[j]$ is a finite and hence semilocal ring. It follows from [23,
Lemma 7.6] that the map
$N:\mathbb{F}_{7}[j]^{\times}\to\mathbb{F}_{7}^{\times}$ has kernel
$SK_{1}(\mathbb{Z}[\zeta_{14},j])$. Since $\langle
j\rangle\leq\operatorname{Ker}(N)$, this implies that $N$ induces an
isomorphism

$\frac{K_{1}(\mathbb{F}_{7}[j])}{K_{1}(\mathbb{Z}[j])\times
K_{1}(\mathbb{Z}[\zeta_{14},j])}\cong\mathbb{F}_{7}^{\times}/\left(\frac{K_{1}(\mathbb{Z}[\zeta_{14},j])}{SK_{1}(\mathbb{Z}[\zeta_{14},j])}\right)\cong\mathbb{F}_{7}^{\times}/\operatorname{Im}(\nu)$

where
$\nu:K_{1}(\mathbb{Z}[\zeta_{14},j])\to\mathbb{Z}[\zeta_{7}+\zeta_{7}]^{\times}$
since $\mathbb{Q}[\zeta_{14},j]$ has centre
$\mathbb{Q}(\zeta_{7}+\zeta_{7}^{-1})$. By [47, Theorem 2 (ii)], we have that
$\operatorname{Im}(\nu)=(\mathbb{Z}[\zeta_{7}+\zeta_{7}^{-1}]^{\times})^{+}$,
the totally positive units. Since $|C(\mathbb{Z}[\zeta_{7}])|=1$ is odd, [41,
Corollary B25] implies that
$(\mathbb{Z}[\zeta_{7}+\zeta_{7}^{-1}]^{\times})^{+}=(\mathbb{Z}[\zeta_{7}+\zeta_{7}^{-1}]^{\times})^{2}$
and this has image $(\mathbb{F}_{7}^{\times})^{2}$ in
$\mathbb{F}_{7}^{\times}$. ∎

Since the inclusion is a bijection, this implies that

$C(\mathbb{Z}Q_{28})\cong
C(\Lambda)\cong\mathbb{F}_{7}^{\times}/(\mathbb{F}_{7}^{\times})^{2}\cong\mathbb{Z}/2$

by combining with Lemma 8.5.

######  Lemma 8.8.

$\mathbb{Z}[j]^{\times}\backslash\mathbb{F}_{7}[j]^{\times}/\mathbb{Z}[\zeta_{14},j]^{\times}=\\{[1],[1+j],[1+2j],[1+4j]\\}$.

###### Proof.

By [23, Lemma 7.5 (b)], we have
$\mathbb{Z}[\zeta_{14},j]^{\times}=\langle\mathbb{Z}[\zeta_{14}]^{\times},j\rangle$.
Now the map $\mathbb{Z}[\zeta_{14}]^{\times}\to\mathbb{F}_{7}[j]^{\times}$
maps $\zeta_{14}\mapsto 1$ and, for all $r\in\mathbb{F}_{7}^{\times}$, the
cyclotomic unit $(\zeta_{7}^{r}-1)/(\zeta_{7}-1)$ maps to
$r\in\mathbb{F}_{7}^{\times}$ and so this has image $\mathbb{F}_{7}^{\times}$.
Now $\mathbb{F}_{7}[j]=\mathbb{F}_{7}[x]/(x^{2}+1)$ and
$\left(\frac{-1}{7}\right)=-1$ implies that $x^{2}+1$ is irreducible and so
$\mathbb{F}_{7}[j]$ is a field and
$\mathbb{F}_{7}[j]^{\times}=\mathbb{F}_{7}[j]\setminus\\{0\\}$. Since
$\mathbb{Z}[j]^{\times}=\langle j\rangle$, this implies that

$\mathbb{Z}[j]^{\times}\backslash\mathbb{F}_{7}[j]^{\times}/\mathbb{Z}[\zeta_{14},j]^{\times}=\mathbb{F}_{7}[j]^{\times}/\mathbb{F}_{7}^{\times}\cdot\langle
j\rangle=(\mathbb{F}_{7}[j]\setminus\\{0\\})/\mathbb{F}_{7}^{\times}\cdot\langle
j\rangle.$

By acting by $\mathbb{F}_{7}^{\times}$ and $j$, each coset has a
representative of the form $1+aj$ for some $a\in\mathbb{F}_{7}$. If $a\neq 0$,
then acting by $j$ shows that $[1+aj]=[1-a^{-1}j]$. Now
$\\{1,2,4\\}\subseteq\mathbb{F}_{7}^{\times}$ represent the class under this
action and so we get:

$(\mathbb{F}_{7}[j]\setminus\\{0\\})/\mathbb{F}_{7}^{\times}\cdot\langle
j\rangle=\\{[1],[1+j],[1+2j],[1+4j]\\}.\qed$

By combining Lemmas 8.7 and 8.8, the map

$\operatorname{Cls}(\mathbb{Z}Q_{28})\cong\operatorname{Cls}(\Lambda)\to
C(\mathbb{Z}Q_{28})\cong C(\Lambda)$

is given by

$N:\\{[1],[1+j],[1+2j],[1+4j]\\}\to\mathbb{F}_{7}^{\times}/(\mathbb{F}_{7}^{\times})^{2}.$

Since $(\mathbb{F}_{7}^{\times})^{2}=\\{1,2,4\\}$ and $N(1)=1$, $N(1+j)=2$,
$N(1+2j)=5$, $N(1+4j)=3$, we have that

$\operatorname{Cls}^{0}(\mathbb{Z}Q_{28})\leftrightarrow\\{[1],[1+j]\\},\quad\operatorname{Cls}^{1}(\mathbb{Z}Q_{28})\leftrightarrow\\{[1+2j],[1+4j]\\}$

where we identify $C(\mathbb{Z}Q_{28})\cong\mathbb{Z}/2$.

Let $\mathcal{R}$ be the Milnor square defined above and let
$\operatorname{Aut}(\mathcal{R})$ denote the set of Milnor square
automorphisms (see [32, Section 7]).

######  Lemma 8.9.

Let $a\in(\mathbb{Z}/{14})^{\times}$, $b\in\mathbb{Z}/{14}$. Then
$\tilde{\theta}_{a,b}\in\operatorname{Aut}(\Lambda)$ extends to a Milnor
square automorphism

$\hat{\theta}_{a,b}=(\tilde{\theta}_{a,b},\theta_{a,b}^{1},\theta_{a,b}^{2},\bar{\theta}_{a,b})\in\operatorname{Aut}(\mathcal{R})$

where $\bar{\theta}_{a,b}:j\mapsto(-1)^{b}j$.

It is easy to see that $\bar{\theta}_{a,b}$ fixes $[1]$ and $[1+j]=[1-j]$ for
all $a\in(\mathbb{Z}/{14})^{\times}$, $b\in\mathbb{Z}/{14}$. However, we also
have:

$\bar{\theta}_{a,b}([1+2j])=\begin{cases}[1+2j],&\text{if $b$ is even}\\\
[1-2j]=[1+4j],&\text{if $b$ is odd}\end{cases}$

By [32, Corollary 7.7], this implies that:

$\operatorname{Cls}^{0}(\mathbb{Z}Q_{28})/\operatorname{Aut}(Q_{28})\leftrightarrow\\{[1],[1+j]\\},\quad\operatorname{Cls}^{1}(\mathbb{Z}Q_{28})/\operatorname{Aut}(Q_{28})\leftrightarrow\\{[1+2j]\\}.$

In particular, if $[P]=1\in\mathbb{Z}/2\cong C(\mathbb{Z}Q_{28})$, then $[P]$
has non-cancellation but $[P]/\operatorname{Aut}(Q_{28})$ has cancellation and
this completes the proof of Theorem 8.4.

In fact, by applying [1, Theorems 2.2 and 3.11], we can give explicit
generators and relations for all projective modules involved:

$\operatorname{Cls}^{0}(\mathbb{Z}Q_{28})\leftrightarrow\\{\mathbb{Z}Q_{28},(1+y,1+x)\\},\quad\operatorname{Cls}^{1}(\mathbb{Z}Q_{28})=\\{(1+2y,1+x),(1+4y,1+x)\\}$

so that $[(1+2y,1+x)]$ has non-cancellation but
$(1+2y,1+x)_{\theta_{1,1}}\cong(1+4y,1+x)$ and so
$[(1+2y,1+x)]/\operatorname{Aut}(Q_{28})$ has cancellation.

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