# Splitting via Noncommutativity

M. L. Lewis, D. V. Lytkina, V. D. Mazurov,
and A. R. Moghaddamfar

###### Abstract

Let $G$ be a nonabelian group and $n$ a natural number. We say that $G$ has a
strict $n$-split decomposition if it can be partitioned as the disjoint union
of an abelian subgroup $A$ and $n$ nonempty subsets
$B_{1},B_{2},\ldots,B_{n}$, such that $|B_{i}|>1$ for each $i$ and within each
set $B_{i}$, no two distinct elements commute. We show that every finite
nonabelian group has a strict $n$-split decomposition for some $n$. We
classify all finite groups $G$, up to isomorphism, which have a strict
$n$-split decomposition for $n=1,2,3$. Finally, we show that for a nonabelian
group $G$ having a strict $n$-split decomposition, the index $|G:A|$ is
bounded by some function of $n$.

Keywords: strict $n$-split decomposition, simple group, commuting graph.

footnotetext: $2010$ Mathematics Subject Classification: 20D05.

##  1 Introduction

Throughout this paper, all groups are finite. We are interested in studying
how to split up a nonabelian group into disjoint subsets where the members of
the subsets do not commute with each other. Let $G$ be a nonabelian group and
$n$ a natural number. An $n$-split decomposition of $G$ is the disjoint union
of an abelian subgroup $A$ and $n$ nonempty subsets $B_{1},B_{2},\ldots,B_{n}$
such that for each $i$ no two distinct elements of the set $B_{i}$ commute.
When $G$ has an $n$-split decomposition, we will denote the decomposition by
$G=A\uplus B_{1}\uplus B_{2}\uplus\cdots\uplus B_{n}$. We will say that the
$n$-split decomposition is strict if $|B_{i}|>1$ for each $i$ with $1\leqslant
i\leqslant n$. When $n=1$, we simply say $G=A\uplus B_{1}$ is a split
decomposition of $G$.

We note that groups having split decompositions have previously been studied
in [2], and this paper arose as a generalization suggested by Prof. Isaacs
when reading that paper. We note that $n$-split decompositions are of interest
when consider the commuting graph of a group. If $G$ is a nonabelian group,
then the commuting graph $\Delta(G)$ of $G$ is the graph whose vertex set is
$G\setminus Z(G)$ and there is an edge between $x$ and $y$ if $x$ and $y$
commute. An $n$-split decomposition of $G$ corresponds to a partition of
$\Delta(G)$ where $A\setminus Z(G)$ is a complete subgraph and the $B_{i}$’s
are disjoint independent subsets. Saying the $n$-split decomposition is strict
is equivalent to requiring each of the independent subsets to have at least
$2$-vertices.

Notice that that every group trivially has a $(|G|-1)$-decomposition by taking
$A=1$ and taking the $B_{i}$’s to be the singleton sets $\\{g\\}$ as $g$ runs
over the nonidentity elements of $G$. To eliminate this trivial decomposition,
it is reasonable to ask which groups have a strict $n$-split decomposition for
some $n$. We will prove that all nonabelian groups have a strict $n$-split
decomposition for some positive integer $n$ and some abelian subgroup $A$.

###### Theorem 1.1

If $G$ is a nonabelian group, then there is an abelian subgroup $A$ and a
positive integer $n$ so that $G$ has a strict $n$-split decomposition with
respect to $A$.

We will see that usually there is no unique (strict) $n$-split decomposition
for a given group. Moreover, we will see that for most groups that have a
(strict) $n$-split decomposition will also have a (strict) $(n+1)$-split
decomposition. The question arises, given a group $G$, what is the smallest
integer $n$ so that $G$ has a (strict) $n$-split decomposition. For some
classes of groups, we are able to answer this question.

###### Theorem 1.2

For each of the following groups, $n$ is the minimal integer so that $G$ has a
(strict) $n$-split decomposition.

  1. 1.

If $G$ is a Frobenius group with abelian Frobenius kernel $A$ and an abelian
Frobenius complement, then $n=|G:A|-1$.

  2. 2.

If $G=L_{2}(2^{m})$, then $n=2^{m}$.

  3. 3.

If $p$ is an odd prime, $q$ is a power of $p$, and $G=L_{2}(q)$, then $n=q-1$.

  4. 4.

If $p$ is an odd prime, $q$ is a power of $p$, and $G={\rm PGL}_{2}(q)$, then
$n=q$.

  5. 5.

If $G={\rm Sz}(2^{2m+1})$, $m\geqslant 1$, then $n=2^{2m+2}-1$.

For small values of $n$, we are able to obtain the structure of groups $G$
which have a strict $n$-split decomposition. In particular, we present the
classification for $n=1,2,3$. Notice that Theorem 1.2 (1) shows that it is not
possible to bound $|A|$ in terms of $n$ when $G$ has an $n$-split
decomposition. On the other hand, the classifications for $n=1,2,3$ suggest
that it may be possible to bound $|G:A|$ in terms of $n$ when $G$ has an
$n$-split decomposition. In our final theorem, we shall show that this is in
fact true.

###### Theorem 1.3

There exists a positive integer valued function $f$ defined on the positive
integers so that if $G$ has an $n$-split decomposition, then $|G:A|\leqslant
f(n)$.

##  2 The existence of a strict $n$-split decomposition

In this section, we consider the problem of the existence of a strict
$n$-split decomposition for each nonabelian group. We begin with the following
observation about when elements commute.

###### Lemma 2.1

Let $A\subset G$ be an abelian subgroup and fix elements $a_{1},a_{2}\in A$
and $g\in G$. Then $a_{1}g$ and $a_{2}g$ commute if and only if
$C_{A}(g)a_{1}=C_{A}(g)a_{2}$.

Proof. This holds by direct calculations. In fact, we have

$\begin{array}[]{lll}a_{1}ga_{2}g=a_{2}ga_{1}g&\Longleftrightarrow&a_{1}ga_{2}=a_{2}ga_{1}\\\\[5.69046pt]
&\Longleftrightarrow&a_{2}^{-1}a_{1}ga_{2}a_{1}^{-1}=g\\\\[5.69046pt]
&\Longleftrightarrow&(a_{2}a_{1}^{-1})^{-1}ga_{2}a_{1}^{-1}=g\\\\[5.69046pt]
&\Longleftrightarrow&a_{2}a_{1}^{-1}\in C_{A}(g)\\\\[5.69046pt]
&\Longleftrightarrow&C_{A}(g)a_{1}=C_{A}(g)a_{2}.\end{array}$

The result follows. $\square$

We next show how to partition the cosets of an abelian group into noncommuting
sets. A set is noncommuting if no pair of distinct elements in the set
commute. Similarly, a set is commuting if all pairs of elements commute.

###### Lemma 2.2

Let $A\subset G$ be an abelian subgroup, and fix an element $g\in G$. Let
$n=|C_{A}(g)|$. Then there exist noncommuting subsets $B_{1},\ldots,B_{n}$
such that $Ag=\uplus_{i=1}^{n}B_{i}$, with $|B_{i}|=|A:C_{A}(g)|$ for all
$i=1,\dots,n$.

Proof. We write $C_{A}(g)=\\{c_{1},\ldots,c_{n}\\}$. Let
$\\{a_{1},\ldots,a_{l}\\}$ be a transversal for $C_{A}(g)$ in $A$. So, we have

$A=\left\\{c_{i}a_{j}\mid i=1,\ldots,n;\ j=1,\ldots,l\right\\}.$

For $i=1,\ldots,n$, set

$B_{i}=\\{c_{i}a_{1}g,\ c_{i}a_{2}g,\ \ldots,\ c_{i}a_{l}g\\}.$

Then, clearly $Ag=\uplus_{i=1}^{n}B_{i}$. Moreover, when, $j_{1}\neq j_{2}$,
it follows that $c_{i}a_{j_{1}}$ and $c_{i}a_{j_{2}}$ lie in different cosets
of $C_{A}(g)$, so $c_{i}a_{j_{1}}g$ and $c_{i}a_{j_{2}}g$ do not commute by
Lemma 2.1. Therefore, we conclude $B_{i}$ is a noncommuting set and
$|B_{i}|=l=|A:C_{A}(g)|$. This completes the proof. $\square$

We are now able to prove that every nonabelian group has a strict $n$-split
decomposition with respect to $A$ for some integer $n$ and some abelian
subgroup $A$.

###### Theorem 2.3

If $G$ is a nonabelian group, then there exists an abelian subgroup $A$ of $G$
and a positive integer $n$ so that $G$ has a strict $n$-split decomposition
with respect to $A$.

Proof. Let $A$ be a maximal abelian subgroup of $G$. It follows that for every
element $g\in G\setminus A$, the group $\langle A,g\rangle$ is not abelian,
and so $C_{A}(g)<A$. Let $\\{g_{1},g_{2},\ldots,g_{m}\\}$ be a transversal for
$A$ in $G$ with $g_{1}\in A$. Clearly, for every value $i$ with $1<i\leqslant
m$, we have $|A:C_{A}(g_{i})|>1$. Furthermore, by Lemma 2.2, for each integer
$i$ with $1<i\leqslant m$, there exist noncommuting subsets
$B_{i,1},\ldots,B_{i,l_{i}}$ so that $Ag_{i}=\uplus_{j=1}^{l_{i}}B_{i,j}$,
where $l_{i}=|C_{A}(g_{i})|$ and $|B_{i,j}|=|A:C_{A}(g_{i})|>1$. It follows
that

$G=A\uplus\biguplus_{i=2}^{m}\biguplus_{j=1}^{l_{i}}B_{i,j}.$

This shows that $G$ has a strict $n$-split decomposition where
$n=\sum_{i=2}^{m}l_{i}$. $\square$

We make several observations here. First, note that our proof can be modified
to say: Let $A$ be a maximal abelian subgroup of $G$, then there is a positive
integer $n$ so that $G$ has a strict $n$-split decomposition with respect to
$A$. We will see that $G$ may also have strict split decompositions with
respect to abelian subgroups that are not maximal. See for example, Lemma 4.4
(4) where we have a strict $2$-split decomposition for $S_{3}$ with the
trivial abelian subgroup which is definitely not a maximal abelian subgroup.
We will also see that there are abelian subgroups for which the group will
have no strict split decompositions. In particular, in Lemma 3.1 (5), we see
that no proper subgroup of the center of a group can have a strict
$n$-decomposition.

Second, note in the proof of Theorem 2.3 that the $n$-split decomposition we
obtain has at least one $B_{i}$ for each coset of $A$ in $G$. This shows that
$G$ will always have an $n$-split decomposition with $n\geqslant|G:A|-1$ for
every maximal abelian subgroup of $G$. This shows that $G$ will always have a
strict $n$-split decomposition where $|G:A|$ is bounded by $n+1$. On the other
hand, the $n$-split decompositions found in this theorem likely do not have
the smallest $n$ that will work. We will see that it is possible for $|G:A|$
to be larger than $n+1$. See Lemma 4.4 (3) and (4) for examples.

##  3 Preliminary Results

If $G$ is a group and $g\in G$, then we write $o(g)$ for the order of $g$. We
start with the following general result:

###### Lemma 3.1

Suppose a group $G$ has an $n$-split decomposition with abelian subgroup $A$.
Then, the following hold:

  1. 1.

If $U$ is an abelian subgroup of $G$ not contained in $A$, then

$|U\cap A|(|U:U\cap A|-1)\leqslant n.$

In particular, $|U\cap A|\leqslant n$, $|U:U\cap A|\leqslant n+1$, and
$|U|\leqslant 2n$.

  2. 2.

If $b\in G\setminus A$ with $\langle b\rangle\cap A=1$, then
$(o(b)-1)|C_{A}(b)|\leqslant n$.

  3. 3.

If $b\in G\setminus A$, then $|C_{A}(b)|\leqslant n$ and $o(b)\leqslant 2n$.

  4. 4.

If $A$ is normal in $G$ and $b\in G\setminus A$, then $o(Ab)\leqslant n+1$.

  5. 5.

If the $n$-split decomposition is strict, then $Z(G)\leqslant A$ and
$|Z(G)|\leqslant n$.

Proof. Let $G=A\uplus B_{1}\uplus B_{2}\uplus\cdots\uplus B_{n}$ be the
$n$-split decomposition of $G$. Suppose that $U$ is an abelian subgroup of $G$
not contained in $A$. Observe that $U\setminus U\cap A$ is a commuting subset
of $G\setminus A$. Hence, each element of $U\setminus U\cap A$ must lie in a
distinct $B_{i}$. This implies that $|U\setminus U\cap A|\leqslant n$. Notice
that $|U\setminus U\cap A|=|U|-|U\cap A|=|U\cap A|(|U:U\cap A|-1)$. This
implies that $|U\cap A|(|U:U\cap A|-1)\leqslant n$. Since $U$ is not contained
in $A$, we see that $(|U:U\cap A|-1)\geqslant 1$, and so, $|U\cap A|\leqslant
n$. Similarly, as $|U\cap A|\geqslant 1$, we have that $(|U:U\cap
A|-1)\leqslant n$ and so, $|U:U\cap A|\leqslant n+1$. We have $|U|=|U\setminus
U\cap A|+|U\cap A|\leqslant n+n=2n$.

Suppose $b\in G\setminus A$. Observe that $C=C_{A}(b)\langle b\rangle$ is an
abelian subgroup not contained in $A$ and $C\cap A=C_{A}(b)$. The results of
(2) and (3) now follow from (1). Note that if $A$ is normal in $G$, then
$o(Ab)=|A\langle b\rangle/A|=|\langle b\rangle:\langle b\rangle\cap A|$, so
(4) also follows from (1).

If we assume that the $n$-split decomposition is strict, then using the fact
that $|B_{i}|>1$, it follows that $B_{i}$ does not contain any central element
of $G$, and so $Z(G)\subseteq A$. It follows that $Z(G)\leqslant C_{A}(b)$ for
each $b\in G\setminus A$. We now have $|Z(G)|\leqslant|C_{A}(b)|\leqslant n$.
$\square$

###### Lemma 3.2

Suppose a group $G$ has an $n$-split decomposition with abelian subgroup $A$,
$p$ is a prime divisor of $|G|$, and $P$ is a $p$-subgroup of $G$. If either
(1) $p\geqslant n+2$ or (2) $p^{2}>2n$ and $|P|\geqslant p^{2}$, then
$P\leqslant A$.

Proof. Let $b\in G$ be an element of order $p$. If $b$ is not contained in
$A$, then $\langle b\rangle\cap A=1$. By Lemma 3.1 (2), we have $o(b)\leqslant
n+1$. Thus, if $p\geqslant n+2$, then we must have $b\in A$, and if $P$ is a
$p$-group, then all of elements lie in $A$, so $P\leqslant A$. We now assume
that $p\leqslant n+1$ and $p^{2}>2n$ and $|P|\geqslant p^{2}$. Suppose that
$P$ is not contained in $A$. Thus, there is an element $b\in P\setminus A$. By
Lemma 3.1 (2), we have that $o(b)\leqslant n+1\leqslant 2n<p^{2}$. This
implies that $b$ has order $p$. If $b\in Z(P)$, then this implies $\langle
b\rangle<C_{P}(b)$. On the other hand, we know $1\neq Z(P)\leqslant C_{P}(b)$,
so if $b\not\in Z(P)$, then $\langle b\rangle<C_{P}(b)$. Let $c$ be an element
of $C_{P}(b)$ that does not lie in $\langle b\rangle$, and let $U=\langle
b,c\rangle$. Note that $U$ is an abelian subgroup of $G$ that is not contained
in $A$ and has order at least $p^{2}$. This is a contradiction to Lemma 3.1
(1), and hence, $P\leqslant A$. $\square$

We show that the bound obtained in Lemma 3.1 (1) cannot be improved.

###### Lemma 3.3

If $G$ is an extra-special group of order $p^{3}$ for some prime $p$, then $G$
has a (strict) $p(p-1)$-split decomposition and every $n$-split decomposition
of $G$ satisfies $n\geqslant p(p-1)$.

Proof. Observe that the centralizer $C_{G}(x)$ of every noncentral element $x$
of $G$ is a maximal subgroup of $G$ of order $p^{2}$ and is abelian. In fact,
$G$ has $p+1$ abelian subgroups of order $p^{2}$. Suppose that $G$ has an
$n$-split decomposition with abelian subgroup $A$, and let $U$ be an abelian
subgroup of order $p^{2}$ that does not equal $A$. It follows that either
$|U\cap A|=|U:U\cap A|=p$ or $|U\cap A|=1$ and $|U:U\cap A|=p^{2}$. We now
apply Lemma 3.1 (1). In the first case, we obtain $n\geqslant|U\cap
A|(|U:U\cap A|-1)=p(p-1)$, and in the second case, we have $n\geqslant|U:U\cap
A|-1=p^{2}-1$. In either case, we have $n\geqslant p^{2}-p$.

We now prove that $G$ has an $(p^{2}-p)$-split decomposition. Let $A$ be an
abelian subgroup of order $p^{2}$, and fix an element $g\in G\setminus A$. We
know that $g,g^{2},\dots,g^{p}$ forms a transversal for $A$ in $G$; moreover,
$g^{i}\not\in A$ for $1\leqslant i\leqslant p-1$. It follows that
$C_{A}(g^{i})=Z(G)$ has order $p$. We now apply Lemma 2.2 to obtain
noncommuting sets $B_{i,1},\dots,B_{i,p}$ so that
$Ag^{i}=B_{i,1}\uplus\cdots\uplus B_{i,p}$ and note that $|B_{i,j}|=p$. We now
obtain $G=A\uplus B_{1,1}\uplus\cdots\uplus B_{p-1,p}$ is a strict
$(p^{2}-p)$-split decomposition of $G$. $\square$

Here we present some examples of nonabelian groups $G$ having a $n$-split
decomposition, for some natural number $n$. The first example concerns a
Frobenius group $G$ with an abelian kernel $A$.

###### Lemma 3.4

If $G$ is a Frobenius group with abelian Frobenius kernel $A$, then the cosets
of $A$ form a strict $(|G:A|-1)$-split decomposition of $G$ with respect to
$A$. If in addition, $G$ has an abelian Frobenius complement, then any
$n$-split decomposition of $G$ with respect to $A$ has $n\geqslant|G:A|-1$.

Proof. It is obvious that $G$ is the disjoint union of the cosets of $A$. If
$B$ is a coset of $A$ that does not equal $A$ and $b\in B$, then we know that
$C_{G}(b)$ is contained in a Frobenius complement of $G$, and so,
$C_{G}(b)\cap B=\\{b\\}$. It follows that $b$ does not commute with any other
element of $B$. This proves that the cosets of $A$ form a strict
$(|G:A|-1)$-split decomposition of $G$ with respect to $A$. Suppose $G$ has an
$n$-split decomposition with respect to $A$ and that $U$ is an abelian
Frobenius complement of $G$. We know that $|U\cap A|=1$, so $|U|=|U:U\cap A|$,
and by Lemma 3.1 (1), we have $|U|\leqslant n+1$. Since $|U|=|G:A|$, we
conclude that $|G:A|-1\leqslant n$. $\square$

In particular, a Frobenius group of order $2m$, where $m$ is odd and the
kernel is abelian of order $m$, has a split decomposition over the Frobenius
kernel.

The following result is proved in [2], and states that these are the only
possible split decompositions. We provide a direct group-theoretic proof for
the sake of completeness, which was inspired by Isaacs.

###### Corollary 3.5

The following conditions on a nonabelian group $G$ are equivalent:

  * $(1)$

$G$ has a split decomposition with respect to an abelian subgroup $A$.

  * $(2)$

$G$ is a Frobenius group of order $2n$, where $n$ is odd, and the Frobenius
kernel is abelian of order $n$ and is $A$.

Proof. $(1)\Rightarrow(2).$ Suppose $G=A\uplus B$ is a split decomposition of
$G$. Let $b$ in $B$. Then $(o(b)-1)|C_{A}(b)|\leqslant 1$ by Lemma 3.1 (2).
This implies that $o(b)=2$ and $C_{A}(b)=1$. Let $P$ be a Sylow $2$-subgroup
of $G$. If $|P|\geqslant 4$, then by Lemma 3.2, we would have $P\leqslant A$
which is a contradiction. Thus, $|P|=2$, and so $P\cap A=1$. Now, we have that
$|A|$ is odd, $|G:A|=2$, and every element outside $A$ has order $2$. This
implies that $G$ is a Frobenius group with abelian Frobenius kernel $A$.

$(2)\Rightarrow(1).$ This is Lemma 3.4 when $|G:A|=2$. $\square$

##  4 Groups having a strict $2$-split decomposition

We now want to understand groups with an $n$-split decomposition where $n>1$.
We begin with the following observation that shows that having an $n$-split
decomposition with respect to $A$ in most cases will yield an $(n+1)$-split
decomposition for the same group $A$. The following lemma follows easily from
the definition.

###### Lemma 4.1

Suppose the group $G$ has a strict $n$-split decomposition $G=A\uplus
B_{1}\uplus B_{2}\uplus\cdots\uplus B_{n}$, and at least one of the following
occurs:

  * (a)

there exists an integer $i$ with $1\leqslant i\leqslant n$ such that
$|B_{i}|\geqslant 4$.

  * (b)

there exist integers $1\leqslant i<j\leqslant n$ with $|B_{i}|\geqslant 3$ and
$|B_{j}|\geqslant 3$, and two elements $x\in B_{i}$ and $y\in B_{j}$ such that
$xy\neq yx$.

Then $G$ has a strict $(n+1)$-split decomposition.

Hence, a group usually will have strict $n$-split decompositions for different
values of $n$. Notice that, a group $G$ can have more than one (strict)
$n$-split decomposition for a fixed $n$. In particular, we have the following
$2$-split decompositions for $S_{3}$: $A=\\{1,(12)\\}$,
$B_{1}=\\{(13),(123)\\}$, $B_{2}=\\{(23),(132)\\}$ and $A=\\{1\\}$,
$B_{1}=\\{(12),(13),(123)\\}$, $B_{2}=\\{(23),(132)\\}$.

We now work to classify the groups having a $2$-split decomposition. Lemma 3.4
implies that the Frobenius groups with an abelian kernel of index $3$ have a
$2$-split decomposition for the Frobenius kernel. Applying Lemma 4.1 and
Corollary 3.5, we see that every Frobenius group of order $2n$ with an abelian
Frobenius kernel of order $n$ except for $D_{6}\cong S_{3}$, also has a
$2$-split decomposition. Note that $S_{3}$ cannot have a strict $2$-split
decomposition over its Frobenius kernel since there are only three elements
outside the Frobenius kernel. On the other hand, the previous paragraph shows
that $S_{3}$ does have strict $2$-split decompositions over other subgroups
$A$. The next result generalizes the fact that Frobenius groups whose
Frobenius kernels have index $2$ and order at least $5$ have a strict
$2$-decompositions.

###### Lemma 4.2

Let $G$ be a nonabelian group of order greater than $6$ containing an abelian
subgroup $A$ of index $2$. Then $G$ has a (strict) $2$-split decomposition for
$A$ if and only if $|Z(G)|\leqslant 2$.

Proof. $(\Rightarrow)$ This follows immediately from Lemma 3.1 (5).

$(\Leftarrow)$ Suppose first that $Z(G)=1$. Then $G$ is a Frobenius group with
kernel $A$ of odd order $\geqslant 5$. We know $G$ has a split decomposition
by Lemma 3.4 and also a $2$-split decomposition by Lemma 4.1(a). So suppose
that $|Z(G)|=2$. Let $t$ be an arbitrary element in $G\setminus A$. Then
$C_{A}(t)=Z(G)$. By Lemma 2.2, there exist noncommuting sets $B_{1}$ and
$B_{2}$ so that $At=B_{1}\uplus B_{2}$. We conclude that $G=A\uplus
B_{1}\uplus B_{2}$ is a $2$-split decomposition of $G$. $\square$

In light of Lemma 4.2, dihedral, semidihedral, and (generalized) quaternion
groups of order at least $8$ have (strict) $2$-split decompositions with
respect to their abelian subgroups of index $2$. We note that the dihedral
group of order $8$ and the quaternion group of order $8$ also have strict
$2$-split decompositions over their centers.

###### Lemma 4.3

Let $G$ be a nonabelian group. Then $G$ has an abelian subgroup of index $2$
and $|Z(G)|\leqslant 2$ if and only if $G$ is the semi-direct product of a
nontrivial $2$-group $P$ acting on a group $Q$ with odd order where $P$ has an
abelian subgroup $B$ of index $2$ and either (1) $|Z(P)|=2$ or (2)
$|P|\leqslant 4$ and $|Q|>1$, so that $B$ centralizes $Q$ and any element of
$P$ outside of $B$ inverts every element of $Q$. In this situation, if
$|P|\geqslant 8$, then $P$ is dihedral, semi-dihedral, quaternion, or
generalized quaternion.

Proof. Suppose that $G$ has an abelian subgroup $A$ of index $2$ and
$|Z(G)|\leqslant 2$. Let $P$ be a Sylow $2$-subgroup of $G$ and let $Q$ be the
Hall $2$-complement $O_{2^{\prime}}(A)$ of $A$, and observe that $Q$ is a
normal Hall $2$-complement of $G$. Thus, $G$ is the semi-direct product of $P$
acting on $Q$. Let $B=P\cap A$ and observe that $B$ is an abelian subgroup of
index $2$ in $P$ and centralizes $Q$. Obviously, $|P|\geqslant 2$. If
$|P|\leqslant 4$, then since $G$ is nonabelian, we must have $|Q|>1$. Thus, we
may assume that $|P|\geqslant 8$. Notice that $|Z(P)|\geqslant 2$. Observe
that both $P$ and $Q$ will centralize $B\cap Z(P)$, so $B\cap Z(P)$ is central
in $PQ=G$, and thus, $B\cap Z(P)\leqslant Z(G)$, and so, $|B\cap
Z(P)|\leqslant 2$. Notice that if $Z(P)$ is not contained in $B$, then
$P=BZ(P)$ and both $B$ and $Z(P)$ centralize $B$; so $B$ is central in $P$.
This implies that $B=B\cap Z(P)$. We deduce that $|B|\leqslant 2$ and
$|P|\leqslant 4$ which contradicts $|P|\geqslant 8$. Thus, $|Z(P)|=|B\cap
Z(P)|\leqslant 2$, and conclude that $|Z(P)|=2$. Suppose $x$ is an element of
$P$ that lies outside $B$. We know that $Z(G)\leqslant C_{G}(x)$. Observe that
$x$, $B$, and $Q$ will all centralize $C_{Q}(x)$, and so, $C_{Q}(x)\leqslant
Z(G)\cap Q$, and thus, $C_{Q}(x)=1$. This implies $xB$ acts Frobeniusly on
$Q$, and this implies that $x$ inverts every element of $Q$.

Suppose now that $|P|\geqslant 8$. Since $|Z(P)|=2$, it follows that $P$ is a
nonabelian group. If $|P|=8$, this implies that $P$ is either the dihedral
group or the quaternion group. Thus, we may assume that $|P|\geqslant 16$.
This implies that $|B|\geqslant 8$. Let $U=\\{b\in B\mid b^{4}=1\\}$. It is
not difficult to see that $U$ is a characteristic subgroup of $B$. By the
Fundamental Theorem of abelian groups, we see that $|U|\geqslant 4$ and
$|U|=4$ if and only if $B$ is cyclic. Suppose $B$ is not cyclic, so
$|U|\geqslant 8$. Fix an element $x\in P\setminus B$. Observe that $\langle
x\rangle\cap B$ and $C_{B}(x)$ will both be centralized by both $x$ and $B$
and so are central in $P$. This implies that $|\langle x\rangle\cap
B|\leqslant 2$ and so, $x$ has order at most $4$. Also, $|C_{B}(x)|=2$, so
$C_{B}(x)=Z(P)$. We can find an $x$-invariant subgroup $V$ of $U$ with
$Z(P)\leqslant V$ and $|V|=8$. Let $D=\langle x\rangle V$, and observe that
$|D|=16$. Notice that $C_{V}(x)=C_{B}(x)\cap V=Z(P)\cap V=Z(P)$. This implies
that $Z(D)=Z(P)$. We deduce that $D^{\prime}>1$, and so $Z(D)\leqslant
D^{\prime}$. If $D^{\prime}=Z(D)$, then $D$ would be extra-special which is a
contradiction since $|D:Z(D)|=8$. Thus, we must have $Z(D)<D^{\prime}$. Since
$D$ is not cyclic, we know that $|D:D^{\prime}|\geqslant 4$, and we conclude
that $|D:D^{\prime}|=4$. This implies that $D$ is either dihedral,
semidihedral, or generalized quaternion. But in this case, the only abelian
subgroup of index $2$ is cyclic since $|D|=16$. We deduce that $V$ is cyclic
which is a contradiction. Therefore, $B$ is cyclic, and hence, $P$ is
dihedral, semidihedral, or generalized quaternion (see Satz I.14.9 of [4]).

Finally, suppose that $G$ is the semi-direct product of a nontrivial $2$-group
$P$ acting on a group $Q$ with odd order where $P$ has an abelian subgroup $B$
of index $2$ and either (1) $|Z(P)|=2$ or (2) $|P|\leqslant 4$ and $|Q|>1$, so
that $B$ centralizes $Q$ and any element of $P$ outside of $B$ inverts every
element of $Q$. Let $A=B\times Q$. Observe that $|G:A|=|PQ:BQ|=|P:P\cap
BQ|=|P:B|=2$. Since every element of $P$ outside $B$ inverts all every element
of $Q$, we conclude that $Z(G)\cap Q=1$, so $Z(G)\leqslant P$. This implies
that $Z(G)\leqslant Z(P)$. If $|Z(P)|=2$, then $|Z(G)|\leqslant 2$. If
$|P|\leqslant 4$, then there exist nontrivial elements of $Q$, so the elements
of $P$ outside of $B$ cannot be central in $G$, so $Z(G)\leqslant B$, and
since $|B|\leqslant 2$, we conclude that $|Z(G)|\leqslant 2$. $\square$

We now classify the groups $G$ that have a $2$-split decomposition with
respect to a normal abelian group $A$.

###### Lemma 4.4

Let $G$ be a group. Then $G$ has a strict $2$-split decomposition for a normal
abelian subgroup $A$ if and only if one of the following occurs:

  1. 1.

$|G/A|=2$ and $|Z(G)|\leqslant 2$ and $|A|\geqslant 4$.

  2. 2.

$|G/A|=3$ and $G$ is a Frobenius group with Frobenius kernel $A$.

  3. 3.

$|G/A|=4$ and $G$ is either the quaternion group of order $8$ or the dihedral
of order $8$ (note that $A$ must be the center of $G$).

  4. 4.

$|G/A|=6$ and $G$ is $S_{3}$ with $A=1$ or $S_{4}$ and $A$ is the normal Klein
$4$-subgroup.

Proof. Suppose that $G$ has a strict $2$-split decomposition for the normal
abelian subgroup $A$. By Lemma 3.1 (4), we know that if $b\in G\setminus A$,
then $o(Ab)\leqslant 3$. This implies that every element of $G/A$ has order at
most $3$.

Assume that $G/A$ has an element of order $3$. Let $T$ be a Sylow $3$-subgroup
of $G$. By Lemma 3.2, we see that if $|T|\geqslant 9$, then $T\leqslant A$
which is a contradiction. Thus, we must have $|T|=3$. This implies that $3$
does not divide $|A|$. If $1\neq x\in T$, then by Lemma 3.1, we have
$|C_{A}(x)|\leqslant 2$. Suppose $|C_{A}(x)|=2$, and let $a$ be the
nonidentity element of $C_{A}(x)$. Then $ax$ is an element of $G\setminus A$
and $o(ax)=6$ which contradicts $o(ax)\leqslant 4$. Thus, $|C_{A}(x)|=1$. This
implies that $AT$ is a Frobenius group when $A>1$. If $G=AT$, then we must
have $A>1$ and so, we have conclusion (2). Note that if $|G/A|$ is odd, then
$G/A$ only contains elements of order $3$, so $G=AT$, and this completes the
result in this case.

Thus, we may assume that $|G/A|$ is even, and so $G/A$ contains an involution,
say $Ai$. By Lemma 3.1 (3), $|C_{A}(i)|\leqslant 2$. Thus, $H=A\langle
i\rangle$ has a normal subgroup $A$ of index $2$. If $|A|\geqslant 3$, then
$i$ is not in $Z(H)$ which implies that $Z(H)\leqslant A$. This implies that
$Z(H)\leqslant C_{A}(i)$ and hence $|Z(H)|\leqslant 2$ when $|A|\geqslant 3$.
If $G=H$, then since the split decomposition is strict, we must have
$|A|\geqslant 4$ which gives conclusion (1). Note that if $|G/A|=2$, then this
completes the result in this case.

We now assume that $|G/A|$ is even and at least $4$. Suppose that $4$ divides
$|G/A|$. Thus, there exist involutions $Ai$ and $Aj$ so that $\langle
Ai,Aj\rangle$ is a subgroup of $G/A$ of order $4$. We may assume that $i$ and
$j$ are $2$-elements. Let $D=\langle i,j\rangle$. Observe that $DA/A=\langle
Ai,Aj\rangle$, so $|D:D\cap A|=|DA/A|=4$, so by Lemma 3.1 (1), we conclude
that $D$ is not abelian. Let $Q$ be the Hall $2$-complement of $A$. Notice
that $D\cap A$ is normalized by both $D$ and $A$, so $D\cap A$ is normal in
$DA$. Notice that $DQ/(D\cap A)$ is the semidirect product of $D/(D\cap A)$
acting on $Q$. Every element of $D/(D\cap A)$ will act fixed-point freely on
$Q$, so if $Q>1$, then $DQ/(D\cap A)$ is a Frobenius group with Frobenius
complement $D/(D\cap A)$. However, $D/(D\cap A)$ is abelian and not cyclic
which is a contradiction to being a Frobenius complement. This implies that
$Q=1$ and $A$ is a $2$-group.

If $|A|\geqslant 8$, then $A$ has index $2$ in $H=A\langle i\rangle$ and
$|Z(H)|\leqslant 2$. By Lemma 4.3, we know that $H$ is dihedral, semi-
dihedral, or generalized quaternion. In these groups, the only abelian
subgroup of index $2$ is cyclic, and so, we deduce that $A$ is cyclic. Let $C$
be the subgroup of $A$ of order $4$. Since $A$ is cyclic, $C$ is
characteristic which implies that $C$ is invariant under the action of $j$. We
let $C=\langle c\rangle$. Since $i$ and $j$ do not centralize $c$, we must
have $c^{i}=c^{3}=c^{j}$. We conclude that $c^{ij^{-1}}=c$, and so, $ij^{-1}$
is element of $G$ outside of $A$ that centralizes $C$. In particular,
$|C_{A}(ij^{-1})|\geqslant 4$ which contradicts Lemma 3.1 (3). Thus, we must
have $|A|\leqslant 4$.

Suppose that $|A|=4$ which implies that $|DA|=16$. Since $DA$ is a
$2$-subgroup, we deduce that $|Z(DA)|\geqslant 2$. Observe that $DA/A$ is
abelian, so $(DA)^{\prime}\leqslant A$. If $(DA)^{\prime}=A$, then
$|DA:(DA)^{\prime}|=4$. This implies that $DA$ is dihedral, semi-dihedral, or
generalized quaternion. It follows that $DA$ contains a cyclic subgroup $C$ of
index $2$ and order $8$ in contradiction to Lemma 3.1 (1). Thus, we must have
$|(DA)^{\prime}|=2$ since we know that $D$ is nonabelian. By Lemma 3.1 (3),
$|C_{A}(i)|=|C_{A}(j)|=|C_{A}(ij)|=2$. This implies that $|Z(DA)|=2$. We know
$DA$ is nonabelian, so $(DA)^{\prime}>1$, and thus, $(DA)^{\prime}\cap
Z(DA)>1$. We see that $Z(DA)=(DA)^{\prime}$. If $g\in DA$, then
$[g^{2},h]=[g,h]^{2}=1$ for all $h\in DA$. It follows that $DA/(DA)^{\prime}$
is elementary abelian, and so, $DA$ is extra-special. This contradicts
$|DA:Z(DA)|=8$. Therefore, $|A|=2$ which implies that $DA$ has order $8$, and
thus, $DA$ is either the dihedral group or the quaternion group. If $|G/A|=4$,
then $G=DA$ and we have conclusion (3).

Continuing to assume that there exists involutions $Ai$ and $Aj$ as above,
suppose that $3$ divides $|G/A|$. Hence, there exists an element $k$ of order
$3$ in $G\setminus A$. It follows that $A\langle k\rangle$ is a cyclic
subgroup of order $6$ that is not contained in $A$ which contradicts Lemma 3.1
(1). Thus, $3$ does not divide $|G/A|$. On the other hand, suppose there is an
involution $Al$ that is not contained in $A\langle i,j\rangle$. Now,
$E=A\langle i,j,l\rangle$ is a group of order $16$. Notice that $Z(E)=A$ and
$E/A$ is abelian, so $A=E^{\prime}$. As in the previous paragraph, this
implies that $E$ is an extra-special group which is a contradiction since
$|E:A|=|E:E^{\prime}|=8$. We conclude that if $4$ divides $|G:A|$ then
$|G:A|=4$.

Finally, we know that the odd part of $|G/A|$ is at most $3$. It follows that
the remaining possibility is that $|G/A|=6$. Let $Q$ be the Hall
$2$-complement of $A$. Let $P$ be the Sylow $2$-subgroup of $A$ and let $T$ be
a Sylow $2$-subgroup. Observe that $P$ is an abelian subgroup of index $2$ in
$T$. Let $i$ be an element of $T\setminus P$, and observe that $i$ lies in
$G\setminus A$. Applying Lemma 3.1 (3), $|C_{A}(i)|\leqslant 2$ which implies
that $|C_{P}(i)|\leqslant 2$ and so, $|Z(T)|\leqslant 2$. In light of Lemma
4.3, we see that $T$ is dihedral, semi-dihedral, or generalized quaternion.
Since the only abelian subgroup of index $2$ in those groups are cyclic, we
conclude that $P$ is cyclic. We know that a cyclic $2$-group of order at least
$8$ has no odd order automorphisms. We conclude that the elements of order $3$
in $G\setminus A$ centralize $P$ which contradicts Lemma 3.1 (3). We determine
that $|P|\leqslant 4$. If $|P|=2$, then the elements of order $3$ in
$G\setminus A$ would centralize $P$. This would imply that $G\setminus A$
would contain an element of order $6$ which contradicts Lemma 3.1 (3). Thus,
we have that either $|P|=1$ or $|P|=4$. Notice that if $|P|=4$, then $P$ is a
Klein $4$-group and $T$ is the dihedral group of order $8$ in this case. Let
$R$ be a Sylow $3$-subgroup of $G$. Since $G/A$ has order $6$, we see that
$AR/A$ is normal in $G/A$. This implies that $PR$ is normal in $G$, and so,
$RT=R(PT)$ is a subgroup of $G$. If $P=1$, then since $RT$ has order $6$ and
intersects $A$ trivially, we cannot have $RT$ abelian. On the other hand, we
see that if $|P|=4$, then $C_{T}(P)=P$, and so, $C_{RT}(P)$ which implies that
$RT/P$ is isomorphic to a subgroup of the automorphism group of $P$ which is
$S_{3}$. It follows that $RT/P$ is isomorphic to $S_{3}$. Let $Q$ be the Hall
$2$-complement of $A$. If $Q>1$, note that Lemma 3.1 (3) implies that $RT/P$
acts Frobeniusly on $Q$ which is a contradiction since $S_{3}$ cannot be a
Frobenius complement. Therefore, we conclude that $Q=1$, and $G$ is either
$S_{3}$ with $A=1$ or $S_{4}$ with $A$ being the normal Klein $4$-subgroup.

Conversely, if $G$ has a normal subgroup $A$ of index $2$ with
$|Z(G)|\leqslant 2$, then Lemma 4.2 implies that $G$ has a $2$-split
decomposition. If $G$ is a Frobenius group with abelian Frobenius complement
$A$ of index $3$, then Lemma 3.4 may be used to show that $G$ has a $2$-split
decomposition with respect to $A$. We have provided above, $2$-split
decompositions for $G$ with respect to $A$ when $G$ is either the dihedral
group or the quaternion group of order $8$ and $A$ is the center of $G$, and
when $G$ is $S_{3}$ and $A=1$. Finally, when $G$ is $S_{4}$ and $A$ is the
normal Klein $4$-subgroup, we have the $2$-split decomposition:

$\begin{array}[]{lcl}A&=&\\{(1),(12)(34),(13)(24),(14)(23)\\}\\\\[2.84544pt]
B_{1}&=&\\{(123),(124),(134),(234),(1234),(1243),(1324),(12),(13),(14)\\}\\\\[2.84544pt]
B_{2}&=&\\{(132),(142),(143),(243),(1432),(1342),(1423),(34),(24),(23)\\}.\\\\[2.84544pt]
\end{array}$

The lemma is proved. $\square$

Next, we classify the groups $G$ that have a strict $2$-split decomposition
with respect to a nonnormal abelian subgroup $A$.

###### Lemma 4.5

Let $G$ be a nonabelian group. Then $G$ has a strict $2$-split decomposition
with respect to a nonnormal abelian group $A$ if and only if one the following
occurs:

  1. 1.

$G\cong S_{3}$ with $|G:A|=3$. I.e., $A$ is a nonnormal subgroup of order $2$.

  2. 2.

$G\cong A_{4}$ with $|G:A|=6$. I.e. $A$ is a nonnormal subgroup of order $2$.

Proof. First of all, for every abelian subgroup $U$ of $G$, either $U\leqslant
A$ or $|U|\leqslant 4$ by Lemma 3.1 (1). Since $A$ is a nonnormal subgroup of
$G$, we have $A^{g}\nleqslant A$, for some $g\in G$, which is an abelian
subgroup of $G$. Since $A^{g}$ is not contained in $A$, we obtain
$|A|=|A^{g}|\leqslant 4$. We conclude that all abelian subgroups of $G$ have
order at most $4$. By Lemma 3.1 (3), the elements outside $A$ have orders at
most $4$. We deduce that $G$ is a $\\{2,3\\}$-group. Since every group of
order $16$ contains an abelian subgroup of order $8$ and every group of order
9 is abelian, this forces $|G|=2^{a}\cdot 3^{b}$ where $a\leqslant 3$ and
$b\leqslant 1$. Now, it is easy to check that $G$ is isomorphic to one of the
following groups: $S_{3}$, $D_{8}$, $Q_{8}$, $A_{4}$, or $S_{4}$. In $S_{3}$,
if $A$ is nonnormal, then $|G:A|=3$. For $A_{4}$, notice that the nonnormal
subgroups have order $2$ or $3$. Notice that for subgroups of order $3$, the
existence of the Klein $4$-subgroup and Lemma 3.1 (1) would imply that
$n\geqslant 3$ which is a contradiction. Thus, we must have $|A|=2$ which
yields $|G:A|=6$. For $S_{4}$, we can find an abelian subgroup $U$ of order
$4$ that intersects $A$ trivially. To see this, observe that if $|A|=3$, then
we can take $U$ to be any abelian subgroup of order $4$. If $A$ intersects the
Klein $4$-subgroup trivially, take $U$ to be the Klein $4$-subgroup.
Otherwise, $A$ will intersect the Klein $4$-subgroup in a subgroup of order
$2$. Take $U$ to be a cyclic subgroup of order $4$ that intersects the Klein-
subgroup in a different subgroup of order $2$, and it follows that $U$ and $A$
will intersect trivially. This implies by Lemma 3.1 (1) that any $n$-split
decomposition for $A$ must have $n\geqslant 3$. To see that $D_{8}$ and
$Q_{8}$ cannot have an $n$-split decomposition with $A$ nonnormal, observe
that since the decomposition is strict, we have $Z(G)\leqslant A$, and this
implies $A$ is normal when $G$ is $D_{8}$ or $Q_{8}$.

Conversely, suppose that $G$ is one of the groups given and $A$ has the given
index. We now show that these groups have a $2$-split decomposition for the
given one of the possible $A$’s. In each case, we can obtain
$2$-decompositions for the other possible $A$’s by conjugating. When
$G=S_{3}$, we take

$A=\\{1,(12)\\},\ \ B_{1}=\\{(13),(123)\\},\ \ B_{2}=\\{(23),(132)\\}.$

For $G=A_{4}$ and $|A|=2$, we present the $2$-split decomposition:

$\begin{array}[]{lcl}A&=&\\{(1),(12)(34)\\};\\\\[2.84544pt]
B_{1}&=&\\{(13)(24),(123),(124),(134),(234)\\},\\\\[2.84544pt]
B_{2}&=&\\{(14)(23),(132),(142),(143),(243)\\}.\\\\[2.84544pt] \end{array}$

The lemma is proved. $\square$

##  5 Groups having a strict $3$-split decomposition

We now work to determine which groups have a strict $3$-split decomposition.

###### Lemma 5.1

If $G$ has a strict $2$-split decomposition over $A$, then $G$ has a strict
$3$-split decomposition over $A$ except for the following: (1) $G\cong S_{3}$
(with $|A|=2$), (2) $G\cong D_{8}$ or $Q_{8}$ and $|A|=4$, or (3) $G\cong
D_{10}$ (with $|A|=5$).

Proof. Let $G$ be a group having a strict $2$-split decomposition over $A$;
say $G=A\uplus B_{1}\uplus B_{2}$. If either of $B_{1}$ or $B_{2}$ contains at
least four elements, then it follows from Lemma 4.1 (a) that $G$ has a strict
$3$-split decomposition, and the result is proved. We assume, therefore, that
$|B_{1}|\leqslant 3$ and $|B_{2}|\leqslant 3$. Since $|A|$ divides
$|B_{1}\uplus B_{2}|$, we have $|A|\leqslant 6$ and so $|G|\leqslant 12$. It
is now easy to check that $G$ has a strict $3$-split decomposition, except the
cases mentioned above. $\square$

Please note that Lemma 5.1 does not imply that the groups $G$ mentioned in the
conclusion do not have strict $3$-split decompositions, just that the strict
$2$-split decompositions for the given $A$’s do not yield strict $3$-split
decompositions. Indeed, the strict $2$-split decompositions for $D_{8}$ and
$Q_{8}$ over their centers do yield strict $3$-split decompositions. On the
other hand, since $S_{3}$ only has $6$ elements, it is not possible for
$S_{3}$ to have a strict $3$-split decomposition. Let $G=D_{10}$ and let $A$
be an abelian subgroup of $G$ that does not have order $5$. Then $A$ will
intersect the subgroup of order $5$ trivially, and if $G$ has a strict
$n$-split decomposition with respect to $A$, then $n\geqslant 4$ by Lemma 3.1
(1).

###### Lemma 5.2

Suppose $G$ has a strict $3$-split decomposition of $G$ with respect to a
normal abelian subgroup $A$. Then the following are true:

  1. 1.

Every element of $G/A$ has order at most $4$.

  2. 2.

$G/A$ is a $\\{2,3\\}$-group.

  3. 3.

If $3$ divides $|G/A|$, then a Sylow $3$-subgroup of $G$ has order $3$.

Proof. Note that $G/A$ does not contain any element of order greater than or
equal to $5$ by Lemma 3.2. Also, by Lemma 3.2, we know that if $3$ divides
$|G/A|$, then a Sylow $3$-subgroup of $G$ has order $3$. $\square$

###### Lemma 5.3

Suppose the nonabelian group $G$ has a strict $3$-split decomposition of $G$
with respect to a normal abelian subgroup $A$ of odd order.

  1. 1.

If a Sylow $2$-subgroup $S$ of $G$ is isomorphic to $D_{8}$, then one of the
following occurs:

    1. (a)

$A=1$ and $G\cong S_{4}$

    2. (b)

$G=AS$, such that $S=\langle x,y\mid x^{4}=y^{2}=x^{y}x=1\rangle(=D_{8})$ and
$A=\langle u,v\mid u^{3}=v^{3}=[u,v]=1\rangle,$ where $u,v$ can be chosen such
that $u^{x}=v$, $v^{x}=u^{-1}$, $u^{y}=u$, $v^{y}=v^{-1}$.

  2. 2.

If a Sylow $2$-subgroup $S$ of $G$ is isomorphic to $Q_{8}$, then $G=AQ_{8}$
is a Frobenius group.

Furthermore, each of the three groups listed have a strict $3$-split
decomposition with respect to $A$.

Proof. Suppose $G/A$ is not a $2$-group and has a Sylow $2$-subgroup that is
nonabelian of order $8$. It follows that $G/A$ has order $24$ by Lemma 5.2. If
$G/A$ has a normal Sylow $2$-subgroup, then an element of order $3$ will
centralize the center of $S$ and this yields an element of order $6$ which is
a contradiction. Thus, a Sylow $2$-subgroup of $G/A$ is not normal, and hence
$G/A$ has three Sylow $2$-subgroups; so the action of $G/A$ on its Sylow
$2$-subgroups yields a homomorphism into $S_{3}$. It is not difficult to see
that this implies that $G/A$ is isomorphic to $S_{4}$ when $G/A$ is not a
$2$-group.

We may assume that $S$ is either $D_{8}$ or $Q_{8}$. Let $t$ be the central
involution of $S$. By Lemma 3.1 (3), we have $|C_{A}(t)|\leqslant 3$. If
$C_{A}(t)\neq 1$, then $|C_{A}(t)|=3$. Observe that $S$ acts on $C_{A}(t)$,
and if $K=C_{S}(C_{A}(t))$, then $|S/K|=2$, so $|K|=4$. Notice that
$C_{A}(t)K$ is an abelian subgroup of $G$ of order $12$ that is not contained
in $A$ and this contradicts Lemma 3.1(1) which shows that such a subgroup has
size at most $2\cdot 3=6$. Thus, $C_{A}(t)=1$ and so, $t$ inverts every
element of $A$.

Now assume that $S$ is isomorphic to $D_{8}=\langle
x,y|x^{4}=y^{2}=x^{y}x=1\rangle$ as above, and this implies that $G/A$ is
either $D_{8}$ or $S_{4}$. Suppose $A=1$, then since $G$ has a strict
$3$-split decomposition, we cannot have $G\cong D_{8}$ by Lemma 3.1 (5).
Suppose that $A>1$. The previous paragraph implies that $|C_{A}(x^{2})|=1$ and
hence $|C_{A}(x)|=1$. Since $S=D_{8}$ is not a Frobenius complement, we must
have $C_{A}(y)\neq 1$ and $|C_{A}(y)|=3$ by Lemma 3.1 (1). Recall that Lemma
3.2 implies that $3$ does not divide $|G/A|$ since $3$ divides $|A|$. We
conclude that $G/A$ is a $2$-group and $G=AD_{8}$. By Fitting’s lemma, we have
$A=C_{A}(y)\times[A,y]$ and $y$ inverts every element of $[A,y]$. Clearly,
$[A,y]\neq 1$ and $yx^{2}$ centralizes $[A,y]$. Thus, $[A,y]\leqslant
C_{A}(yx^{2})$ which implies that $|[A,y]|=3$ and $|A|=9$. Let $1\neq u\in
C_{A}(y)$ and $v=u^{x}$. Then $A=\langle u,v\rangle$, and we have

$v^{y}=u^{xy}=u^{yx^{-1}}=u^{x^{-1}}=(u^{x})^{x^{2}}=v^{x^{2}}=v^{-1},$

and $v^{x}=u^{x^{2}}=u^{-1}$.

Assume $S$ is isomorphic to $Q_{8}$. Since a Sylow $2$-subgroup of $S_{4}$ is
dihedral, we do not have $G/A\cong S_{4}$, and so, $G/A\cong Q_{8}$ which
implies that $G=AS$. Lemma 3.1 (5) implies that $A>1$. Let $t$ be the
involution in $S$. From the second paragraph, we have $|C_{A}(t)|=1$, and so,
$\langle t\rangle$ acts fixed-point-freely on $A$. This implies that $Q_{8}$
acts fixed-point-freely on $A$; hence $G=AS$ is a Frobenius group with abelian
kernel $A$.

Conversely, we now show that the three groups named have strict $3$-split
decompositions for the given $A$.

For $G=S_{4}$ and $A=1$, we obtain a $3$-split decomposition as follows:

$\begin{array}[]{ccl}B_{1}&=&\\{(12)(34),(13),(14),(1234),(1243),(123),(142),(234)\\},\\\\[2.84544pt]
B_{2}&=&\\{(13)(24),(12),(23),(1324),(1342),(132),(134),(243)\\},\\\\[2.84544pt]
B_{3}&=&\\{(14)(23),(34),(24),(1423),(1432),(124),(143)\\}.\\\\[2.84544pt]
\end{array}$

Suppose $G=AS$, such that $S=\langle x,y\mid
x^{4}=y^{2}=x^{y}x=1\rangle(=D_{8})$ and $A=\langle u,v\mid
u^{3}=v^{3}=[u,v]=1\rangle,$ where $u,v$ can be chosen such that $u^{x}=v$,
$v^{x}=u^{-1}$, $u^{y}=u$, $v^{y}=v^{-1}$. Observe that $C_{A}(y)=\langle
u\rangle$, $C_{A}(x^{2}y)=\langle v\rangle$, $C_{A}(xy)=\langle
uv^{2}\rangle$, and $C_{A}(x^{3}y)=\langle uv\rangle$. We have

$Ay=[A,y]y\cup[A,y]uy\cup[A,y]u^{2}y.$

Similarly,

$Ax^{2}y=[A,x^{2}y]x^{2}y\cup[A,x^{2}y]vx^{2}y\cup[A,x^{2}y]v^{2}x^{2}y,$
$Axy=[A,xy]xy\cup[A,xy]uv^{2}xy\cup[A,xy]u^{2}vxy,\ \ $

and

$\ \ \ \ \
Ax^{3}y=[A,x^{3}y]x^{3}y\cup[A,x^{3}y]uvx^{3}y\cup[A,x^{3}y]u^{2}v^{2}x^{3}y.$

Note first that $x^{2}$ inverts both $u$ and $v$, so $x^{2}$ inverts every
element of $A$. This implies that $A\langle x\rangle$ is a Frobenius group.
Thus, no pair of elements in any of the cosets $Ax$, $Ax^{2}$, and $Ax^{3}$
will commute with each other, and $Ax$ is contained in the conjugacy class of
$x$, $Ax^{2}$ is contained in the conjugacy class of $x^{2}$, and $Ax^{3}$ is
contained in the conjugacy class of $x^{3}$. Note that $x^{y}=x^{3}$; so
$Ax\cup Ax^{3}$ is contained in the conjugacy class of $x$. Observe that
$|{\rm cl}(x)|\geqslant 2|A|=18$. On the other hand, $\langle
x\rangle\leqslant C_{G}(x)$, and thus, $4\leqslant|C_{G}(x)|=|G|/|{\rm
cl}(x)|\leqslant 72/18=4$. We conclude that $C_{G}(x)=\langle x\rangle$ and
${\rm cl}(x)=Ax\cup Ax^{3}$. Also, we know that $|{\rm
cl}(x^{2})|\geqslant|A|=9$ and $S\leqslant C_{G}(x)$, so
$8=|S|\leqslant|C_{G}(x^{2})|=|G|/|{\rm cl}(x^{2})|\leqslant 72/9=8$. We
deduce that $C_{G}(x^{2})=S$ and ${\rm cl}(x^{2})=Ax^{2}$.

We see that $C_{G}(x)$ contains one element from each of $Ax$, $Ax^{2}$ and
$Ax^{3}$. In particular, $C_{G}(x)$ does not intersect

$[A,y]y,\ [A,xy]xy,\ [A,x^{2}y]vx^{2}y,\ {\rm or}\ [A,x^{3}y]uvx^{3}y.$

For any element $g$ of $Ax$, the centralizer of $g$ in $G$ will be conjugate
to $C_{G}(x)$, and thus, we deduce that $g$ commutes with no element in these
four cosets. It follows that no element of $Ax$ commutes with any element in
these four cosets. Similarly, no element of $Ax^{3}$ commutes with any element
in $[A,y]uy$, $[A,xy]u^{2}vxy$, $[A,x^{2}y]x^{2}y$, or $[A,x^{3}y]x^{3}y$.

Note that $[A,y]\langle y\rangle$ will be a Frobenius group, so $y$ is
conjugate to all the elements in $[A,y]y$ and no two elements in $[A,y]y$
commute with each other. Similarly, $xy$, $x^{2}y$, and $x^{3}y$
(respectively) are conjugate to all of the elements in the cosets $[A,xy]xy$,
$[A,x^{2}y]x^{2}y$, and $[A,x^{3}y]x^{3}y$ (respectively) and no two elements
in any of those cosets will commute. We see that $y^{x}=x^{2}y$, so
$[A,y]y\cup[A,x^{2}y]x^{2}y\subseteq{\rm cl}(y)$. This implies that
$|[A,y]|+|[A,x^{2}y]|=3+3=6\leqslant|{\rm cl}(y)|$. On the other hand, we know
that $\langle u\rangle\langle y,x^{2}\rangle\leqslant C_{G}(y)$. This implies
that $12\leqslant|C_{G}(y)|=|G|/|{\rm cl}(y)|\leqslant 72/6=12$. We determine
that ${\rm cl}(y)={\rm cl}(x^{2}y)=[A,y]y\cup[A,x^{2}y]x^{2}y$ and
$C_{G}(y)=\langle u\rangle\langle y,x^{2}\rangle$. In a similar fashion, one
can see that ${\rm cl}(xy)={\rm cl}(x^{3}y)=[A,xy]xy\cup[A,x^{3}y]x^{3}y$ and
$C_{g}(xy)=\langle uv^{2}\rangle\langle xy,x^{2}\rangle$.

We have now that $C_{G}(x^{2})$ consists of elements in $Ax$ $Ax^{2}$,
$Ax^{3}$, $[A,y]y$, $[A,xy]xy$, $[A,x^{2}y]x^{2}y$, $[A,x^{3}y]$ and does not
intersect $[A,y]u^{2}y$, $[A,xy]u^{2}vxy$, $[A,x^{2}y]v^{2}x^{2}y$, or
$[A,x^{3}y]u^{2}v^{2}x^{3}y$. Using conjugacy, we see that no element in
$Ax^{2}$ will commute with any element in $[A,y]u^{2}y$, $[A,xy]u^{2}vxy$,
$[A,x^{2}y]v^{2}x^{2}y$, or $[A,x^{3}y]u^{2}v^{2}x^{3}y$.

Since $u$ commutes with $y$ and the elements in $[A,y]$, we see that all the
elements in $[A,y]uy$ are conjugate as are the elements in $[A,y]u^{2}y$ and
no two elements in either of those two cosets will commute. Observe that

$(uy)^{x^{2}}=u^{2}y,\ \ (uy)^{x}=vx^{2}y\ \ {\rm and}\ \
(uy)^{x^{3}}=v^{2}x^{2}y.$

Arguing as above, we can show that

${\rm
cl}(uy)=[A,y]uy\cup[A,y]u^{2}y\cup[A,x^{2}y]vx^{2}y\cup[A,x^{2}y]v^{2}x^{2}y,$

and $C_{G}(uy)=\langle uy\rangle$ is cyclic of order $6$. Similarly, we can
obtain

${\rm
cl}(u^{2}vxy)=[A,xy]u^{2}vxy\cup[A,xy]uv^{2}xy\cup[A,x^{3}y]uvx^{3}y\cup[A,x^{3}y]u^{2}v^{2}x^{3}y,$

and $C_{G}(u^{2}vxy)=\langle u^{2}vxy\rangle$ is cyclic of order $6$.

We have that
$C_{G}(y)=\\{1,u,u^{2},y,x^{2},x^{2}y,uy,ux^{2},ux^{2}y,u^{2}y,u^{2}x^{2},u^{2}x^{2}y\\}$.
Notice that $u^{x^{2}y}=u^{-1}$, so $[A,x^{2}y]=\\{1,u,u^{2}\\}$. It follows
that $C_{G}(y)$ consists of three elements of $A$, three elements of $Ax^{2}$,
the coset $[A,x^{2}y]x^{2}y$ and one element in each of $[A,y]y$, $[A,y]uy$,
and $[A,y]u^{2}y$. So $y$ does not commute with any elements of the cosets
$[A,x^{3}y]x^{3}y$, $[A,x^{2}y]vx^{2}y$, and $[A,x^{3}y]uvx^{3}y$. Noting that
conjugating, we see that this applies to all the elements in the coset
$[A,y]y$.

Observe that $C_{G}(uy)=\\{1,uy,u^{2},y,u,u^{2}y\\}$, and so, $C_{G}(uy)$
contains three elements of $A$, and one element in each of $[A,y]y$,
$[A,y]uy$, and $[A,y]u^{2}y$. Working the same way, we can see that
$C_{G}(vuy)$ and $C_{G}(v^{2}uy)$ are composed from the same number of
elements in the same sets. Thus, we see that no element in $[A,y]uy$ commutes
with any element in the cosets $[A,x^{2}y]v^{2}x^{2}y$, $[A,xy]u^{2}vxy$, and
$[A,x^{3}y]u^{2}v^{2}x^{3}y$.

Using similar arguments, we can show that the following sets form a strict
$3$-split decomposition of $G$ with respect to $A$:

$\begin{array}[]{ccl}B_{1}&=&Ax\cup[A,y]y\cup[A,xy]xy\cup[A,x^{2}y]vx^{2}y\cup[A,x^{3}y]uvx^{3}y,\\\\[5.69046pt]
B_{2}&=&Ax^{2}\cup[A,y]uy\cup[A,xy]u^{2}vxy\cup[A,x^{2}y]v^{2}x^{2}y\cup[A,x^{3}y]u^{2}v^{2}x^{3}y,\\\\[5.69046pt]
B_{3}&=&Ax^{3}\cup[A,y]u^{2}y\cup[A,xy]uv^{2}xy\cup[A,x^{2}y]x^{2}y\cup[A,x^{3}y]x^{3}y.\\\\[2.84544pt]
\end{array}$

Finally, suppose that $G=AS$ is a Frobenius group with abelian Frobenius
kernel $A$ and Frobenius complement $Q_{8}=\\{\pm 1,\pm i,\pm j,\pm k\\}$. We
claim to obtain a strict $3$-split decomposition for $G$ with respect to $A$
by taking

$B_{1}=A(-1),\ \ B_{2}=Ai\cup Aj\cup Ak\ \ {\rm and}\ \ B_{3}=A(-i)\cup
A(-j)\cup A(-k).$

To see this, observe that since $G$ is a Frobenius group, we know that no two
elements in any coset of $A$ that is not $A$ will commute. Suppose $g\in Ai$
and $h\in Aj$. When $g$ and $h$ lie in different Frobenius complements, then
they do not commute. However, if they lie in the same Frobenius complement,
they will be two elements of order $4$ that are not inverses. In $Q_{8}$, this
implies that they do not commute. Similarly, we can show that no two elements
of either $B_{2}$ or $B_{3}$ commute proving our claim. $\square$

###### Lemma 5.4

Let $G$ be a nonabelian group, and let $S$ a Sylow $2$-subgroup of $G$. Then
$G$ has a strict $3$-split decomposition with respect to a normal abelian
subgroup $A$ of odd order if and only if one of the following holds:

  1. 1.

$|S|=1$ and $G$ is a Frobenius group, with Frobenius kernel $A$ satisfying
$|G:A|=3$.

  2. 2.

$|S|=2$ and $G$ is a Frobenius group with Frobenius kernel $A$ satisfying
$|A|\geqslant 7$ and $|G:A|=2$.

  3. 3.

$|S|=2$ and $G=Z_{3}\times F$ where $F$ is a Frobenius group with Frobenius
kernel $A\cap F$ and Frobenius complement $S$.

  4. 4.

$S$ is cyclic of order $4$ and $G$ is a Frobenius group with kernel $A$
satisfying $|G:A|=4$.

  5. 5.

$S$ is cyclic of order $4$ and $G=AS$ such that $A=[A,S]$ and
$A=C_{A}(t)\times[A,t]$, where $t$ is the involution in $S$, $|C_{A}(t)|=3$,
and $[A,t]>1$.

  6. 6.

$S$ is a Klein $4$-group, $A=1$, and $G\cong A_{4}$.

  7. 7.

$S$ is a Klein $4$-group and $G=AS$, where $A$ is elementary abelian of order
$9$ or $27$, and either $G=S_{3}\times S_{3}$ where $A=A_{3}\times A_{3}$ or
$G=AS$ satisfies $A=C_{A}(s_{1})\times C_{A}(s_{2})\times C_{A}(s_{3})$ where
$S=\\{1,s_{1},s_{2},s_{3}\\}$ and $|C_{A}(s_{i})|=3$ for $i=1,2,3$.

  8. 8.

$S\cong D_{8}$ and $G$ satisfies one of the groups in Conclusion (1) of Lemma
5.3.

  9. 9.

$S\cong Q_{8}$ and $G=AQ_{8}$ is a Frobenius group.

Proof. We suppose first that $G$ has a strict $3$-split decomposition with
respect to $A$. If $|S|=1$, then $|G|$ is odd. We deduce that $G/A$ has order
$3$ and $3$ does not divide $|A|$ by Lemma 5.2. Let $b\neq 1$ be a $3$-element
in $G\backslash A$. Then $|C_{A}(b)|\leqslant 3$ by Lemma 3.1 (3). Since $|A|$
is odd and not divisible by $3$, we conclude that $C_{A}(b)=1$. Thus
$G=A\langle b\rangle$ is a Frobenius group.

Let $|S|=2$ and consider the involution $t$ in $S$. By Lemma 3.1 (3), we know
that $|C_{A}(t)|\leqslant 3$. Since $|A|$ is odd, either $|C_{A}(t)|=1$ or
$|C_{A}(t)|=3$. If $|C_{A}(t)|=3$, then by Lemma 5.2, we see that $G/A$ is a
$2$-group. Thus, $G=AS$ and by Fitting’s lemma, we have
$A=C_{A}(t)\times[A,t]$. It is not difficult to see that
$G=C_{A}(t)\times[A,t]S$ where $C_{A}(t)\cong Z_{3}$ and $F=[A,t]S$ is a
Frobenius group with Frobenius kernel $[A,t]=F\cap A$ and Frobenius complement
$S$. We now consider the case $|C_{A}(t)|=1$. If $G=AS$, then it is not
difficult to see that $G$ is a Frobenius group with Frobenius kernel $A$ and
Frobenius complement $S$. If $G\neq AS$, then $G/A$ is a $\\{2,3\\}$-group and
not a $2$-group. Thus, $3$ divides $|G:A|$ which implies that $|G:A|=6$ and
$3$ does not divide $|A|$. We can find an element $b\in G$ of order $3$ so
that $B=\langle b,t\rangle$ is a Hall $\\{2,3\\}$-subgroup of $G$. Notice that
$B\cong G/A$, so $B$ is not cyclic. On the other hand, arguing as in the first
paragraph of this proof, we have $C_{A}(b)=1$ which would imply that $B$ is a
Frobenius complement and this is a contradiction since $B$ not abelian implies
that $B\cong S_{3}$ and $S_{3}$ cannot be a Frobenius complement. This
completes the case when $|S|=2$.

Suppose $S$ is cyclic of order $4$. Let $t$ be the involution in $S$. As in
the case $|S|=2$, one can prove that $|C_{A}(t)|$ equals either $1$ or $3$. If
$|C_{A}(t)|=3$, then $3$ does not divide $|G:A|$. This implies that $G=AS$. By
Lemma 3.1 (2), we have $(|S|-1)|C_{A}(S)|\leqslant 3$ which implies that
$|C_{A}(S)|=1$ and hence, $A=[A,S]$. By Fitting’s lemma, we have
$A=[A,t]\times C_{A}(t)$. Notice that if $[A,t]=1$, then $t$ will be central
in $G$, and this violates Lemma 3.1 (5), so we must have $[A,t]>1$.

We now assume that $C_{A}(t)=1$. This implies that $AS$ is a Frobenius group.
If $3$ divides $|G:A|$, then we can choose an element $b$ of order $3$ so that
$\langle S,b\rangle$ is a Hall $\\{2,3\\}$-subgroup of $G$. Notice that $b$
and $t$ will have to commute which gives $G/A$ an element of order $6$ which
is a contradiction. Thus, $G=AS$ as desired.

Let $S$ be an elementary abelian subgroup of order $4$. Since $S$ cannot be a
Frobenius complement, either $|A|=1$ or there exists an involution $t$ in $S$
such that $|C_{A}(t)|=3$. Suppose that $|A|=1$. Since $G$ is not abelian, we
must have $|G|=12$, and since $A_{4}$ is the only nonabelian group of order
$12$ that does not have an element of order $6$, we deduce that $G\cong
A_{4}$.

Suppose now that there exists an involution $t$ so that $|C_{A}(t)|=3$. By
Lemma 3.2, $A$ contains a Sylow $3$-subgroup of $G$ and hence $G=AS$. Let $r$
be an involution in $S$ such that $r\neq t$. If $C_{A}(r)\leqslant C_{A}(t)$
then $C_{[A,t]}(r)=1$. It follows that $[A,t]\leqslant[A,r]$. Since both $r$
and $t$ invert all the elements of $[A,t]$, it follows that $rt$ centralizes
$[A,t]$. In particular, we have $[A,t]\leqslant C_{A}(rt)$. Applying Lemma 3.1
(3) to $rt$, we obtain $|C_{A}(rt)|\leqslant 3$ so $|[A,t]|\leqslant 3$ and
$|A|\leqslant 3^{2}$. If $|A|=3$, then since $|C_{A}(t)|=3$, we have
$A=C_{A}(t)$. It follows that $t$ is central in $G$, and since the
decomposition is strict, this contradicts Lemma 3.1 (5). If $|A|=3^{2}$, then
$C_{A}(rt)=[A,t]$. Since $C_{A}(r)\leqslant C_{A}(t)$, we see that every
element of $C_{A}(r)$ is centralized by both $r$ and $t$ and so is centralized
by $rt$. Since $C_{A}(t)\cap C_{A}(rt)=1$, we deduce that $C_{A}(r)=1$. This
implies that $r$ inverts every element of $A$. It follows that $rt$ inverts
every element of $C_{A}(t)$, and so, $G=[A,t]\langle t\rangle\times
C_{A}(t)\langle rt\rangle\cong S_{3}\times S_{3}$.

We may assume that $C_{A}(t)$, $C_{A}(r)$, and $C_{A}(rt)$ are three distinct
subgroups of order $3$. Applying Fitting’s lemma, we have
$A=C_{A}(t)\times[A,t]$. Since $r$ and $t$ commute, it follows that $C_{A}(r)$
is $t$-invariant. By Fitting’s lemma, we have $C_{A}(r)=[C_{A}(r),t]\times
C_{C_{A}(r)}(t)$. Since $|C_{A}(r)|=3$ and $C_{A}(r)\neq C_{A}(t)$, we see
that $C_{A}(r)=[C_{A}(r),t]\leqslant[A,t]$. Using Fitting’s lemma once more,
we have $[A,t]=C_{A}(r)\times[[A,t],r]$. Again, both $r$ and $t$ will invert
all the elements of $[[A,t],r]$ so $rt$ centralizes every element in
$[[A,t],r]$. Notice that $rt$ will invert every element in
$C_{A}(t)C_{A}(r)=C_{A}(t)\times C_{A}(r)$. Since $C_{A}(rt)$ is nontrivial,
we must have $[[A,t],r]=C_{A}(rt)$. We conclude that $A=C_{A}(t)\times
C_{A}(r)\times C_{A}(rt)$ and $|A|=27$.

If $|S|=8$, then since $S$ is nonabelian, $S\cong Q_{8}$ or $S\cong D_{8}$ and
the conclusion follows from Lemma 5.3.

Finally, if $|S|>8$, then $S$ contains an abelian subgroup of order $8$ which
is impossible by Lemma 3.1.

Conversely, we show that if $G$ is one of the groups mentioned, then $G$ has a
strict $3$-split decomposition. If $G$ is a Frobenius group with abelian
Frobenius kernel of odd order and abelian Frobenius complement of order $2$,
$3$, or $4$, then the result follows from Lemma 3.4 and Lemma 5.1.

Suppose that $G=Z\times F$ where $Z\cong Z_{3}$ and $F$ is a Frobenius group
with abelian Frobenius kernel $B$ and Frobenius complement $S$ with $|S|=2$.
We write $Z$ for $Z\times 1$ and $F$ for $1\times F$. Notice that $Z(G)=Z\cong
Z_{3}$. Take $A=ZB$. Let $t$ be the involution in $S$ and let $z$ be a
generator of $Z$. Since $F$ is a Frobenius group, we see that the elements of
$Bt$ do not commute. Since $z$ and $z^{2}$ are central in $G$, it will follow
that the elements of $Bzt$ and $Bz^{2}t$ do not commute. It is not difficult
to see that $G=A\uplus Bt\uplus Bzt\uplus Bz^{2}t$ is a strict $3$-split
decomposition for $G$.

Next, suppose that $S$ is cyclic of order $4$ and $G=AS$ such that $A=[A,S]$
and $A=C_{A}(t)\times[A,t]$, where $t$ is the involution in $S$,
$|C_{A}(t)|=3$, and $[A,t]>1$. Let $S=\langle s\rangle$ and $C_{A}(t)=\langle
z\rangle$, and note that $t=s^{2}$. Notice that $C_{A}(s)=1$, so $C_{G}(s)=S$.
By Lemma 2.1, no two elements in $As$ commute, and in a similar fashion, no
two elements in $As^{3}$ commute. Applying Lemma 2.1 to $t=s^{2}$, we see that
no two elements of $[A,t]t$ commute. Since $z$ and $z^{2}$ centralize $t$ and
$[A,t]$, we conclude that no two elements in $[A,t]zt$ and no two elements in
$[A,t]z^{2}t$ commute. Since $A$ is partitioned by
$[A,t]\cup[A,t]z\cup[A,t]z^{2}$, it follows that the coset $At$ is partitioned
by $[A,t]t\cup[A,t]zt\cup[A,t]z^{2}t$.

The conjugacy class of $s$ in $G$ has size $|G:C_{G}(s)|=|G:S|=|A|$. On the
other hand, it is easy to see that $G^{\prime}\leqslant A$ and ${\rm
cl}(s)\subseteq G^{\prime}s\subseteq As$. We see that $|A|=|{\rm
cl}(s)|\leqslant|G^{\prime}|\leqslant|A|$. This implies that $A=G^{\prime}$
and ${\rm cl}(s)=As$. Similarly, we obtain ${\rm cl}(s^{3})=As^{3}$. We have
that $C_{G}(t)=C_{A}(t)\times S$. This implies that $|{\rm cl}(t)|=|[A,t]|$.
Since $[A,t]\langle t\rangle$ is a Frobenius group, we have
$[A,t]t\subseteq{\rm cl}(t)$, and hence, we conclude that ${\rm
cl}(t)=[A,t]t$. The centralizer of $s$ has the form $\\{1,s,s^{2},s^{3}\\}$
and every element in $As$ is conjugate to $s$. For any element in $As$, its
centralizer will consist of $1$, itself, a conjugate of $s^{2}$, and a
conjugate of $s^{3}$. Since the conjugacy classes of $s$, $s^{2}$, and $s^{3}$
are $As$, $[A,t]t$, and $As^{3}$ respectively, we conclude that no element in
$As$ will commute with any element of $[A,t]zt$. Similarly, no element in
$As^{3}$ will commute with an element of $[A,t]z^{2}t$. Thus, if we take
$B_{1}=As\cup[A,t]zt$, $B_{2}=[A,t]t$, and $B_{3}=As^{3}\cup[A,t]z^{2}t$, then
we obtain a strict $3$-split decomposition of $G$.

When $G=A_{4}$, we get a strict $3$-split decomposition with $A=1$ by taking
$B_{1}=\\{(12)(34),(123),(142),(234)\\}$,
$B_{2}=\\{(13)(24),(132),(134),(243)\\}$, and
$B_{3}=\\{(14)(23),(124),(143)\\}$.

For $G=S_{3}\times S_{3}$, we obtain a strict $3$-split decomposition for
$A=A_{3}\times A_{3}$ by taking

$\begin{array}[]{ccl}B_{1}&=&\\{((12),(1)),((13),(1)),((23),(1)),((123),(12)),((123),(13)),((123),(23))\\},\\\
B_{2}&=&\\{((12),(123)),((13),(123)),((23),(123)),((1),(12)),((1),(13)),((1),(23))\\},\\\
B_{3}&=&\\{((12),(132)),((13),(132)),((23),(132)),((132),(12)),((132),(13)),\\\
&&((132),(23)),((12),(12)),((12),(13)),((12),(23)),((13),(12)),((13),(13)),\\\
&&((13),(23)),((23),(12)),((23),(13)),((23),(23))\\}.\end{array}$

Next, suppose $S$ is a Klein $4$-group and $G=AS$, where $A$ is elementary
abelian of order $27$, and $G=AS$ satisfies $A=C_{A}(s_{1})\times
C_{A}(s_{2})\times C_{A}(s_{3})$ where $S=\\{1,s_{1},s_{2},s_{3}\\}$ and
$|C_{A}(s_{i})|=3$ for $i=1,2,3$. For $i=1,2,3$, let $C_{A}(s_{i})=\langle
a_{i}\rangle$. Notice that $s_{1}$ will normalize $C_{A}(s_{2})$ and since
$C_{A}(s_{1})\cap C_{A}(s_{2})=1$, we have that
$C_{A}(s_{2})=[C_{A}(s_{2}),s_{1}]\leqslant[A,s_{1}]$. Similarly,
$C_{A}(s_{3})\leqslant[A,s_{1}]$. By Fitting’s lemma,
$A=C_{A}(s_{1})\times[A,s_{1}]$, and so, we determine that
$[A,s_{1}]=C_{A}(s_{2})\times C_{A}(s_{3})$. Similarly, we have
$[A,s_{2}]=C_{A}(s_{1})\times C_{A}(s_{3})$ and $[A,s_{3}]=C_{A}(s_{1})\times
C_{A}(s_{2})$. By Lemma 2.1, we have that no two elements in $[A,s_{1}]s_{1}$
commute. Since $a_{1}$ commutes with $s_{1}$ and $[A,s_{1}]$, we see that no
two elements in $[A,s_{1}]a_{1}s_{1}$ will commute and no two elements in
$[A,s_{1}]a_{1}^{2}s_{1}$ will commute.

We see that $C_{G}(s_{1})=C_{A}(s_{1})S$, so the conjugacy class of $s_{1}$
has size $|G:C_{G}(s_{1})|=|A:C_{A}(s_{1})|=|[A,s_{1}]|$. Notice that
$[A,s_{1}]\langle s_{1}\rangle$ is a Frobenius group, so
$[A,s_{1}]s_{1}\leqslant{\rm cl}(s_{1})$. By consideration of sizes, we have
${\rm cl}(s_{1})=[A,s_{1}]s_{1}$. Similarly, ${\rm cl}(s_{2})=[A,s_{2}]s_{2}$
and ${\rm cl}(s_{3})=[A,s_{3}]s_{3}$. Since
$C_{A}(s_{1})\leqslant[A,s_{2}]\cap[A,s_{3}]$, we have
$C_{A}(s_{1})s_{2}\subseteq[A,s_{2}]s_{2}$ and
$C_{A}(s_{1})s_{3}\subseteq[A,s_{3}]s_{3}$. Hence, $C_{A}(s_{1})$ contains no
elements in $[A,s_{2}]a_{2}s_{2}$, $[A,s_{2}]a_{2}^{2}s_{2}$,
$[A,s_{3}]a_{3}s_{3}$, and $[A,s_{3}]a_{3}^{2}a_{3}$. Since every element in
$[A,s_{1}]s_{1}$ is conjugate to $s_{1}$, we conclude that no element in
$[A,s_{1}]s_{1}$ commutes with any element in $[A,s_{2}]a_{2}s_{2}$ or
$[A,s_{3}]a_{3}s_{3}$. Similarly, no element in $[A,s_{2}]s_{2}$ commutes with
any element in $[A,s_{1}]a_{1}s_{1}$ or $[A,s_{3}]a_{3}^{2}s_{3}$ and no
element in $[A,s_{3}]s_{3}$ commutes with any element in
$[A,s_{1}]a_{1}^{2}s_{1}$ or $[A,s_{2}]a_{2}^{2}s_{2}$.

Notice that $\langle a_{1}s_{1}\rangle\leqslant C_{G}(a_{1}s_{1})$, so
$6\leqslant|C_{G}(a_{1}s_{1})|$. Since $a_{1}$ commutes with $s_{1}$ and
$[A,s_{1}]$, we conclude that all the elements $[A,s_{1}]a_{1}s_{1}$ are
conjugate as are all the elements in $[A,s_{1}]a_{1}^{2}s_{1}$. On the hand,
since $a_{1}\in C_{A}(s_{1})\leqslant[A,s_{2}]$, we know that $s_{2}$ inverts
$a_{1}$, so $(a_{1}s_{1})^{s_{2}}=a_{1}^{s_{1}}s_{2}^{s_{1}}=a_{1}^{2}s_{1}$.
Hence, $a_{1}s_{1}$ and $a_{1}^{2}s_{1}$ are conjugate. This implies that
$[A,s_{1}]a_{1}s_{1}\cup[A,s_{1}]a_{1}^{2}s_{1}\subseteq{\rm cl}(a_{1}s_{1})$.
This implies that $18\leqslant|{\rm cl}(a_{1}s_{1})|$, and thus,
$|C_{G}(a_{1}s_{1})|\geqslant|G|/|{\rm cl}(a_{1}s_{1})|=108/18=6$. We deduce
that $C_{G}(a_{1}s_{1})=\langle a_{1}s_{1}\rangle$. Now, $C_{G}(a_{1}s_{1})$
contains no elements in $[A,s_{3}]a_{3}^{2}s_{3}$. Using conjugacy, we
conclude that no element in $[A,s_{1}]a_{1}s_{1}$ commutes with any element in
$[A,s_{3}]a_{3}^{2}s_{3}$. Similarly, no element in $[A,s_{1}]a_{1}^{2}s_{1}$
will commute with any element in $[A,s_{2}]a_{2}^{2}s_{2}$, and no element in
$[A,s_{2}]a_{2}s_{2}$ commutes with any element in $[A,s_{3}]a_{3}s_{3}$.
Hence, we have a strict $3$-split decomposition for $G$ with respect to $A$ by
taking

$B_{1}=[A,s_{1}]s_{1}\cup[A,s_{2}]a_{2}s_{2}\cup[A,s_{3}]a_{3}s_{3},$
$B_{2}=[A,s_{2}]s_{2}\cup[A,s_{1}]a_{1}s_{1}\cup[A,s_{3}]a_{3}^{2}s_{3},$

and

$B_{3}=[A,s_{3}]s_{3}\cup[A,s_{1}]a_{1}^{2}s_{1}\cup[A,s_{2}]a_{2}^{2}s_{2}.$

The remaining groups were handled in Lemma 5.3. $\square$

###### Lemma 5.5

Let $G$ be a nonabelian group with a normal abelian subgroup $A$ with even
order. Then $G$ has a strict $3$-split decomposition with respect to $A$ if
and only if one of the following holds:

  1. 1.

$G$ is a Frobenius group with Frobenius kernel $A$ and a Frobenius complement
of order $3$.

  2. 2.

$A$ has index $2$ in $G$, $|A|\geqslant 6$, and $|Z(G)|=2$.

  3. 3.

$G$ is either $D_{8}$ or $Q_{8}$ and $A=Z(G)$.

  4. 4.

$G\cong S_{4}$ and $A$ is the Klein $4$-subgroup.

Proof. We first suppose that $G$ has a strict $3$-split decomposition with
respect to $A$. Suppose $b$ is an element of order $3$ in $G\setminus A$. By
Lemma 5.2 (3), $3$ does not divide $|A|$. Moreover, Lemma 3.1 (2) implies that
$(o(b)-1)|C_{A}(b)|\leqslant 3$. Since $o(b)-1=2$, we conclude that
$|C_{A}(b)|=1$. This implies that $A\langle b\rangle$ is a Frobenius group. If
$|G:A|$ is odd, then by Lemma 5.2 (3), we have $G=A\langle b\rangle$ and
conclusion (1) holds.

Let $|G/A|$ be even and $tA$ is an involution in $G/A$ where $t$ is an
$2$-element. Observe that $C_{A}(t)\langle t\rangle$ is an abelian group. By
Lemma 3.1 (1), we have that $|C_{A}(t)|\leqslant 3$. Since $|A|$ is even, we
see that $C_{A}(t)$ is not trivial. Thus $|C_{A}(t)|=2$. If $G=A\langle
t\rangle$, then $A$ will have index $2$ and $Z(G)=C_{A}(t)$ will have order
$2$. Since the decomposition is strict, we see that $G\setminus A$ must have
three subsets each having size at least $2$, so $|A|=|G\setminus A|\geqslant
6$. This gives conclusion (2).

We now assume that $|G/A|\geqslant 4$. Let $T$ be the Sylow $2$-subgroup of
$A$. By Lemma 4.3, we know that $\langle T,t\rangle$ either has order $4$ or
is isomorphic to a dihedral, semidihedral, quaternion, or generalized
quaternion group. It follows that $T$ is cyclic or $|T|=4$.

Suppose that $|T|>2$. Let $S$ be a Sylow $2$-subgroup of $G$ and observe that
$S\cap A=T$. If $C_{S}(T)>T$, then there exists $s\in C_{S}(T)\setminus T$.
Observe that $T\langle s\rangle$ is abelian and not contained in $A$. Since
$|T|>3$, this would violate Lemma 3.1 (1). Thus, we have $C_{S}(T)=T$. Let
$T_{0}$ be a subgroup of order $4$ in $T$. If $C_{G}(T_{0})\neq A$, then for
$b\in C_{G}(T_{0})\backslash A$, $T_{0}\langle b\rangle$ is abelian so it
would also violate Lemma 3.1 (1). Thus, $C_{G}(T_{0})=A$. We know that $G/A$
is isomorphic to a subgroup of the automorphism group of $T_{0}$. Since
$|G:A|\geqslant 4$, we must have that $T_{0}$ is a Klein $4$-group and
$G/A\cong S_{3}$. Observe that $G/A$ acts fixed-point-freely on the Hall
$2$-complement $N$ of $A$. Since $S_{3}$ is not a Frobenius complement, we
have $N=1$. Notice that $T_{0}$ being not cyclic implies that $T=T_{0}$, and
so, $G\cong S_{4}$ where $A$ is the Klein $4$-subgroup, and $(4)$ holds.

Now, let $|T|=2$. Then $T\leqslant Z(G)$ and in light of the first paragraph,
we see that $G/A$ is a $2$-group. If $G/A$ contains an element of order $4$,
then $G$ contains an abelian subgroup of order $8$ that is not contained in
$A$ and this violates Lemma 3.1 (1). Thus $G/A$ is elementary abelian. Letting
$N$ be the Hall 2-complement of $A$, we see that $G/A$ acts fixed-point-freely
on $N$; so $N\neq 1$ would imply that $|G/A|=2$ and conclusion (2) holds.
Thus, we may assume that $N=1$; so $A=T=Z(G)$. This implies that $G$ is a
$2$-group. We know that if $G$ has order greater than $8$, then $G$ has an
abelian subgroup of order at least $8$ and this violates Lemma 3.1 (2). We
conclude that $G$ is nonabelian of order $8$ and conclusion (3) holds.

Conversely, notice that each of the groups mentioned has a strict $2$-split
decomposition, and so, we obtain a strict $3$-split decomposition by appealing
to Lemma 5.1. The lemma is proved. $\square$

###### Lemma 5.6

Let $G$ be a nonabelian group and let $A$ be a nonnormal abelian subgroup of
$G$. Then $G$ has a strict $3$-split decomposition with respect to $A$ if and
only if either $G\cong A_{4}$ and $A$ is a subgroup of order $2$ or $3$ or
$G\cong S_{4}$ and $A$ is a subgroup of order $2$, $3$, or $4$.

Proof. Suppose that $G$ has a strict $3$-split decomposition with respect to
$A$. Let $A_{1}=A^{g}\neq A$. Then $|A_{1}\cap A|\cdot(|A:A_{1}\cap
A|-1)\leqslant 3$ by Lemma 3.1 (1), and hence one of the following holds: (1)
$|A_{1}\cap A|=1$ and $|A:(A_{1}\cap A)|\leqslant 4$, (2) $|A_{1}\cap A|=2$
and $|A:A_{1}\cap A|\leqslant 2$, or (3) $|A_{1}\cap A|=3$ and $|A:A_{1}\cap
A|=2$. Notice that in case (1) we have $|A|=2$, $3$, or $4$, in case (2), we
have $|A|$ is either $2$ or $4$, and in case (3), we have $|A|=6$. If $x$ is
any element of $G$ outside $A$, then $o(x)\leqslant 6$ by Lemma 3.1 (3). Note
that if $o(x)=5$, then $\langle x\rangle\cap A=1$, and applying Lemma 3.1 (2)
we obtain $o(x)\leqslant 4$ which is a contradiction. Thus, every nonidentity
element of $G$ has order $2$, $3$, $4$, or $6$. Thus, $|G|=2^{a}\cdot 3^{b}$
for nonnegative integers $a,b$, and hence $G$ is solvable.

By Lemma 3.1 (1), we see that any abelian subgroup of $G$ has order at most
$6$. Since $G$ is nonabelian, we see that if $G$ were a $3$-group, then
$|G|\geqslant 27$ and $G$ would have an abelian subgroup of order $9$ which is
a contradiction. If $G$ is a $2$-group of order at least $16$, then $G$ would
have an abelian subgroup of $8$ which is a contradiction. This would force $G$
to have order $8$. By Lemma 3.1 (5), we see that $G^{\prime}=Z(G)\leqslant A$
which implies $A$ is normal, a contradiction. We conclude that $G$ is neither
a $2$-group nor a $3$-group.

Let $Q$ be a Sylow $3$-group of $G$. We have just shown that $Q$ is
nontrivial. On the other hand, if $|Q|\geqslant 9$, then $G$ will have an
abelian subgroup of order $9$, and we have seen that this is a contradiction.
Thus, we have that $|Q|=3$. Similarly, if $P$ is a Sylow $2$-subgroup of $G$,
then if $|P|\geqslant 16$, then $G$ will have an abelian subgroup of $8$ which
is not allowed. Thus, we have that $|P|\leqslant 8$. We see that the
possibilities for $|G|$ are $6$, $12$, and $24$. Since $G$ has a strict
$3$-split decomposition, there must be at least $6$ elements of $G$ outside of
$A$. This rules out $|G|=6$.

Notice that if $Q$ is normal in $G$, then $|G:QC_{P}(Q)|\leqslant 2$ since
$G/QC_{P}(Q)$ is isomorphic to a subgroup of ${\rm Aut}(Q)\cong Z_{2}$. Since
$C_{P}(Q)>1$, we have that $Z=Z(P)\cap C_{P}(Q)>1$. Notice that both $P$ and
$Q$ will centralize $Z$, so $Z$ is normal in $G$. By Lemma 3.1 (5), we know
that $Z\leqslant A$, and since $A$ is not normal, we have $Z<A$. From the
available orders for $A$, this implies that either $A=ZQ$ or $|A|=4$. Since
$A$ is not normal, we have $A\neq ZQ$, so $|A|=4$. Also, $ZQ$ is an abelian
subgroup of $G$ and $ZQ\cap A=Z$. This implies that $(|ZQ:ZQ\cap A|-1)|ZQ\cap
A|=(3-1)2=4$ which violates Lemma 3.1 (1). Thus, the case $Q$ is normal cannot
occur.

Now, we have that $Q$ is not normal. Since the number of Sylow $3$-subgroups
in $G$ is congruent to $1$ mod $3$, $Q$ must have $4$ conjugates in $G$. Let
$K$ be the kernel of the action of $G$ on the Sylow $3$-subgroups of $G$.
Observe that $G/K$ has order at least $12$ and is isomorphic to a subgroup of
$S_{4}$. If $K>1$, then since $|G|\leqslant 24$, we must have $|K|=2$ and
$|G:K|=12$. Notice that $K$ will be central in $G$, so $K\leqslant A$. Notice
that $KQ$ is an abelian subgroup. Replacing $Q$ by a conjugate if necessary,
we may assume that $QK\cap A=K$. We have $(|QK:QK\cap A|-1)|QK\cap
A|=(3-1)2=4$ which violates Lemma 3.1 (1). Thus, $K=1$. This implies that $G$
is either $A_{4}$ or $S_{4}$. If $G$ is $A_{4}$, then $A$ can have order $2$
or $3$. When $G$ is $S_{4}$, then $A$ can have order $2$, $3$, or $4$.

Note that the $2$-split decompositions for $A_{4}$ when $A$ is a subgroup of
order $2$ in the Klein $4$-subgroup that appeared in Lemma 4.5 yields strict
$3$-split decomposition. The following gives a strict $3$-split decomposition
for $A_{4}$ when $A$ is a Sylow $3$-subgroup:

$\begin{array}[]{lcl}A&=&\\{1,(123),(132)\\};\\\\[2.84544pt]
B_{1}&=&\\{(12)(34),(124),(234)\\},\\\\[2.84544pt]
B_{2}&=&\\{(13)(24),(142),(143)\\},\\\\[2.84544pt]
B_{3}&=&\\{(14)(23),(134),(243)\\}.\\\\[2.84544pt] \end{array}$

Now, we present $3$-split decompositions for $S_{4}$ with respect to various
possible $A$’s. We present an example for one representative of each of the
conjugacy classes of the possible $A$’s.

$\begin{array}[]{lcl}A&=&\\{(1),(12)(34)\\};\\\\[2.84544pt]
B_{1}&=&\\{(13)(24),(123),(134),(234),(1243),(1324),(12),(14)\\},\\\\[2.84544pt]
B_{2}&=&\\{(14)(23),(132),(142),(143),(1432),(1423),(34),(24)\\},\\\\[2.84544pt]
B_{3}&=&\\{(1234),(1342),(124),(243),(13),(23)\\}.\\\\[2.84544pt] \hline\cr
A&=&\\{(1),(24)\\};\\\\[2.84544pt]
B_{1}&=&\\{(13)(24),(123),(134),(234),(1243),(1324),(12),(14)\\},\\\\[2.84544pt]
B_{2}&=&\\{(14)(23),(132),(142),(143),(1432),(1423),(34)\\},\\\\[2.84544pt]
B_{3}&=&\\{(12)(34),(1234),(1342),(124),(243),(13),(23)\\}.\\\\[2.84544pt]
\hline\cr A&=&\\{(1),(123),(132)\\};\\\\[2.84544pt]
B_{1}&=&\\{(13)(24),(134),(234),(1243),(1324),(12),(14)\\},\\\\[2.84544pt]
B_{2}&=&\\{(14)(23),(142),(143),(1432),(1423),(34),(24)\\},\\\\[2.84544pt]
B_{3}&=&\\{(12)(34),(1234),(1342),(124),(243),(13),(23)\\}.\\\\[2.84544pt]
\hline\cr A&=&\\{(1),(12),(34),(12)(34)\\};\\\\[2.84544pt]
B_{1}&=&\\{(13)(24),(123),(134),(234),(1243),(1324),(14)\\},\\\\[2.84544pt]
B_{2}&=&\\{(14)(23),(132),(142),(143),(1432),(1423),(24)\\},\\\\[2.84544pt]
B_{3}&=&\\{(1234),(1342),(124),(243),(13),(23)\\}.\\\\[2.84544pt] \hline\cr
A&=&\\{1,(1234),(13)(24),(1432)\\};\\\\[2.84544pt]
B_{1}&=&\\{(13),(23),(123),(124),(1243),(1324)\\},\\\\[2.84544pt]
B_{2}&=&\\{(34),(14),(12)(34),(132),(243),(143),(1342)\\},\\\\[2.84544pt]
B_{3}&=&\\{(12),(24),(14)(23),(134),(234),(142),(1423)\\}.\\\\[2.84544pt]
\end{array}$

The lemma is proved. $\square$

##  6 Decompositions of $L_{2}(q)$, ${\rm Sz}(q)$, and ${\rm PGL}_{2}(q)$

Let $G$ be a nonabelian group and $\min(G)$ the minimal number $n$ for which
the group $G$ has a strict $n$-split decomposition for some abelian group $A$.
Clearly, if $H$ and $K$ are two maximal abelian subgroups of $G$ with $H\cap
K=1$, then $\min(G)\geqslant\min\\{|H|,|K|\\}-1$. In particular, if $G$ has a
maximal abelian subgroup $A$ which intersects trivially with some of its
conjugates, then $\min(G)\geqslant|A|-1$. We now investigate $\min(G)$ for two
families of groups.

In what follows, we restrict our attention to the almost simple groups (recall
that $G$ is almost simple if $S\leqslant G\leqslant{\rm Aut}(S)$ for some
nonabelian simple group $S$). We begin with the following result:

###### Lemma 6.1

Suppose that $q=p^{m}\geqslant 4$, with $p$ a prime and $m\geqslant 1$ an
integer. Then the following assertions hold.

  * $(1)$

If $p=2$ and $G=L_{2}(q)\cong{\rm PGL}(2,q)$, then $\min(G)=q$.

  * $(2)$

When $p$ is an odd prime, then $\min(G)=q-1$ if $G=L_{2}(q)$, and $\min(G)=q$,
if $G={\rm PGL}(2,q)$.

In particular, we have the following:

  * •

If $S\cong A_{5}\cong L_{2}(4)\cong L_{2}(5)$, then $\min(S)=4$.

  * •

If $S\cong L_{3}(2)\cong L_{2}(7)$, then $\min(S)=6$.

  * •

If $S\cong A_{6}\cong L_{2}(9)$, then $\min(S)=8$.

  * •

If $S\cong S_{5}\cong{\rm PGL}_{2}(5)$, then $\min(S)=5$.

Proof. It is known that $G$ contains abelian subgroups $C$, $D$, $F$, of
orders $(q-1)/k$, $q$, and $(q+1)/k$, respectively where $k=1$ if either $p=2$
or $G={\rm PGL}(2,q)$ when $p$ is odd and $k=2$ if $G=L_{2}(q)$ when $p$ is
odd; and every two distinct conjugates of any of these groups intersect
trivially. Furthermore, every element of $G$ is a conjugate of an element in
$C\cup D\cup F$ (see Satz II.8.5 of [4]). Label the elements
$C=\\{1,c_{1},\ldots,c_{(q-1)/k-1}\\}$, $D=\\{1,d_{1},\ldots,d_{q-1}\\}$, and
$F=\\{1,f_{1},\ldots,f_{(q+1)/k-1}\\}$. Let $\\{a_{1},\dots,a_{r}\\}$ be a
transversal for $N_{C}=N_{G}(C)$ in $G$, let $\\{b_{1},\dots,b_{s}\\}$ be a
transversal for $N_{D}=N_{G}(D)$ in $G$, and let $\\{g_{1},\dots,g_{t}\\}$ be
a transversal for $N_{F}=N_{G}(F)$ in $G$. Note that each of $C$, $D$, and $F$
are the centralizers for the nonidentity element they contain. So the only way
two nonidentity elements of $G$ can commute is if they lie in the same
conjugate of one of these three subgroups. Now, we can define the sets

  * $R_{i}=\\{c_{i}^{a_{1}},\ldots,c_{i}^{a_{r}}\\}$, $i=1,2,\ldots,(q-1)/k-1$,

  * $S_{j}=\\{d_{j}^{b_{1}},\ldots,d_{j}^{b_{s}}\\}$, $j=1,2,\ldots,q-1$,

  * $T_{k}=\\{f_{k}^{g_{1}},\ldots,f_{k}^{g_{t}}\\}$, $k=1,2,\ldots,(q+1)/k-1$.

We take $B_{i}=R_{i}\cup S_{i}\cup T_{i}$ when $1\leqslant
i\leqslant(q-1)/k-1$. Note that $(q-1)/k=(q+1)/k-1$ for both choices of $k$.
We take $B_{(q-1)/k}=S_{(q-1)/k}\cup T_{(q-1)/k}$. Finally, we take
$B_{j}=T_{j}$ where $j=q$ if $p$ is even and $(q+1)/k+1\leqslant j\leqslant
q-1$ when $p$ is odd. Set $A=F$. Then $G=A\uplus B_{1}\uplus\cdots\uplus
B_{q}$ is a strict $q$-split decomposition of $G$ when $p$ is $2$ or when
$G={\rm PGL}_{2}(q)$ when $p$ is odd and $G=A\uplus B_{1}\uplus\cdots\uplus
B_{q-1}$ is a strict $(q-1)$-split decomposition when $G=L_{2}(q)$ when $p$ is
odd.

Suppose now that $G$ has a strict $u$-split decomposition: $G=A^{\prime}\uplus
B_{1}^{\prime}\uplus\cdots\uplus B_{u}^{\prime}$. Then we can assume that
$A^{\prime}\cap F=1$ and hence $u+1\geqslant|F:A^{\prime}\cap F|=|F|=(q+1)/k$.
When $p$ is even or when $G={\rm PGL}_{2}(q)$, we conclude that $u\geqslant
q$, and so, $\min(G)=q$. In a similar fashion, we may assume that
$A^{\prime}\cap D=1$, and hence, $u+1\geqslant|D|=q$. We deduce that
$u\geqslant q-1$. Therefore, we obtain $\min(G)=q-1$ when $G=L_{2}(q)$ and $p$
is odd. $\square$

###### Lemma 6.2

Let $S$ be a simple group with $\min(S)=n$. Then $|S|\leqslant(n+1)^{n+1}$. In
particular, $A_{5}\cong L_{2}(4)\cong L_{2}(5)$ is the only simple group $S$
with $\min(S)\leqslant 4$.

Proof. Let $S=A\uplus B_{1}\uplus\cdots\uplus B_{n}$ be a strict $n$-split
decomposition of $S$, where $n=\min(S)$. From [11] one can deduce that
$|S|\leqslant|B|^{|B|}$ where $B$ is an abelian subgroup of maximal order. By
[12], there exists $x\in S$ such that $A\cap B^{x}=1$ and hence $|B|\leqslant
n+1$. This gives the desired estimate.

Suppose that ${\rm min}(S)\leqslant 4$. By the previous paragraph
$|S|\leqslant 5^{5}=3125$, and hence $S$ is isomorphic to $A_{7}$ or
$L_{2}(q)$, $q=4,8,9,11,13,17$. By Lemma 6.1 (1) and (2), $\min(L_{2}(4))=4$,
while $\min(L_{2}(q))\geqslant 8$, for $q=8,9,11,13,17$. Finally, if $S\cong
A_{7}$, then $S$ has a maximal abelian subgroup of order $7$, which intersects
trivially with some of its conjugates, and hence $\min(S)\geqslant 6$. This
completes the proof. $\square$

The Suzuki groups ${\rm Sz}(q)$, an infinite series of simple groups of Lie
type, were defined in [9, 10] as subgroups of the groups $L_{4}(q)$ with
$q=2^{2n+1}$ elements and set $r=2^{n+1}$.

By [10, Theorem 7], the order of ${\rm Sz}(q)$ is

$|{\rm Sz}(q)|=q^{2}(q-1)(q^{2}+1)=q^{2}(q-1)(q+r+1)(q-r+1),$

note that these factors are mutually coprime. We are now ready to find a
strict $k$-split decomposition for the Suzuki groups.

###### Lemma 6.3

If $G={\rm Sz}(q)$, where $q=2^{2n+1}\geqslant 8$, then ${\rm min}(G)=2q-1$.

Proof. By Lemma XI.11.6 of [6], $G$ is partitioned by its Sylow $2$-subgroups
and its cyclic subgroups of order $q-1$, $q-r+1$, and $q+r+1$. Looking at the
proof of Lemma XI.11.6, we see that the cyclic subgroups of order $q-1$,
$q-r+1$, and $q+r+1$ are the centralizers of their nonidentity elements. We
obtain the sets $C_{1},\dots,C_{q-2}$ so that each $C_{i}$ contains one
nonidentity element from each of the cyclic subgroups of order $q-1$, the sets
$D_{1},\dots,D_{q-r}$ so that each $D_{i}$ contains one nonidentity element
from each of the the cyclic subgroups of order $q-r+1$, and the sets
$E_{1},\dots,E_{q+r}$ so that each $E_{i}$ contains one nonidentity element
from each of the the cyclic subgroups of order $q+r$. Note that $r<q-1$, so
$q+r<2q-1$.

Let $P$ be a Sylow $2$-subgroup of $G$. By Theorem VIII.7.9 of [5] and Lemma
XI.11.2 of [6], $Z(P)$ is an elementary abelian $2$-group of order
$q=2^{2n+1}$ and every element outside $Z(P)$ has order $4$. Observe that $P$
is the centralizer in $G$ of all of the nontrivial elements of $Z(P)$. Label
elements in $Z(P)=\\{z_{0}=1,z_{1},\dots,z_{q-1}\\}$. If $x\in P\setminus
Z(P)$, then $P_{0}=\langle Z(P),x\rangle\leqslant C_{G}(x)$. In the proof of
Lemma XI.11.7 of [6], we see that the elements of order $4$ in $G$ lie in two
conjugacy classes. It follows that $|C_{G}(x)|=2|Z(P)|$. This implies that
$C_{G}(x)=P_{0}$. It follows that if $\\{x_{1},\dots,x_{q}\\}$ is a
transversal for $Z(P)$ in $P$, then $C_{G}(x_{i}z_{j})=\langle
Z(P),x_{i}\rangle$ for $i=1,\dots,q$ and $j=1,\dots,q-1$. Let
$\\{a_{1},\dots,a_{t}\\}$ for a transversal for $N_{G}(P)$ in $G$. We define
$F_{i}$ as follows. For $1\leqslant i\leqslant q-1$, set
$F_{i}=\\{z_{i}^{a_{1}},\dots,z_{i}^{a_{t}}\\}$. For $q\leqslant i\leqslant
2q-1$, we define

$F_{i}=\bigcup_{j=1}^{q}\left\\{(x_{j}z_{i-q})^{a_{1}},\dots,(x_{j}z_{i-q})^{a_{r}}\right\\}.$

Finally, we define the $B_{i}$’s. For $1\leqslant i\leqslant q-r$, set
$B_{i}=C_{i}\cup D_{i}\cup E_{i}\cup F_{i}$; for $q-r+1\leqslant i\leqslant
q-2$, set $B_{i}=C_{i}\cup E_{i}\cup F_{i}$; for $q-1\leqslant i\leqslant
q+r$, set $B_{i}=E_{i}\cup F_{i}$; and for $q+r+1\leqslant i\leqslant 2q-1$,
set $B_{i}=F_{i}$. Take $A=1$. Then $G=A\uplus B_{1}\uplus\cdots\uplus
B_{2q-1}$ is a strict $(2q-1)$-split decomposition of $G$.

Note that $G$ has a maximal abelian subgroup $A$ of order $2q$ and $A$ has a
conjugate that it intersects trivially, so we know that $\min(G)\geqslant
2q-1$. This gives the conclusion that $\min(G)=2q-1$. $\square$

##  7 Some upper bounds

Suppose $G$ has an $n$-split decomposition with respect to an abelian subgroup
$A$. In what follows, we show that the index $|G:A|$ is bounded by some
function of $n$. Since for every positive integer $n$ we can find Frobenius
groups with arbitrarily large abelian Frobenius kernels whose Frobenius
complements are cyclic of order $n+1$, we can use Theorem 3.4 to see that it
is not possible to bound $|G|$, particularly $|A|$, in terms of $n$. However,
we now show that we can bound the index $|G:A|$ in terms of $n$. We have not
worked to obtain optimal bounds, and in fact, we are sure that the bounds
obtained are far from optimal. We see from the first couple of paragraphs of
the proof that when $A$ is not normal in $G$, then it is possible to bound
$|G|$ in terms of $n$.

###### Theorem 7.1

There exists a positive integer valued function $f$ defined on the positive
integers so that if $G$ has an $n$-split decomposition with respect an abelian
subgroup $A$, then $|G:A|$ is bounded by $f(n)$.

Proof. Notice that we have the result when $n=2$ and $3$. Thus, we may assume
that $n\geqslant 4$. We begin by noting that it suffices to prove that the
size of all the abelian subgroups of $G$ are bounded by a function of $n$. In
fact, if all of the abelian subgroups of $G$ have order at most $m$, then
$|G|\leqslant m!$, see Problem 1D.11 in [7].

Suppose first that $A$ is not normal. Let $U$ be an abelian subgroup of $G$
that is not contained in $A$. We know that $|U|\leqslant 2n$ by Lemma 3.1 (1).
Also, since $A$ is not normal, there is some conjugate of $A$ that does not
contain $A$. The above work shows that the size of the conjugate is bounded by
$2n$, and so $|A|$ is bounded by $2n$. Therefore, we conclude that the size of
every abelian subgroup of $G$ is bounded by $2n$, and we see that
$|G|\leqslant(2n)!$. Since $|G:A|\leqslant|G|$, this gives the result

We now assume that $A$ is normal in $G$. If $U$ is an abelian subgroup of $G$
that is not contained in $A$, then $|U:U\cap A|\leqslant n+1$ and
$|U|\leqslant 2n$ by Lemma 3.1. If $x$ lies in $G\setminus A$, then
$o(Ax)\leqslant n+1$ by Lemma 3.1. Thus, the only primes that can divide
$|G:A|$ must be less than or equal to $n+1$. In particular, the number of
distinct prime divisors of $|G:A|$ is at most the number of primes less than
or equal to $n+1$ which is certainly bounded by $n$.

Let $P$ be a Sylow $p$-subgroup of $G$. It suffices to show that $|PA:A|$ is
bounded in terms of $n$. Suppose $p$ does not divide $|A|$. If $U$ is an
abelian subgroup of $P$, then we have $U\cap A=1$, so $|U|\leqslant n+1$.
Applying the observation in the first paragraph of the proof, we have
$|P|\leqslant(n+1)!$.

Thus, we may assume that $p$ divides $|A|$. We see that $P$ will be a
$p$-group that has the $k$-split decomposition $(A\cap P)\uplus(B_{1}\cap
P)\uplus\cdots\uplus(B_{n}\cap P)$ where $k$ is the number of the sets
$B_{i}\cap P$ that are not empty. Note that we have not required that the
decomposition be strict, so sets of size one are allowed. We will prove that
$|P:P\cap A|\leqslant(n^{2})!$. If $k<n$, then $(k^{2})!\leqslant(n^{2})!$, so
working by induction on $n$, we may assume that $k=n$.

For the rest of this proof, we assume that $G$ is a $p$-group for some prime
$p\leqslant n$. Suppose $B$ is an abelian normal subgroup of $G$ that is not
contained in $A$ and $C_{G}(B)=B$. Then as above $|B|\leqslant 2n$. Also, by
the $N/C$-theorem, we have that

$G/B=N_{G}(B)/C_{G}(B)\leqslant{\rm Aut}(B)\leqslant{\rm Sym}(B),$

so $|G:B|\leqslant(2n)!$. It follows that

$|G:A|<|G|\leqslant(2n)(2n)!\leqslant(n^{2})!,$

which yields the desired conclusion.

Suppose $U$ is an abelian normal subgroup of $G$ that is not contained in $A$.
We claim that there exists a subgroup $B$ normal in $G$ so that $U\leqslant
B$, $B$ is abelian, and $B=C_{G}(B)$. Observe that $U\leqslant C_{G}(U)$ and
$C_{G}(U)$ is normal in $G$. If $U=C_{G}(U)$, then take $B=U$, and we are
done. Thus, we may assume $U<C_{G}(U)$. Since $G$ is a $p$-group, we can find
$V$ normal in $G$ so that $U<V\leqslant C_{G}(U)$ and $|V:U|=p$. Notice that
$U$ is central in $V$ and $V/U$ has order $p$, so $V/U$ is cyclic. This
implies that $V$ is abelian. Also, $C_{G}(V)\leqslant C_{G}(U)$, so
$|C_{G}(V):V|<|C_{G}(U):U|$. Working by induction, on $|C_{G}(U):U|$, we
obtain the conclusion. Using the existence of $B$ and the previous paragraph,
we see that $|G:A|\leqslant(n^{2})!$.

Thus, we may assume that $A$ contains every normal abelian subgroup of $G$ and
that $A=C_{G}(A)$. Suppose $U$ is a subgroup of $A$ that is maximal such that
it is normal in $G$ and $C_{G}(U)$ is not contained in $A$. Notice that such a
subgroup $U>1$ must exist since $1<Z(G)\leqslant A$ and $G=C_{G}(Z(G))$ is not
contained in $A$. Thus, there is an element $g$ in $G\setminus A$ that
centralizes $U$. Thus, $U\langle g\rangle$ is an abelian subgroup of $G$ that
is not contained in $A$. Observe that $U=U\langle g\rangle\cap A$, and by
Lemma 3.1 (1), we know that $|U|=|U\langle g\rangle\cap A|\leqslant n$. Since
$A=C_{G}(A)$, we know that $U<A$. Hence, we can find $V$ so that $U<V\leqslant
A$, $V$ is normal in $G$, and $|V:U|=p$. Since $p\leqslant n$, we have
$|V|\leqslant n^{2}$. Now, $G/C_{G}(V)$ is a subgroup of ${\rm Aut}(V)$, so
$|G:C_{G}(V)|$ is bounded by $(n^{2})!$. Now, the choice of $V$ implies that
$C_{G}(V)\leqslant A$ and since $V\leqslant A$ and $A$ is abelian, we have
$C_{G}(V)=A$. Thus, we now have that $|G:A|\leqslant(n^{2})!$. $\square$

We do have some cases where we can obtain a better bound.

###### Lemma 7.2

Suppose $G$ has an $n$-split decomposition with respect to $A=1$, then
$|G|\leqslant n!$.

Proof. Let $p$ be a prime divisor of $|G|$, and let $P\in{\rm Syl}_{p}(G)$.
Let $M$ be a maximal abelian normal subgroup of $P$ and set $m=|M|$. Then
$M=C_{P}(M)$. Thus, $P/M$ is isomorphic to a subgroup of ${\rm
Aut}(M)\leqslant S_{m}$, which forces $|P:M|$ to divide $m!$. Since
$m\leqslant n$, we conclude that $|P|$ divides the $p$-part of $n!$. The claim
follows from the fact that $|G|$ is the least common multiple of the orders of
Sylow subgroups of $G$. $\square$

###### Lemma 7.3

If $G$ has an $n$-split decomposition with respect to $A=Z(G)$ where $|A|=n$,
then $|G:A|\leqslant 2n$.

Proof. Let $k=|G:A|$. As the cosets of the center are commuting subsets of
$G$, no $B_{i}$ can contain more than one element of any coset of $A$, and it
cannot contain an element of $A$. We see that $|B_{i}|\leqslant k-1$. On the
one hand, we have $|G|=|G:A||A|=kn$, and on the other hand, we obtain

$|G|=|A|+\sum_{i=1}^{n}|B_{i}|\leqslant n+(k-1)n=kn.$

We must have equality throughout this inequality, so $|B_{i}|=k-1$ for all
$i$. Thus each $B_{i}$ contains representatives of every nontrivial coset of
$A$. This shows that the complementary graph of $\Delta(G)$, which is called
noncommuting graph of $G$ and denoted by $\nabla(G)$, is a complete
$(k-1)$-partite graph. Now, by Proposition 3(ii) in [1], $G/A$ is an
elementary abelian $2$-group and the size of the class of $g$, for every
noncentral element $g\in G$, is $k/2$. Finally, since $G/A$ is abelian, the
entire conjugacy class of $g$ is contained in the coset $Ag$, which has size
$n$. Thus $k/2\leqslant n$, so $k\leqslant 2n$. $\square$

Some examples. Both $D_{8}$ and $Q_{8}$ have $2$-split decompositions with
respect to their centers who have order $2$. There are also many examples
where $n$ is large. One family of examples is the Suzuki $2$-groups (see [3]).
We omit the details here.

Acknowledgments

This work was done during the fourth author had a visiting position at the
Department of Mathematical Sciences, Kent State University, USA. He would like
to thank the hospitality of the Department of Mathematical Sciences of KSU. He
would like also to express his sincere thanks to Prof. I. M. Isaacs for
helpful comments. The work of D. V. Lytkina and V. D. Mazurov is supported by
grant 17-51-560005 of Russian Foundation of Basic Research.

We would like to thank the referee for a thorough reading of this paper and
the resulting helpful suggestions.

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M. L. Lewis
Department of Mathematical Sciences, Kent State University,
Kent, Ohio $44242$, United States of America
E-mail address<EMAIL_ADDRESS>
D. V. Lytkina
Siberian State University of Telecommunications and Information Sciences,
Novosibirsk State University, Novosibirsk, Russia
E-mail address<EMAIL_ADDRESS>
V. D. Mazurov
Sobolev Institute of Mathematics and Novosibirsk State University,
Novosibirsk, Russia,
E-mail address<EMAIL_ADDRESS>
A. R. Moghaddamfar
Faculty of Mathematics, K. N. Toosi University of Technology, P. O. Box
$16315$–$1618$, Tehran, Iran,
and
Department of Mathematical Sciences, Kent State University,
Kent, Ohio $44242$, United States of America
E-mail addresses<EMAIL_ADDRESS>and<EMAIL_ADDRESS>