# Finite groups in which every commutator has prime power order

Mateus Figueiredo  Mateus Figueiredo: Department of Mathematics, University of
Brasilia, Brasilia DF, Brazil<EMAIL_ADDRESS>and  Pavel Shumyatsky
Pavel Shumyatsky: Department of Mathematics, University of Brasilia, Brasilia
DF, Brazil<EMAIL_ADDRESS>
###### Abstract.

Finite groups in which every element has prime power order (EPPO-groups) are
nowadays fairly well understood. For instance, if $G$ is a soluble EPPO-group,
then the Fitting height of $G$ is at most 3 and $|\pi(G)|\leqslant 2$ (Higman,
1957). Moreover, Suzuki showed that if $G$ is insoluble, then the soluble
radical of $G$ is a 2-group and there are exactly eight nonabelian simple
EPPO-groups.

In the present work we concentrate on finite groups in which every commutator
has prime power order (CPPO-groups). Roughly, we show that if $G$ is a CPPO-
group, then the structure of $G^{\prime}$ is similar to that of an EPPO-group.
In particular, we show that the Fitting height of a soluble CPPO-group is at
most 3 and $|\pi(G^{\prime})|\leqslant 3$. Moreover, if $G$ is insoluble, then
$R(G^{\prime})$ is a 2-group and $G^{\prime}/R(G^{\prime})$ is isomorphic to a
simple EPPO-group.

###### Key words and phrases:

Finite groups, commutators

###### 2020 Mathematics Subject Classification:

20F12, 20E34

This work was supported by CNPq.

##  1\. Introduction

Finite groups in which every element has prime power order (EPPO-groups for
short) were first studied by Higman in [6]. In the literature these groups are
sometimes called CP-groups. Higman showed that a soluble EPPO-group has
Fitting height at most 3. Moreover, the order of a soluble EPPO-group is
divisible by at most two primes. Suzuki classified simple EPPO-groups in his
celebrated work [12], finding that only eight simple EPPO-groups exist. These
are the groups $PSL(2,q)$ $(q=4,7,8,9,17)$, $PSL(3,4)$, $Sz(8)$, $Sz(32)$.
Moreover Suzuki showed in [11] that if $G$ is an insoluble EPPO-group, then
the soluble radical $R(G)$ is a 2-group. Further clarifications of the
structure of EPPO-groups were obtained in [2] and [1].

More recently, also infinite groups in which every element has prime power
order have attracted attention. In particular, the reader can check the papers
[10] and [4] for the study of profinite and locally finite groups with that
property.

In this paper we focus on finite groups in which the commutators have prime
power orders (CPPO-groups). By a commutator we mean any element $a$ of a group
$G$ for which there are $x,y\in G$ such that $a=[x,y]=x^{-1}y^{-1}xy$. As
usual, $G^{\prime}$ stands for the commutator subgroup of $G$. It is well-
known that elements of $G^{\prime}$ need not be commutators. On the other
hand, by the celebrated verification of the Ore conjecture [7], every element
of a nonabelian finite simple group is a commutator. Therefore any simple
CPPO-group is an EPPO-group (from the above list determined by Suzuki). We do
not know if the commutator subgroup of any CPPO-group is necessarily EPPO.
This seems unlikely. Our main results can be summarized as follows.

######  Theorem 1.1.

Let $G$ be a soluble CPPO-group. Then

  1. (a)

The Fitting height of $G$ is at most $3$;

  2. (b)

The order of $G^{\prime}$ is divisible by at most $3$ primes.

As usual, we say that a group $K$ is perfect if $K=K^{\prime}$. We write
$R(K)$ to denote the maximal soluble normal subgroup of $K$.

######  Theorem 1.2.

Let $G$ be an insoluble CPPO-group. Then

  1. (a)

$G^{\prime}$ is perfect;

  2. (b)

$R(G^{\prime})=[G^{\prime},R(G)]$ is a $2$-group; and

  3. (c)

$G^{\prime}/R(G^{\prime})$ is a simple EPPO-group.

Thus, our results show that the structure of the commutator subgroup of a
CPPO-group is roughly similar to that of an EPPO-group.

##  2\. Preliminaries

If $A$ is a group of automorphisms of a group $G$, the subgroup generated by
all elements of the form $g^{-1}g^{\alpha}$ with $g\in G$ and $\alpha\in A$ is
denoted by $[G,A]$. It is well known that the subgroup $[G,A]$ is an
$A$-invariant normal subgroup in $G$. We write $C_{G}(A)$ for the centralizer
of $A$ in $G$. If $G$ and $A$ are finite and $(|G|,|A|)=1$, we say that $A$ is
a group of coprime automorphisms of $G$. Throughout, $\pi(G)$ denotes the set
of prime divisors of the order of $G$.

We start with a lemma which lists some well known properties of coprime
actions (see for example [5, Ch. 5 and 6]). In the sequel the lemma will often
be used without explicit references.

######  Lemma 2.1.

Let $A$ be a group of coprime automorphisms of a finite group $G$. Then

  1. (i)

$G=[G,A]C_{G}(A)$. If $G$ is abelian, then $G=[G,A]\oplus C_{G}(A)$.

  2. (ii)

$[G,A,A]=[G,A]$.

  3. (iii)

$C_{G/N}(A)=NC_{G}(A)/N$ for any $A$-invariant normal subgroup $N$ of $G$.

  4. (iv)

If $[G/\Phi(G),A]=1$, then $[G,A]=1$.

  5. (v)

If $G$ is nilpotent and $A$ is a noncyclic abelian group, then $G=\prod_{1\neq
a\in A}C_{G}(a)$.

  6. (vi)

$G$ contains an $A$-invariant Sylow $p$-subgroup for each prime $p\in\pi(G)$.

The next lemma is immediate from the previous one.

######  Lemma 2.2.

Let $A$ be a group acting coprimely on a finite abelian group $V\neq 1$.
Suppose that $C_{V}(a)=1,~{}\text{for every}~{}1\neq a\in A$. Then $A$ is
cyclic if at least one of the following conditions holds:

  1. (i)

$A$ is abelian;

  2. (ii)

$A$ is a $p$-group for an odd prime $p$;

  3. (iii)

$A$ is a $2$-group but not a quaternion group.

######  Lemma 2.3.

Let $A$ be a noncyclic abelian group acting coprimely on an abelian group $V$.
Then

$\bigcap_{1\neq a\in A}[V,a]=1.$

###### Proof.

We argue by induction on $|V|$. The result is clear if $|V|=1$. Assume
$|V|>1$. As $A$ is a noncyclic group, by Lemma 2.2 there exists an element
$1\neq a\in A$ for which $C_{V}(a)\neq 1$. Since $A$ is abelian, $C_{V}(a)$ is
$A$-invariant. Applying the induction hypothesis to the quotient group
$V/C_{V}(a)$ we obtain

$\bigcap_{1\neq b\in A}\big{[}V/C_{V}(a),b\big{]}=1.$

This means that $C_{V}(a)$ contains the subgroup

$\bigcap_{1\neq b\in A}[V,b],$

which clearly is also contained in $[V,a]$. Since the action is coprime,

$[V,a]\cap C_{V}(a)=1,$

whence

$\bigcap_{1\neq b\in A}[V,b]=1,$

as claimed. ∎

Throughout this article we write $G=\langle X\rangle$ to mean that the group
$G$ is generated by a set $X$.

######  Lemma 2.4.

Let $G=V\langle a\rangle$ be a finite group, which is a product of a normal
subgroup $V$ and a cyclic subgroup $\langle a\rangle$ such that $(|V|,|a|)=1$.
Suppose that $v\in V$ has the property that $(|V|,|av|)=1$. Then $v\in[V,a]$.

###### Proof.

We can pass to the quotient group $G/[V,a]$ and without loss of generality
assume that $[V,a]=1$. But then it is clear that $(|V|,|av|)=1$ if and only if
$v=1$. Hence the result. ∎

We remind the reader that a $p$-group $P$ is called extraspecial if the centre
$Z(P)$ has order $p$, and the quotient $P/Z(P)$ is a nontrivial elementary
abelian $p$-group.

######  Lemma 2.5.

Let $\varphi$ be a coprime automorphism of a finite extraspecial $p$-group $P$
such that $C_{P}(\varphi)=\Phi(P)$. Then every element of $P\setminus\Phi(P)$
is conjugate to an element of the form $[x,\varphi]$ for some $x\in P$.

###### Proof.

Since $C_{P}(\varphi)=\Phi(P)$, it follows that the map
$\overline{g}\mapsto[\overline{g},\varphi]$ is surjective on the group
$\overline{P}=P/\Phi(P)$. Thus, for an element $g\in P\setminus\Phi(P)$ we can
find elements $x\in P\setminus\Phi(P)$ and $z\in\Phi(P)$ for which
$g=[x,\varphi]z$. Let $y\in P$ be an element such that $[x,\varphi,y]\neq 1$.
Since $\Phi(P)=Z(P)$ has prime order, it is generated by $[x,\varphi,y]$.
Therefore there exists an integer $r$ satisfying $[x,\varphi,y]^{r}=z^{-1}$.
It follows that

$g^{y^{r}}=([x,\varphi]z)^{y^{r}}=[x,\varphi]^{y^{r}}z=[x,\varphi][x,\varphi,y^{r}]z=[x,\varphi][x,\varphi,y]^{r}z=[x,\varphi].$

This shows that $g$ is conjugate to $[x,\varphi]$, as required. ∎

An important role in this paper is played by the concept of towers as
introduced by Turull (see [13]).

######  Definition 2.6.

Let $G$ be a group. A sequence $(P_{i})_{i=1,\ldots,h}$ of subgroups of $G$ is
said to be a tower of height $h$ if the following are satisfied:

  1. (i)

$P_{i}$ is a $p_{i}$-group for all $i=1,\ldots,h$.

  2. (ii)

$P_{i}$ normalizes $P_{j}$ for all $i<j$.

  3. (iii)

Put $\overline{P_{h}}=P_{h}$ and
$\overline{P_{i}}={P_{i}}/{C_{P_{i}}(\overline{P_{i+1}})},\ i=1,\ldots,h-1.$
Then $\overline{P_{i}}$ is nontrivial for all $i$.

  4. (iv)

$p_{i}\neq p_{i+1}$ for all $i=1,\ldots,h-1$.

In what follows, whenever we have a tower $(P_{i})_{i=1,\ldots,h}$, we write
$p_{i}$ to denote the prime divisor of the order of $P_{i}$.

Recall that the Fitting height of a finite soluble group is the minimal number
$h=h(G)$ such that $G$ possesses a normal series of length $h$ all of whose
factors are nilpotent. The next lemma lemma is taken from [13].

######  Lemma 2.7.

If $G$ is a finite soluble group, then $h(G)$ is the maximum of heights of
towers of $G$.

The following lemma will be useful. It is taken from [3, Lema 3.7].

######  Lemma 2.8.

Let $(P_{i})_{i=1,\ldots,h}$ be a tower of a group $G$ with $h\geqslant 2$.
Let $N$ be anormal subgroup of $G$ such that

$N\cap P_{i}\leqslant C_{P_{i}}(P_{h}),~{}i=1,\ldots,h-1.$

Then $(P_{i}N/N)_{i=1,\ldots,h-1}$ is a tower of $G/N$.

The next lemma is given without a proof because it is pretty similar to [3,
Lema 3.6].

######  Lemma 2.9.

Let $(P_{i})_{i=1,\ldots,h}$ be a tower of a group $G$. Then
$C_{P_{i}}(P_{j})\leq C_{P_{i}}(\overline{P_{i+1}})$ for each $1\leqslant
i<j\leqslant h$.

We will now record the almost obvious but important fact that only the first
two factors of a tower can be cyclic.

######  Lemma 2.10.

Let $(P_{i})$ be a tower of height $h\geqslant 3$ in a group $G$. Then for any
$i\geqslant 3$ the subgroup $P_{i}$ is not cyclic.

###### Proof.

Suppose that $P_{i}$ is a cyclic group for some $i\geqslant 3$. Then any two
automorphisms of $P_{i}$ commute. Consequently, $[P_{i-1},P_{i-2}]\leq
C_{P_{i-1}}(P_{i})\leq C_{P_{i-1}}(\overline{P_{i}})$. It follows that
$\overline{P_{i-2}}=1$, which is a contradiction. ∎

We will require Turull’s concept of an irreducible tower.

######  Definition 2.11.

Let $G$ be a group and let $(P_{i})$ be a tower of height $h$ in $G$. The
tower $(P_{i})$ is said to be irreducible if the following are satisfied:

  1. (i)

$\Phi(\Phi(\overline{P_{i}}))=1$, $\Phi(\overline{P_{i}})\leq
Z(\overline{P_{i}})$ and if $p_{i}\neq 2$ then $exp(\overline{P_{i}})=p_{i}$
for $i=1,\ldots,h$ and $P_{i-1}$ centralizes $\Phi(\overline{P_{i}})$ for all
$i=2,\ldots,h$.

  2. (ii)

$P_{1}$ is cyclic and $\overline{P_{1}}$ has prime order.

  3. (iii)

There exists an elementary abelian subgroup $H_{i}$ of $\overline{P_{i-1}}$
such that $[H_{i},\overline{P_{i}}]=\overline{P_{i}}$ for $i=2,\ldots,h$.

  4. (iv)

If $H$ is a $P_{1}\cdots P_{i-1}$-invariant subgroup of $P_{i}$ whose image on
$\overline{P_{i}}$ is not contained in $\Phi(\overline{P_{i}})$, then
$H=P_{i}$.

Let $(P_{i}^{(1)})$ and $(P_{j}^{(2)})$ be towers of a group $G$ of heights
$h_{1}$ and $h_{2}$, respectively. We will say that $(P_{i}^{(1)})$ is
contained in $(P_{j}^{(2)})$ if there exists an increasing map
$f:\\{1,\ldots,h_{1}\\}\longrightarrow\\{1,\ldots,h_{2}\\}$ such that
$P_{i}^{(1)}\subseteq P_{f(i)}^{(2)}$ for every $i=1,\ldots,h_{1}$.

Note that our definition of irreducible towers differs from that given by
Turull in [13] only by the item (ii). However, the difference is inessential.
Indeed, if $(P_{1},\ldots,P_{h})$ is a tower of a group $G$, we may take an
element $a\in P_{1}\setminus C_{P_{1}}(\overline{P_{2}})$ so that $(\langle
a\rangle,P_{2},\ldots,P_{h})$ is also a tower of $G$. By [13, Lemma 1.4], this
last tower contains an irreducible one which now satisfies Definition 2.11.
Therefore we have

######  Lemma 2.12.

Let $G$ be a group and let $(P_{i})_{i=1,\ldots,h}$ be a tower of $G$. Then
$(P_{i})$ contains an irreducible tower of same height.

##  3\. Soluble CPPO-groups

Observe that subgroups and quotient groups of a CPPO-group are again CPPO-
groups. In the sequel, this will be used throughout the paper without being
mentioned explicitly.

######  Lemma 3.1.

Let $G$ be a finite group containing a tower $(P_{1},P_{2},P_{3})$ of abelian
subgroups such that $P_{1}$ is cyclic, $P_{2}$ is noncyclic and
$P_{2}=[P_{2},P_{1}]$. Then $G$ has a commutator whose order is not a prime
power.

###### Proof.

Suppose the lemma is false and $G$ is a CPPO-group.

Write $P_{1}=\langle a\rangle$. For elements $1\neq b\in P_{2}$ and $c\in
P_{3}$, the equality

$[c,a][a,b]=[cb^{-1},a]^{b}$

implies that $[c,a][a,b]$ has prime power order. Therefore, Lemma 2.4
guarantees that $[c,a]\in[P_{3},[a,b]]$, which is equivalent to
$[c,a]\in[P_{3},[b,a]]$. Since $c$ can be chosen in $P_{3}$ arbitrarily, we
get

$[P_{3},P_{1}]=[P_{3},a]\leq[P_{3},[b,a]],~{}\text{for any}~{}1\neq b\in
P_{2}.$ (3.1)

On the other hand, $P_{2}$ is an abelian group and $C_{P_{2}}(a)=1$. It
follows that the map $b\mapsto[b,a]$ is surjective on $P_{2}$. Therefore the
containment 3.1 implies that

$[P_{3},P_{1}]\leq\bigcap_{1\neq b\in P_{2}}[P_{3},b].$ (3.2)

We conclude from (3.2) and from Lemma 2.3 that $[P_{3},P_{1}]=1$, that is,
$P_{1}=C_{P_{1}}(P_{3})$. In view of Lemma 2.9 we conclude that
$\overline{P_{1}}=1$, which contradicts the definition of tower. ∎

######  Lemma 3.2.

Let $G$ be a CPPO-group containing a tower $(P_{1},P_{2},P_{3})$ with the
following properties:

  1. (i)

$P_{1}$ is cyclic;

  2. (ii)

$P_{2}$ is extraspecial and $C_{P_{2}}(P_{1})=\Phi(P_{2})$;

  3. (iii)

$P_{3}$ is abelian and $P_{3}=[P_{3},\Phi(P_{2})]$.

Then $p_{2}=2$ and $P_{2}$ is isomorphic to the quaternion group $Q_{8}$.

###### Proof.

Write $P_{1}=\langle a\rangle$. For any elements $b\in
P_{2}\setminus\Phi(P_{2})$ and $c\in P_{3}$, the element
$[c,a][a,b]=[cb^{-1},a]^{b}$ has prime power order, so by Lemma 2.4 we have

$[c,a]\in[P_{3},[b,a]].$ (3.3)

Moreover, for a nontrivial element $z\in\Phi(P_{2})$, the element
$[c,z][c,a]^{z}[a,b]=[cb^{-1},az]^{b}$ has prime power order and using again
Lemma 2.4 we see that

$[c,z][c,a]^{z}\in[P_{3},[b,a]].$ (3.4)

Since $P_{2}$ is extraspecial, $z$ commutes with $[b,a]$ and it follows from
(3.3) and (3.4) that

$[c,z]\in[P_{3},[b,a]].$

By hypothesis we have $P_{3}=[P_{3},\Phi(P_{2})]$. Since the element $c$ can
be chosen in $P_{3}$ arbitrarily, we conclude that

$P_{3}=[P_{3},[b,a]],~{}\text{for any}~{}b\in P_{2}\setminus\Phi(P_{2}).$
(3.5)

Lemma 2.5 guarantees that any element of $P_{2}\setminus\Phi(P_{2})$ may be
writen in the form $[b,a]^{x}$ for some $b\in P_{2}\setminus\Phi(P_{2})$ and
$x\in P_{2}$. We deduce from (3.5) and the equality
$P_{3}=[P_{3},\Phi(P_{2})]$ that

$C_{P_{3}}(b)=1,~{}\text{for any}~{}1\neq b\in P_{2}.$

By virtue of Lemma 2.2 we deduce that $p_{2}=2$ and $P_{2}$ is a quaternion
group. Taking into account that $Q_{8}$ is the only extraspecial quaternion
group, $P_{2}$ is isomorphic to $Q_{8}$. ∎

######  Lemma 3.3.

Let $G$ be a CPPO-group and suppose that $G$ contains an irreducible tower
$(P_{1},P_{2},P_{3},P_{4})$ with the following properties:

  1. (i)

$C_{P_{3}}(P_{4})=1$;

  2. (ii)

$P_{4}$ is elementary abelian.

Let $i\in\\{2,3\\}$. Then $\overline{P_{i}}$ is either abelian or
extraspecial.

###### Proof.

Note that, as the tower is irreducible, it is sufficient to check that
$\Phi(\overline{P_{i}})$ is cyclic. In fact, the inverse image of
$Z(\overline{P_{i}})$ is a $P_{1}\cdots P_{i-1}$-invariant subgroup of
$P_{i}$. Thus if $Z(\overline{P_{i}})\nleq\Phi(\overline{P_{i}})$, Definition
2.11 ensures that $\overline{P_{i}}=Z(\overline{P_{i}})$ is an abelian group.
On the other hand, $\Phi(\Phi(\overline{P_{i}}))=1$ and
$\Phi(\overline{P_{i}})\leq Z(\overline{P_{i}})$. In short,
$\Phi(\overline{P_{i}})$ is elementary abelian and either
$\Phi(\overline{P_{i}})=Z(\overline{P_{i}})$ or $\overline{P_{i}}$ is abelian.

First, we prove that $\Phi(\overline{P_{2}})$ is cyclic. Let $K$ be the kernel
of the action of $P_{1}P_{2}$ on $P_{3}/\Phi(P_{3})$. Observe that
$P_{1}P_{2}/K$ acts faithfully and irreducibly on $P_{3}/\Phi(P_{3})$, so that
$P_{1}P_{2}/K$ has cyclic centre. We will show that $\Phi(\overline{P_{2}})$
is a central subgroup of $P_{1}P_{2}/K$. This would guarantee that
$\Phi(\overline{P_{2}})$ is cyclic. Let $Q$ be the inverse image of
$\Phi(\overline{P_{2}})$ in $P_{2}$. From Definition 2.11 we know that

$[Q,P_{1}P_{2}]\leq C_{P_{2}}(P_{3})\leq K,$

and consequently,

$[P_{1}P_{2}/K,\Phi(P_{2}K/K)]=1.$

This means that $\Phi(\overline{P_{2}})$ is a central subgroup of
$P_{1}P_{2}/K$, as claimed.

Now we will prove that $\Phi(P_{3})$ is cyclic. Let $L$ be the kernel of the
action of $P_{1}P_{2}P_{3}$ over $P_{4}$. Observe that $P_{1}P_{2}P_{3}/L$
acts faithfully and irreducibly on $P_{4}$ and consequently has cyclic centre.
Therefore, as $P_{3}\cap L=1$, it is sufficient to check that $\Phi(P_{3})\leq
Z(P_{1}P_{2}P_{3})$. But if this is not the case, since $\Phi(P_{3})$ is
centralized by both $P_{2}$ and $P_{3}$, we can find elements $a\in P_{1}$ and
$c\in\Phi(P_{3})$ for which $[c,a]\neq 1$. Clearly we may assume that
$P_{1}=\langle a\rangle$ and find an element $b\in P_{2}$ such that $[b,a]\neq
1$. It follows that $[cb,a]=[c,a][b,a]$ has order divisible by $p_{2}$ and
$p_{3}$, which is impossible because $G$ is a CPPO-group. The proof is now
complete. ∎

######  Lemma 3.4.

Let $G$ be a CPPO-group having an irreducible tower
$(P_{1},P_{2},P_{3},P_{4})$ such that

  1. (i)

$\overline{P_{2}}$ is extraspecial;

  2. (ii)

$C_{P_{3}}(P_{4})=1$;

  3. (iii)

$P_{4}$ is elementary abelian.

Then $p_{2}=2$ and $\overline{P_{2}}$ is isomorphic to $Q_{8}$.

###### Proof.

Set $N=C_{P_{2}}(P_{3})\Phi(P_{3})$ and observe that

$C_{P_{2}N/N}(P_{3}N/N)=1.$

It follows from Lemma 2.8 that $(P_{i}N/N)_{i=1,2,3}$ is a tower of
$P_{1}P_{2}P_{3}/N$. Since

$[P_{2}N/N,P_{1}N/N]=P_{2}N/N,$

it follows that

$C_{P_{2}N/N}(P_{1}N/N)\leq\Phi(P_{2}N/N).$ (3.6)

Let $Q$ be the inverse image of $\Phi(\overline{P_{2}})$. By the
irreducibility of the tower $(P_{1},P_{2},P_{3},P_{4})$ we have

$[Q,P_{1}]\leq C_{P_{2}}(P_{3})\leq N,$

and so

$[\Phi(P_{2}N/N),P_{1}N/N]=1.$ (3.7)

It follows from (3.6) and (3.7) that

$C_{P_{2}N/N}(P_{1}N/N)=\Phi(P_{2}N/N).$ (3.8)

Observe that $Q\unlhd P_{1}P_{2}$ and hence $[P_{3},Q]$ is a
$P_{1}P_{2}$-invariant subgroup of $P_{3}$ which is not contained in
$\Phi(P_{3})$ since $\overline{P_{2}}$ is extraspecial. The irreducibility of
the tower $(P_{1},P_{2},P_{3},P_{4})$ now shows that $P_{3}=[P_{3},Q]$.
Therefore we have

$P_{3}N/N=[P_{3}N/N,\Phi(P_{2}N/N)].$

Note that $P_{2}N/N\cong\overline{P_{2}}$ and so the tower
$(P_{i}N/N)_{i=1,2,3}$ satisfies the hypotheses of Lemma 3.2. Hence $p_{2}=2$
and $P_{2}N/N$ is isomorphic to $Q_{8}$. The proof is complete. ∎

The following elementary observation will be helpful later on.

######  Lemma 3.5.

Let $\varphi$ be an involutory automorphism of a group $G$ isomorphic to
$Q_{8}$. There exists an element $u\in G$ such that $[u,\varphi]$ is the
involution of $G$.

###### Proof.

Suppose that this is false. Let $u\in G\setminus\Phi(G)$ be such that
$u^{\varphi}\Phi(G)=u\Phi(G)$. We have $[u,\varphi]\in\Phi(G)$ and so
$[u,\varphi]=1$, that is, $\langle u\rangle\leq C_{G}(\varphi)$. Since
$\varphi$ is nontrivial, we get $\langle u\rangle=C_{G}(\varphi)$. Choose an
element $x\in G\setminus C_{G}(\varphi)$. We certainly have $x^{\varphi}\neq
x$. The assumption that the lemma is false implies that $x^{\varphi}\neq
x^{-1}$. Let $y=x^{\varphi}$. As $\varphi$ is of order two, we have that
$y^{\varphi}=x$. On the one hand, $(xy)^{\varphi}=x^{\varphi}y^{\varphi}=yx$.
On the other hand, as $G=C_{G}(\varphi)\cup\\{x,y,x^{-1},y^{-1}\\}$, we have
$xy\in C_{G}(\varphi)$ and so $xy=yx$. It follows that $x\in Z(G)$. This is a
contradiction since $|Z(G)|=2$. ∎

We are now ready to prove that the Fitting height of a soluble CPPO-group is
at most three.

######  Proposition 3.6.

Let $G$ be a soluble CPPO-group. Then $h(G)\leqslant 3$.

###### Proof.

Assume that the result is false and let $G$ be a counterexample of minimal
possible order. By minimality, $h(G)=4$. Moreover, $\Phi(G)=1$ since
$h(G/\Phi(G))=h(G)$.

Using Lemmas 2.7 and 2.12 we may write $G=P_{1}P_{2}P_{3}P_{4}$ where
$(P_{1},P_{2},P_{3},P_{4})$ is an irreducible tower. Here $P_{4}$ is normal in
$G$ and so $\Phi(P_{4})\leq\Phi(G)=1$, that is, $P_{4}$ is an elementary
abelian $p_{4}$-group. Putting $N=C_{P_{3}}(P_{4})$, observe that $N$ is
normal and

$C_{P_{3}/N}(P_{4}N/N)=1.$

It follows from Lemma 2.8 that $(P_{i}N/N)_{i=1,2,3,4}$ is a tower of $G/N$.
Because of Lemma 2.7 and the minimality of $G$ we conclude that $N=1$. Hence,
$P_{3}=\overline{P_{3}}$. Moreover, Lemma 3.3 shows that $\overline{P_{i}}$ is
either abelian or extraspecial for $i\in\\{2,3\\}$ and $\Phi(P_{3})\leq
Z(P_{1}P_{2}P_{3})$.

The remaining part of the proof consists in analysis of the following three
possibilities for the group $\overline{P_{2}}$: either $\overline{P_{2}}$ is
cyclic, or abelian noncyclic, or extraspecial. The proof will be complete once
we show that in all these cases $G$ contains a commutator whose order is not a
prime power.

Case 1. Assume that $\overline{P_{2}}$ is a cyclic group and write
$\overline{P_{2}}=\langle\overline{b}\rangle$. Observe that $P_{3}=[P_{3},b]$
and $C_{P_{3}}(b)=\Phi(P_{3})$. Thus if $P_{3}$ is an abelian group, Lemma 3.1
shows that the subgroup $P_{2}P_{3}P_{4}$ contains a commutator whose order is
not a prime power, a contradiction. Therefore $P_{3}$ is not abelian. Lemma
3.3 now shows that $P_{3}$ is an extraspecial $p_{3}$-group. Observe that
$[P_{4},\Phi(P_{3})]$ is a normal subgroup contained in $P_{4}$. Moreover,
$[P_{4},\Phi(P_{3})]\neq 1$ since $C_{P_{3}}(P_{4})=1$. It follows from the
irreducibility of the tower that $P_{4}=[P_{4},\Phi(P_{3})]$. Now, applying
Lemma 3.2 to the tower $(P_{2},P_{3},P_{4})$ we deduce that $p_{3}=2$ and
$P_{3}\cong Q_{8}$. Consequently, we have $p_{2}=3$ and $p_{1}=2$.

Set $K=C_{P_{1}P_{2}}(P_{3})$. As neither $P_{1}$ nor $P_{2}$ acts trivially
on $P_{3}$, we have that $P_{1}P_{2}/K$ is isomorphic to a subgroup of $S_{4}$
whose order is at least 6. Since the Sylow 2-subgroup of $P_{1}P_{2}/K$ is
cyclic, it follows that $|P_{1}P_{2}/K|=6$, and in particular, $P_{1}^{2}\leq
K$. Let $a\in P_{1}$ be a generator of $P_{1}$. Note that $a$ induces an
involutory automorphism of $P_{3}$. Lemma 3.5 now shows that there exists an
element $u\in P_{3}$ for which $[u,a]$ is the involution of $P_{3}$. Since
$\overline{P_{2}}=\langle\overline{b}\rangle$, we have $[b,a]\neq 1$ and so
$[ub,a]=[u,a][b,a]$ has order divisible by 6, a contradiction.

Case 2. Now we deal with the case where $\overline{P_{2}}$ is an abelian
noncyclic group. Set $M=C_{P_{2}}(P_{3})\Phi(P_{3})$ and observe that

$C_{P_{2}M/M}(P_{3}M/M)=1.$

In view of Lemma 2.8 we deduce that $(P_{i}M/M)_{i=1,2,3}$ is a tower of
$P_{1}P_{2}P_{3}/M$. The irreducibility of the tower
$(P_{1},P_{2},P_{3},P_{4})$ shows that

$P_{2}M/M=[P_{1}M/M,P_{2}M/M].$

It follows from Lemma 3.1 that $P_{1}P_{2}P_{3}/M$ has a commutator whose
order is not a prime power, a contradiction.

Case 3. It remains to handle the case where $\overline{P_{2}}$ is an
extraspecial $p_{2}$-group. Lemma 3.4 shows that $p_{2}=2$ and
$\overline{P_{2}}$ is isomorphic to $Q_{8}$.

Choose $b\in P_{2}$ such that
$\Phi(\overline{P_{2}})=\langle\overline{b}\rangle$. We have $[P_{3},b]=P_{3}$
and $C_{P_{3}}(b)=\Phi(P_{3})$. Lemma 3.1 shows that $P_{3}$ is not abelian
and it follows from Lemma 3.3 that $P_{3}$ is an extraspecial $p_{3}$-group.
Therefore $P_{4}=[P_{4},\Phi(P_{3})]$ and thus the tower $(\langle
b\rangle,P_{3},P_{4})$ satisfies the hypotheses of Lemma 3.2. Since $p_{3}\neq
2$, it follows that $\langle b\rangle P_{3}P_{4}$ contains a commutator whose
order is not a prime power. This is the final contradiction. ∎

As usual, if $\pi$ is a set of primes, we write $O_{\pi}(G)$ to denote the
maximal normal $\pi$-subgroup of $G$.

######  Lemma 3.7.

Let $G$ be a CPPO-group and let $N$ be a nilpotent normal subgroup of $G$.
Then there is a prime $p\in\pi(N)$ such that $O_{p^{\prime}}(N)\leq Z(G)$.

###### Proof.

If $N\leq Z(G)$ we have nothing to prove. Assume $N\nleq Z(G)$. There exists a
prime number $p$ for which the Sylow $p$-subgroup $P$ of $N$ is not central in
$G$. Then $G=C_{G}(P)\cup C_{G}(O_{p^{\prime}}(N))$. Indeed, suppose $G\neq
C_{G}(P)\cup C_{G}(O_{p^{\prime}}(N))$ and choose $g\in
G\setminus(C_{G}(P)\cup C_{G}(O_{p^{\prime}}(N)))$. There are $a\in P$ and
$b\in O_{p^{\prime}}(N)$ such that $[a,g]\neq 1$ and $[b,g]\neq 1$. The
equality $[ab,g]=[a,g][b,g]$ shows that $G$ contains a commutator whose order
is not a prime power, a contradiction. Hence, $G=C_{G}(P)\cup
C_{G}(O_{p^{\prime}}(N))$. It is well-known that a nontrivial group cannot be
a union of two proper subgroups. Since $P$ is not central,
$G=C_{G}(O_{p^{\prime}}(N))$, that is, $O_{p^{\prime}}(N)\leq Z(G)$. ∎

In what follows we write $\gamma_{\infty}(G)$ to denote the intersection of
the lower central series of a group $G$.

######  Proposition 3.8.

Let $G$ be a soluble CPPO-group. Then $|\pi(G^{\prime})|\leqslant 3$.

###### Proof.

If $G$ is a nilpotent group, Lemma 3.7 allows us to write $G=P\times
O_{p^{\prime}}(G)$ where $P$ is Sylow $p$-subgroup, for some prime $p$, and
$O_{p^{\prime}}(G)\leq Z(G)$. Then $G^{\prime}\leq P$ is a $p$-group.

Assume that $h(G)=2$. In this case, $\gamma_{\infty}(G)$ is a nilpotent
nontrivial normal subgroup of $G$. Using Lemma 3.7 we write
$\gamma_{\infty}(G)=P\times O_{p^{\prime}}(\gamma_{\infty}(G))$ where $P$ is a
Sylow $p$-group of $\gamma_{\infty}(G)$ and
$O_{p^{\prime}}(\gamma_{\infty}(G))\leq Z(G)$. It follows that

$\gamma_{\infty}(G)=[\gamma_{\infty}(G),G]\leq P,$

that is, $\gamma_{\infty}(G)$ is a $p$-group. Moreover, $G/\gamma_{\infty}(G)$
is a nilpotent CPPO-group and applying the previous case we get that
$G^{\prime}/\gamma_{\infty}(G)$ is a $q$-group for some prime $q$. Thus
$G^{\prime}$ is a $\\{p,q\\}$-group.

The case $h(G)=3$ is obtained by applying twice the argument of the previous
case. In fact, in this case we have $h(\gamma_{\infty}(G))=2$ and so
$\gamma_{\infty}(\gamma_{\infty}(G))$ is a $p$-group for some prime $p$. But
on the other hand we have $h(G/\gamma_{\infty}(\gamma_{\infty}(G)))=2$ and so
$G^{\prime}/\gamma_{\infty}(\gamma_{\infty}(G))$ has order divisible by at
most two primes. Hence, the order of $G^{\prime}$ is divisible by at most
three primes. ∎

Note that the combination of Propositions 3.6 and 3.8 yields Theorem 1.1.

##  4\. Insoluble CPPO-groups

The goal of this section is to establish Theorem 1.2. Our first lemma is
almost obvious so we omit the proof.

######  Lemma 4.1.

Let $G=K\times L$ be a CPPO-group, where the subgroups $K$ and $L$ are both
nonabelian. Then $G^{\prime}$ is a $p$-group for some prime number $p$.

Recall that the generalized Fitting subgroup $F^{*}(G)$ of a finite group $G$
is the product of the Fitting subgroup $F(G)$ and all subnormal quasisimple
subgroups; here a group is quasisimple if it is perfect and its quotient by
the centre is a nonabelian simple group. In any finite group $G$ we have
$C_{G}(F^{*}(G))\leq F^{*}(G)$. Therefore the following lemma holds.

######  Lemma 4.2.

Let $G$ be a nontrivial finite group with $R(G)=1$. Then $F^{*}(G)\neq 1$ and
$C_{G}(F^{*}(G))=1$.

Recall that a finite group is almost simple if it has a unique minimal normal
subgroup (the socle) which is nonabelian simple. The Mathieu group $M_{10}$ is
an example of an almost simple EPPO-group, which is not simple.

######  Lemma 4.3.

Let $G$ be a nontrivial CPPO-group with $R(G)=1$. Then $G$ is almost simple.

###### Proof.

Since $R(G)=1$, it follows that $F^{*}(G)$ is a direct product of nonabelian
simple groups. Lemma 4.1 shows that $F^{*}(G)$ is simple. ∎

The next theorem is the famous result, due to Liebeck et al, verifying the Ore
conjecture.

######  Proposition 4.4.

Every element of a nonabelian simple group is a commutator.

It follows that a simple group is a CPPO-group if and only if it is an EPPO-
group. Combining this with Suzuki’s classification of simple EPPO-groups we
obtain.

######  Proposition 4.5.

A nonabelian simple CPPO-group is isomorphic to one of the following groups:
$PSL(2,q)$, with $q\in\\{4,7,8,9,17\\}$, $PSL(3,4)$, $Sz(8)$, $Sz(32)$.

As usual, $Out(G)$ denotes the outer automorphism group of $G$. The next lemma
is now immediate (cf [14]).

######  Lemma 4.6.

Let $G$ be a nonabelian simple CPPO-group. Then one of the following
conditions holds:

  1. (i)

$Out(G)$ is cyclic;

  2. (ii)

$G\cong PSL(2,9)\cong A_{6}$ and $Out(G)$ is the Klein four-group;

  3. (iii)

$G\cong PSL(3,4)$ and $Out(G)$ is the dihedral group $D_{12}$, of order $12$.

######  Lemma 4.7.

Let $G$ be an almost simple group with the socle $H\cong PSL(3,4)$. Then $G$
is a CPPO-group if, and only if, $G/H$ is abelian.

Before we embark on the proof, fix some notation.

Let $F=\\{0,1,a,a^{2}\\}$ be the field with 4 elements and let $L=SL(3,F)$.
Denote by $\varphi$ the Frobenius automorphism of $L$ and note that $\varphi$
has order 2.

Let $\delta$ stand for the automorphism of $L$ induced by conjugation by the
matrix

$\begin{bmatrix}a&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix}\in GL(3,F).$

Note that $\delta$ has order 3.

Denote by $\beta$ the diagonal automorphism of $L$, that is, the map taking a
matrix $A\in L$ to the transpose of the inverse $A^{-T}$. Then $\beta$ has
order 2. Observe that $\delta^{\varphi}=\delta^{\beta}=\delta^{-1}$.

Write $\overline{\varphi},\overline{\delta},\overline{\beta}$ for the
automorphisms of $H\cong PSL(3,4)$ induced by $\varphi,\delta$ and $\beta$,
respectively.

Set $G_{1}=H\langle\overline{\delta},\overline{\varphi}\rangle$ and
$G_{2}=H\langle\overline{\delta},\overline{\beta}\rangle$. Thus, $G_{1}$ and
$G_{2}$ are the two almost simple subgroups of $Aut\,H$ whose images in
$Out\,H$ are nonabelian of order six.

We will now prove Lemma 4.7.

###### Proof of Lemma 4.7.

Note that if $G/H$ is an abelian group, then $G^{\prime}=H$. As $H$ is an
EPPO-group, $G$ is a CPPO-group.

Therefore it suffices to show that if $G/H$ is not abelian, then $G$ is not a
CPPO-group. In this case, $G$ contains a subgroup isomorphic to either $G_{1}$
or $G_{2}$. Consequently, it suffices to show that $G_{1}$ and $G_{2}$ are not
CPPO-groups. In the sequel we write $\overline{X}$ for the image of $X$ in $H$
whenever $X\subseteq L$.

We will show first that $G_{1}$ is not a CPPO-group.

Let $A_{1}\in L$ be the matrix $\begin{bmatrix}1&0&0\\\ 0&1&a\\\
0&0&1\end{bmatrix}.$ Observe that $A_{1}^{2}=1$ and

$A_{1}^{\varphi}=\begin{bmatrix}1&0&0\\\ 0&1&a^{2}\\\ 0&0&1\end{bmatrix}.$

We have

$[A_{1},\varphi]=\begin{bmatrix}1&0&0\\\ 0&1&1\\\ 0&0&1\end{bmatrix}.$

So $[A_{1},\varphi]$ has order 2. Taking into account that $[A_{1},\varphi]\in
C_{L}(\delta)$, remark that in $G_{1}$ the commutator
$[\overline{A_{1}}\overline{\delta},\overline{\varphi}]=\overline{[A_{1},\varphi]}\overline{\delta}$
has order 6. This proves that $G_{1}$ is not a CPPO-group.

Now we show that $G_{2}$ is not a CPPO-group.

Let $A_{2}\in L$ be the matrix $\begin{bmatrix}1&0&0\\\ 0&a&1\\\
0&0&a^{2}\end{bmatrix}.$ Note that

$A_{2}^{\beta}=\begin{bmatrix}1&0&0\\\ 0&a^{2}&0\\\ 0&1&a\end{bmatrix}.$

Compute

$[A_{2},\beta]=\begin{bmatrix}1&0&0\\\ 0&a^{2}&1\\\
0&0&a\end{bmatrix}\begin{bmatrix}1&0&0\\\ 0&a^{2}&0\\\
0&1&a\end{bmatrix}=\begin{bmatrix}1&0&0\\\ 0&a^{2}&a\\\
0&a&a^{2}\end{bmatrix}.$

Now it is easy to check that $[A_{2},\beta]$ has order 2. Keeping in mind that
$[A_{2},\beta]\in C_{Q}(\delta)$ we conclude that in $G_{2}$ the commutator

$[\overline{A_{2}}\overline{\delta},\overline{\beta}]=\overline{[A_{2},\beta]}\overline{\delta}$

has order 6. Therefore $G_{2}$ is not a CPPO-group. This completes the proof.
∎

######  Lemma 4.8.

Let $G$ be a nontrivial CPPO-group with $R(G)=1$. Then $F^{*}(G)=G^{\prime}$
and $G/F^{*}(G)$ is either cyclic or the Klein four-group.

###### Proof.

By Lemma 4.3 $F^{*}(G)$ is a nonabelian simple group. We have
$C_{G}(F^{*}(G))=1$ and, identifying $F^{*}(G)$ with its group of inner
automorphisms, assume that $F^{*}(G)\leq G\leq Aut(F^{*}(G))$. If
$F^{*}(G)\not\cong PSL(3,4)$, in view of Lemma 4.6 we deduce that
$G^{\prime}=F^{*}(G)$ and $G/F^{*}(G)$ is either cyclic or the Klein four-
group. On the other hand, if $F^{*}(G)\cong PSL(3,4)$, it follows from Lemma
4.7 that $G^{\prime}=F^{*}(G)$. Moreover, as $Out(PSL(3,4))\cong D_{12}$, we
again deduce that $G/F^{*}(G)$ is either cyclic or the Klein four-group. ∎

The next observation is now straightforward.

######  Corollary 4.9.

Let $G$ be a nontrivial perfect CPPO-group with $R(G)=1$. Then $G$ is a simple
EPPO-group.

The following result is an immediate consequence of [8, Theorem 1].

######  Lemma 4.10.

Let $G$ be a quasisimple group such that $G/Z(G)$ is a CPPO-group. One of the
following statements holds.

  1. (i)

Every element of $G$ is a commutator;

  2. (ii)

$G/Z(G)\cong A_{6}$, $Z(G)$ is cyclic of order $3$ or $6$ and the noncentral
elements of $G$ which are not commutators have orders in the set
$\\{12,15,24\\}$;

  3. (iii)

$G/Z(G)\cong PSL(3,4)$, $Z(G)\neq 1$, $\pi(Z(G))\subseteq\\{2,3\\}$ and the
noncentral elements of $G$ which are not commutators have orders divisible by
$6$.

Observe that the groups listed in Parts (ii) and (iii) of Lemma 4.10 are not
CPPO-groups. This is because in all cases there is an element $z\in Z(G)$ of
order 2 or 3 and an element $g\in G$ of order 5 such that $zg$ is a
commutator. Therefore we have the following result.

######  Lemma 4.11.

If $G$ is a quasisimple CPPO-group, then $G$ is a simple EPPO-group.

###### Proof.

Suppose $G$ is a quasisimple CPPO-group. In this case every element of $G$ is
a commutator so that $G$ is actually an EPPO-group. Since in an EPPO-group
every centralizer has a prime power order, the existence of a nontrivial
centre implies that the group has prime power order. Thus, $Z(G)=1$ and so $G$
is simple. ∎

######  Lemma 4.12.

Let $G$ be a finite group containing a normal subgroup $N$ such that $G/N$ is
a nonabelian simple CPPO-group. Let $q\in\pi(G)\setminus\pi(N)$. There exists
an elementary abelian $q$-subgroup $Q$ of $G$ and an element $a\in G$, whose
order is a power of a prime different from $q$, such that $Q=[Q,a]$.

###### Proof.

Remark that $G$ does not have a normal $q$-complement. Indeed, if $K$ is a
normal $q$-complement and $G/K$ is a $q$-group, then $N\leq K$ and we get a
contradiction since $G/N$ is simple.

By the Frobenius theorem [5, Theorem 7.4.5], the group $G$ possesses a
$q$-subgroup $H$ such that the group $N_{G}(H)/C_{G}(H)$ is not a $q$-group.
Let $p\in\pi(N_{G}(H)/C_{G}(H))\setminus\\{q\\}$ and pick a $p$-element $a\in
N_{G}(H)\setminus C_{G}(H)$. Thus $[H,a]\neq 1$. Since $G/N$ is an EPPO-group,
$a$ acts fixed-point-freely on $Z(H)$. Let $Q$ be the subgroup generated by
elements of order $q$ of $Z(H)$. Note that $Q$ is elementary abelian and
$Q=[Q,a]$. ∎

Let $S_{n}$ and $A_{n}$ stand respectively for the symmetric and alternating
groups on $n$ symbols. It is well-known that for $n\neq 6$ every automorphism
of $S_{n}$ is inner. On the other hand, $S_{6}$ admits a nontrivial outer
automorphism, often called exceptional. Note that the exceptional automorphism
is not unique in the sense that if $\phi\in Aut\,S_{6}$ is exceptional and
$y\in S_{6}$, then $\phi y\in Aut\,S_{6}\setminus Inn\,S_{6}$. Furthermore,
$Aut\,S_{6}$ can be naturally identified with $Aut\,A_{6}$ and $Out\,A_{6}$ is
isomorphic to the Klein four-group. Slightly abusing terminology any
automorphism of $S_{6}$, which is not inner, will be called exceptional.

######  Lemma 4.13.

Let $\phi\in Aut\,S_{6}$ be an exceptional automorphism. There are elements
$x\in S_{6}\setminus A_{6}$ and $y\in A_{6}$ such that $[x,\phi y]$ has odd
order.

###### Proof.

Let $x=(123456)\in S_{6}\setminus A_{6}$. Up to a conjugation, we have
$x^{\phi}=(142)(56)$. Let $y=(456)$. Observe that $[x,\phi
y]=x^{-1}y^{-1}x^{\phi}y=(143)(256),$ and therefore $[x,\phi y]$ has order 3,
as desired. ∎

######  Proposition 4.14.

The commutator subgroup of an insoluble CPPO-group is perfect.

###### Proof.

Suppose that the result is false and let $G$ be a counterexample of minimal
order. Clearly, $G$ is not perfect so the minimality of $|G|$ implies that
$H=G^{\prime\prime}$ is a perfect group. Moreover, the minimality of $|G|$
also implies that $R(H)=1$. In short, $H$ is a nontrivial perfect CPPO-group
with trivial soluble radical. Corollary 4.9 shows that $H$ is a nonabelian
simple group.

Observe that $[R(G),H]\leq R(G)\cap H=1$, that is, $R(G)\leq C_{G}(H)$. We
claim that $R(G)=Z(G)$. Suppose that this is not the case and take elements
$a\in G$ and $b\in R(G)$ for which $[a,b]$ is a nontrivial $r$-element for
some prime number $r$.

Note that if $g\in H$ has the property that $ag$ normalizes an
$r^{\prime}$-subgroup $K$ of $H$, then $ag$ centralizes $K$. Indeed, for any
$h\in K$ the equality

$[bh,ag]=[b,a][h,ag]$ (4.1)

holds since $R(G)\leq C_{G}(H)$. It follows from (4.1) that $[bh,ag]$ is an
$r$-element. This implies that $[h,ag]$ is an $r$-element. On the other hand,
$ag$ normalizes $K$ so that $[h,ag]\in K$ is an $r^{\prime}$-element. It
follows that $[h,ag]=1$ and so $[K,ag]=1$.

We will now prove that $R(G)=Z(G)$. Let $q\in\pi(H)\setminus\\{r\\}$, and let
$K$ be a nontrivial $q$-subgroup of $H$. Choose a Sylow $q$-subgroup $Q$ of
$H$ containing $K$. By the Frattini argument there is an element $g\in H$ such
that $ag\in N_{G}(Q)$. In view of the above we conclude that $ag\in C_{G}(Q)$
and so $ag\in C_{G}(K)$. For an arbitrary element $x\in N_{H}(K)$ we have
$agx\in N_{G}(K)$ and the above argument shows that $agx\in C_{G}(K)$. It
follows that $x\in C_{H}(K)$. Thus, we proved that $N_{H}(K)=C_{H}(K)$ for any
$q$-subgroup $K$ of $H$. The Frobenius theorem now tells us that $H$ contains
a normal $q$-complement, which is impossible since $H$ simple. This
contradiction shows that $R(G)=Z(G)$, as claimed.

Taking into account Lemma 4.8 observe that $R(G)\neq 1$. Let $M$ be a soluble
minimal normal subgroup of $G$. The minimality of $|G|$ implies that
$G^{\prime}M=HM$. Since $H<G^{\prime}$, it follows that $M\leq G^{\prime}$ and
$G^{\prime}=H\times M$. Since $M$ is central, the order of $M$ is a prime, say
$p$. The minimality of $|G|$ implies that $M$ is a unique minimal soluble
normal subgroup. We deduce that $Z(G)$ is a $p$-group. Moreover, as $M\leq
Z(G)$ we have $G^{\prime}/H\leq Z(G/H)$, that is, $G/H$ is nilpotent of class
2.

Set $\overline{G}=G/Z(G)$. Observe that $R(\overline{G})=1$ and so by Lemma
4.8 $\overline{G^{\prime}}=\overline{H}$. Moreover,
$\overline{G}/\overline{H}$ is either cyclic or the Klein four-group.

If $\overline{G}/\overline{H}$ is cyclic, then there is an element $g\in G$
such that $G=\langle g\rangle Z(G)H$. In this case $G/H$ is abelian and so
$G^{\prime}=H$, a contradiction.

Therefore we assume that $\overline{G}/\overline{H}$ is the four-Klein group.
It follows from Lemma 4.6 that either $H\cong A_{6}$ or $H\cong PSL(3,4)$. Let
$a,b\in G$ be elements for which $G=\langle a,b\rangle Z(G)H$. Here
$[a,b]\notin H$ since $G/H$ is nonabelian. Thus, the equality
$G^{\prime}=H\times M$ implies that $[a,b]$ is a $p$-element. This happens for
every choice of $a$ and $b$. On the other hand, $G/H$ is nilpotent of class 2.
Since $\overline{G}/\overline{H}$ is the Klein four-group, it follows that
$p=2$. We can replace the element $b$ by $bh$, where $h\in H$. Therefore
$[a,bh]$ is a 2-element for any $h\in H$.

If $H\cong A_{6}$, by Lemma 4.13 $a$ and $b$ can be chosen in such a way that
there exists $h\in H$ with $[\overline{a},\overline{b}\overline{h}]$ of odd
order. This rules out the case $H\cong A_{6}$.

Therefore $H$ is isomorphic to $PSL(3,4)$. According to [9] in this case
$Aut\,H$ splits over $H$, that is,
$\overline{G}=\overline{H}\rtimes\langle\overline{a},\overline{b}\rangle$ with
$\langle\overline{a},\overline{b}\rangle$ isomorphic to the Klein group. The
Baer-Suzuki theorem [5, Theorem 3.8.2] guarantees that there is an element
$\overline{g}\in\overline{G}$ such that
$\langle\overline{a},\overline{a}^{\overline{g}}\rangle$ is not nilpotent.
Therefore there exists an odd order element
$\overline{y}\in\langle\overline{a},\overline{a}^{\overline{g}}\rangle$. Since
$\overline{G}/\overline{H}$ is a 2-group, it follows that $y$ can be chosen in
$H$. Observe that $\langle\overline{a},\overline{a}^{\overline{g}}\rangle$ is
a dihedral group, whence $\overline{y}^{\overline{a}}=\overline{y}^{-1}$.
Consequently, $1\neq[\overline{y},\overline{a}]=\overline{y}^{-2}$ has odd
order. On the other hand, $[a,b]\in Z(G)$ is a $2$-element. The equality
$[a,by]=[a,y][a,b]$ implies that $[a,y]$ is a 2-element. This is a
contradiction. The proof is complete. ∎

######  Corollary 4.15.

Let $G$ be an insoluble CPPO-group. Then $R(G^{\prime})=[G^{\prime},R(G)]$.

###### Proof.

By Proposition 4.14 $G^{\prime}$ is a perfect group. Thus,
$G^{\prime}/R(G^{\prime})$ is a nontrivial perfect CPPO-group with trivial
soluble radical. Corollary 4.9 shows that $G^{\prime}/R(G^{\prime})$ is a
simple group. On the other hand, observe that $G^{\prime}/[G^{\prime},R(G)]$
is a quasisimple CPPO-group. Lemma 4.11 shows that
$G^{\prime}/[G^{\prime},R(G)]$ is simple and so
$R(G^{\prime})=[G^{\prime},R(G)]$. ∎

The following lemma will be useful.

######  Lemma 4.16.

Let $G$ be a perfect group and $N\leq G$ a soluble normal subgroup such that
$G/N$ is simple. Let $Q$ be a nontrivial subgroup of $G$ such that $Q\not\leq
N$. Then $G=[G,Q]$.

###### Proof.

Note that $[G,Q]$ is a normal subgroup of $G$. Since $QN/N\neq 1$ and $G/N$ is
simple, it follows that $G=[G,Q]N$. We deduce that $G/[G,Q]$ is a perfect
soluble group, that is $G/[G,Q]=1$. Hence, $G=[G,Q]$. ∎

######  Proposition 4.17.

Let $G$ be an insoluble CPPO-group. Then $R(G^{\prime})\leq O_{2}(G)$ and
$G^{\prime}/R(G^{\prime})$ is a simple EPPO-group.

###### Proof.

We already know that $G^{\prime}/R(G^{\prime})$ is a simple EPPO-group.
Therefore, it suffices to show that $R(G^{\prime})\leq O_{2}(G)$. Assume that
this is false and let $G$ be a counterexample of minimal order. By Proposition
4.14 $G^{\prime}$ is a perfect group so the minimality of $|G|$ shows that $G$
is perfect. It follows that $G/R(G)$ is a nonabelian simple group.

Let $M$ be a soluble minimal normal subgroup of $G$. Then $M$ is an elementary
abelian $p$-subgroup for some prime number $p$. The minimality of $|G|$ yields
that $R(G)/M$ is a 2-group and, in particular, $p\neq 2$. Hence, $M=O_{p}(G)$
is a unique soluble minimal normal subgroup of $G$. In particular, $M=F(G)$.

Suppose first that $M=R(G)$. Lemma 4.12 says that there exists an elementary
abelian 2-subgroup $Q$ and an element $a\in G$ of odd prime power order such
that $Q=[Q,a]$. By virtue of Lemma 4.16 we deduce that $G=[G,Q]$.
Consequently, $Q$ does not centralize $R(G)$, otherwise we would have
$R(G)=Z(G)$ and Lemma 4.11 would imply that $R(G)=1$, a contradiction. Observe
that $(\langle a\rangle,Q,R(G))$ is a tower with abelian factors. Moreover,
$Q$ is an elementary abelian noncyclic group. Combining this with Lemma 3.1 we
deduce that $G$ has a commutator whose order is not a prime power, a
contradiction. We conclude that $M$ is a proper subgroup of $R(G)$ and hence
$R(G)/M$ is a nontrivial 2-group.

Since $O_{2}(G)=1$, the subgroup $M$ is not central in $R(G)$. The minimality
of $M$ implies that $M=[R(G),M]=[G,M]$. Again, using that $G$ is perfect and
that $M\nleq Z(G)$ we conclude that $C_{G}(M)=M$.

Let $q\in\pi(G)\setminus\\{2,p\\}$. By Lemma 4.12, there exists an elementary
abelian $q$-subgroup $Q$ and a prime power order element $a\in G$, whose order
is prime to $q$, such that $Q=[Q,a]$. Set $L=\langle a\rangle QR(G)$. Since
$M$ is self-centralizing, it follows that $(\langle a\rangle,Q,M)$ is a tower
of $L$. Now, Lemma 2.7 and Theorem 1.1 show that $h(L)=3$.

Taking into account that $Q=[Q,a]$ we observe that
$Q\leq\gamma_{\infty}(L)\leq F_{2}(L)$. Moreover, $M\leq F(L)$ and so $F(L)$
is a $p$-group. Since $R(G)/F(L)$ is a 2-group, $R(G)\leq F_{2}(L)$. Here
$F_{2}(L)$ denotes the second term of the upper Fitting series of $L$. Let $S$
be a Sylow 2-subgroup of $R(G)$. The subgroups $Q$ and $S$ are both contained
in $F_{2}(L)$ and so $[Q,S]\leq F(L)$. Hence, $[Q,S]\leq M$. This means that
$Q$ centralizes $R(G)$ modulo $M$. Keeping in mind that in view of Lemma 4.16
$G=[G,Q]$ we now deduce that $G$ centralizes $R(G)$ modulo $M$. In other
words, we proved that $R/M=Z(G/M)$. By Lemma 4.11 we conclude that $R/M=1$,
that is, $R=M$, a contradiction. The proof is complete. ∎

Remark that the combination of Proposition 4.14, Corollary 4.15, and
Proposition 4.17 establishes Theorem 1.2.

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