# Conjugacy classes of maximal cyclic subgroups

M. Bianchi, R.D. Camina, Mark L. Lewis & E. Pacifici

( )

###### Abstract

In this paper, we set $\eta(G)$ to be the number of conjugacy classes of
maximal cyclic subgroups of $G$. We consider $\eta$ and direct and semi-direct
products. We characterize the normal subgroups $N$ so that
$\eta(G/N)=\eta(G)$. We set $G^{-}=\\{g\in G\mid\langle
g\rangle{\rm~{}is~{}not~{}maximal~{}cyclic}\\}$. We show if $\langle
G^{-}\rangle<G$, then $G/\langle G^{-}\rangle$ is either (1) an elementary
abelian $p$-group for some prime $p$, (2) a Frobenius group whose Frobenius
kernel is a $p$-group of exponent $p$ and a Frobenius complement has order $q$
for distinct primes $p$ and $q$, or (3) isomorphic to $A_{5}$.
Keywords: direct product, exponent $p$, covering
2020 Mathematics Subject Classification: 20D15, 20E34

##  1 Introduction

Unless otherwise stated, all groups in this paper are finite. A covering of a
group $G$ is a set of proper subgroups $\\{H_{i}\\}$, called components, such
that $G\subseteq\bigcup_{i}H_{i}$. A covering is called irredundant if
removing any component means the set is no longer a covering. Identifying
coverings of groups and the minimal size of such a covering, called the
covering number, has a long mathematical history. It is an easy problem to
show that a group cannot be covered by $2$ subgroups. In 1926, Scorza
considered groups with a covering of size $3$, [16]; then, in 1994, Cohn
continued the investigation, considering groups with a minimal covering of
sizes $4$, $5$, and $6$ amongst other things, [8]. The case of $7$ is
considerably more difficult and surprising in that no such covering exists.
This was conjectured by Cohn and proved by Tomkinson. Tomkinson also
determined the coverings of solvable groups in [20]. Since then, many authors
have written on the theme; for an overview see [2] or [13]. At this time, the
research appears to be on determining the coverings of nonabelian simple
groups and the symmetric groups.

Recently, authors have considered normal coverings, that is coverings which
are invariant under $G$-conjugation (not coverings by normal subgroups). The
normal covering number is the smallest number of conjugacy classes of proper
subgroups in a normal covering of $G$. In [4], [5] and [6], the authors
consider the normal covering number for symmetric and alternating groups,
while in [11] the authors consider groups for which the normal covering number
is equal to the covering number.

Also, there has been interest in coverings where certain requirements are
imposed on the components of the covering. In this paper, we consider
coverings by cyclic subgroups up to conjugacy. For finite groups, it is not
difficult to see that every cyclic subgroup is contained in a maximal cyclic
subgroup. Furthermore, every covering by cyclic subgroups has to contain the
maximal cyclic subgroups since the only cyclic subgroups containing the
generators of the maximal cyclic subgroups will be the maximal cyclic
subgroups. Therefore, the set of the maximal cyclic subgroups of $G$ form the
only irredundant covering of $G$ by cyclic subgroups. Hence, it makes sense to
study the set of maximal cyclic subgroups of $G$. In particular, the set of
conjugacy classes of maximal cyclic subgroups form an irredudant normal
covering where all components are cyclic.

We note that in infinite groups cyclic subgroups need not be contained in
maximal cyclic subgroups (see the Prüfer group) so the question of irredundant
coverings of infinite groups by cyclic subgroups may be independent of the
question of maximal cyclic subgroups. This is being explored by the second and
third authors with Yiftach Barnea and Mikhail Ershov in [1].

Coverings of abelian groups by maximal cyclic subgroups have been studied by
Rogério in [15]. For nonabelian groups, von Puttkamer initiated the study of
the number of conjugacy classes of maximal cyclic subgroups in Section 5 of
his dissertation ([21]). In particular, von Puttkamer considers finite groups
$G$ that have two conjugacy classes of maximal cyclic subgroups. He proves
that such groups must be solvable, and in addition, he proves that they have
derived length at most $4$. In fact, he essentially classifies them. We pick
up the study of groups in terms of the number of conjugacy classes of maximal
cyclic subgroups in this paper. With this in mind, we set the following
definition.

######  Definition.

Let $G$ be finite group, we denote the number of conjugacy classes of maximal
cyclic subgroups of $G$ by $\eta(G)$.

In Section 2, we will consider $\eta(G)$ when $G$ is a direct product. We also
determine $\eta(G)$ when $G$ is a Frobenius group. For a group $G$, we define
$G^{-}$ to be the set of $g\in G$ such that $\langle g\rangle$ is not maximal
cyclic. In Section 4, we will show that there is a strong connection between
$\eta(G)$ and $G^{-}$. In Section 3, we consider some properties of $G^{-}$ as
a set.

In particular, we prove the following:

######  Theorem 1.1.

If $G$ is a group and $\langle G^{-}\rangle<G$, then $G/\langle G^{-}\rangle$
is either (1) a $p$-group with exponent $p$, (2) a Frobenius group whose
Frobenius kernel is a $p$-group with exponent $p$ and whose Frobenius
complements have order $q$ for some prime $q\neq p$, or (3) $A_{5}$.

We will see that we have stronger control on $G^{-}$ when $G$ is a $p$-group.
When $G$ is a $p$-group and $N$ is a normal subgroup so that $G/N$ has
exponent $p$, then $G^{-}\subseteq N$.

Let $N$ be a normal subgroup of $G$ then it is not hard to see that
$\eta(G/N)\leq\eta(G)$. In Theorem 4.1 we give criteria for when equality
occurs. We note that the criteria is in terms of $G^{-}$. We then show that
when $G$ is a $p$-group, there exists a characteristic subgroup $X(G)$ of $G$
such that $\eta(G)=\eta(G/X(G))$ and $X(G)$ is maximal under this condition.
We will give an example to see that we cannot find such a subgroup in general.

We close by proving a couple of results regarding the structure of $N$ when
$\eta(G/N)=\eta(G)$. The first considers the relationship between $N$ and the
derived subgroup, $G^{\prime}$.

######  Theorem 1.2.

Suppose $G$ is a group and $N$ is a subgroup so that $\eta(G)=\eta(G/N)$. If
every nontrivial Sylow subgroup of $G/G^{\prime}$ is noncyclic, then $N\leq
G^{\prime}$.

Finally, when $G$ is a $p$-group and a maximal cyclic subgroup is normal in
$G$, then we can prove that $X(G)$ is cyclic.

######  Theorem 1.3.

Suppose $G$ is a $p$-group for some prime $p$. If some maximal cyclic subgroup
of $G$ is normal, then $X(G)$ is cyclic.

##  2 $\eta$ and products

In this section, we consider the relationship between $\eta$ and direct
products and semi-direct products. In this next result we give some idea of
the relationship between $\eta$ and direct products.

######  Lemma 2.1.

Suppose $G=H\times K$, then the following are true:

  1. (i)

$\eta(G)\geq\eta(H)\eta(K)$.

  2. (ii)

If $(|H|,|K|)=1$, then $\eta(G)=\eta(H)\eta(K)$.

  3. (iii)

If $p$ is a prime so that $p$ divides $|K|$ and $H$ is a nontrivial $p$-group,
then $\eta(G)\geq\eta(H)\eta(K)+\eta_{p}(K)>\eta(H)\eta(K)$ where
$\eta_{p}(K)$ is the number of conjugacy classes of maximal cyclic subgroups
of $K$ whose order is divisible by $p$.

  4. (iv)

If $H$ is nilpotent and $p$ divides both $|H|$ and $|K|$, then
$\eta(G)>\eta(K)$.

  5. (v)

If $H$ and $K$ are both nontrivial $p$-groups, then
$\eta(G)\geq\eta(H)\eta(K)+\eta(H)+\eta(K)$.

Proof. Suppose $\langle h\rangle$ is a maximal cyclic subgroup of $H$ and
$\langle k\rangle$ is a maximal cyclic subgroup of $K$. Suppose that
$\langle(h,k)\rangle\leq\langle g\rangle$ for some element $g\in G$. We can
write $g=(h_{1},k_{1})$ for $h_{1}\in H$ and $k_{1}\in K$. Since
$(h,k)\in\langle g\rangle$, there is an integer $n$ so that $g^{n}=(h,k)$, and
hence, $h_{1}^{n}=h$ and $k_{1}^{n}=k$. It follows that $\langle
h\rangle\leq\langle h_{1}\rangle$ and $\langle k\rangle\leq\langle
k_{1}\rangle$. By maximality, we have $\langle h\rangle=\langle h_{1}\rangle$
and $\langle k\rangle=\langle k_{1}\rangle$, and so $\langle
g\rangle=\langle(h,k)\rangle$. Thus, $\eta(H)\eta(K)\leq\eta(G)$.

We now assume that $(|H|,|K|)=1$. Let $\langle g\rangle$ be a maximal cyclic
subgroup of $G$. We know $g=(h,k)$ for unique $h\in H$ and $k\in K$. Suppose
$\langle h\rangle\leq\langle h_{1}\rangle$ for some $h_{1}\in H$ and $\langle
k\rangle\leq\langle k_{1}\rangle$ for some $k_{1}\in K$. Since $o(h_{1})$ and
$o(k_{1})$ are coprime, it follows that $(h,1)$ and $(1,k)$ are powers of
$(h_{1},k_{1})$. Thus, $g=(h,1)(1,k)\in\langle(h_{1},k_{1})\rangle$, and so,
$\langle g\rangle\leq\langle(h_{1},k_{1})\rangle$. Maximality of $\langle
g\rangle$ implies that $\langle g\rangle=\langle(h_{1},k_{1})\rangle$, and so,
$\langle h\rangle=\langle h_{1}\rangle$ and $\langle k\rangle=\langle
k_{1}\rangle$. We conclude that $\langle h\rangle$ and $\langle k\rangle$ are
maximal cyclic subgroups. This implies that $\eta(G)\leq\eta(H)\eta(K)$, and
so we have $\eta(G)=\eta(H)\eta(K)$.

Next, we have that $H$ is a nontrivial $p$-group and $p$ divides $|K|$.
Consider an element $k\in K$ so that $\langle k\rangle$ is maximal cyclic in
$K$ and $p$ divides $o(k)$. We claim that $\langle(1,k)\rangle$ will be a
maximal cyclic subgroup of $H\times K$. Suppose there exists an element $g\in
G$ so that $g^{q}=(1,k)$ for some prime $q$ that divides $o(g)$. We can write
$g=(h^{*},k^{*})$ for $h^{*}\in H$ and $k^{*}\in K$. We see that
$(1,k)=g^{q}=((h^{*})^{q},(k^{*})^{q})$ and so, $(k^{*})^{q}=k$. We know that
$o(g)$ is the least common multiple of $o(h^{*})$ and $o(k^{*})$. Notice that
$o(k)$ divides $o(k^{*})$ and so, $p$ divides $o(k^{*})$. Since $o(h^{*})$ is
a power of $p$, we see that $q$ divides $o(k^{*})$ and this contradicts the
choice of $\langle k\rangle$ as a maximal cyclic subgroup of $K$. Notice that
$(1,k)$ is not conjugate to any of the subgroups included in the count in part
(i). This yields $\eta(G)\geq\eta(H)\eta(K)+\eta_{p}(K)$. By Cauchy’s theorem,
we know that $K$ has an element whose order is divisible by $p$, and so, $K$
has a maximal cyclic subgroup whose order is divisible by $p$, and so,
$\eta_{p}(K)\geq 1$. This proves (iii).

We now prove conclusion (iv). We can write $H=P\times Q$ where $P$ is the
Sylow $p$-subgroup of $H$ and $Q$ is the Hall $p$-complement. We then have
$G=(K\times Q)\times P$. By conclusion (iii), we have $\eta(G)>\eta(K\times
Q)$ and by conclusion (i), we have $\eta(K\times Q)\geq\eta(K)$. Combining
these, we have conclusion (iv). For conclusion (v), notice that every maximal
cyclic subgroup in either $H$ or $K$ will have order divisible by $p$. Thus,
if $h\in H$ and $k\in K$ satisfy that $\langle h\rangle$ is a maximal cyclic
subgroup of $H$ and $\langle k\rangle$ is a maximal cyclic subgroup of $K$,
then arguing as in (iii), we see that $\langle(h,1)\rangle$ and
$\langle(1,k)\rangle$ are maximal cyclic subgroups of $G$. Since these cannot
be conjugate to the maximal cyclic subgroups counted in (i), we obtain
$\eta(G)\geq\eta(H)\eta(K)+\eta(H)+\eta(K)$. $\Box$

We note that equality can hold in Lemma 2.1 even if $|H|$ and $|K|$ are not
relatively prime. Take $H=S_{3}$ and $K$ to be the dihedral group of order
$10$. Then $\eta(H\times K)=4$ and $\eta(H)=\eta(K)=2$. The various cases that
arise show that computing $\eta$ for direct products is not trivial when the
factors do not have coprime orders.

We expect the situation for computing $\eta$ of semi-direct products to be
even more complicated. We begin with the case of Frobenius groups which is
relatively straightforward.

######  Proposition 2.2.

Let $G=NH$ be a Frobenius group with Frobenius kernel $N$ and Frobenius
complement $H$. Then $\eta(G)=\eta^{*}(N)+\eta(H)$ where $\eta^{*}(N)$ is the
number of $H$-orbits on the $N$-conjugacy classes of maximal cyclic subgroups
of $N$.

Proof: Let $C$ be a maximal cyclic subgroup of $G$. We know that either $C\leq
N$ or some conjugate of $C$ lies in $H$. Furthermore, if $C\leq N$, then $C$
is a maximal cyclic subgroup of $G$ and if $C\leq H$, then $C$ is maximal
cyclic in $H$. Also, we know that if $C$ and $C^{g}$ lie in $H$, then $1<H\cap
H^{g}$ and so, $g\in H$. It follows that the number conjugacy classes of
maximal cyclic subgroups of $G$ will equal the number of $H$-conjugacy classes
of maximal cyclic subgroups in $H$ plus the number of $H$-orbits on the
$N$-conjugacy classes of maximal cyclic subgroups of $N$. $\Box$

In general, when $G$ is a semi-direct product of $H$ acting on $N$ where
$(|N|,|H|)=1$, we would like to bound $\eta(G)$ in terms of $\eta^{*}(N)$ and
$\eta(H)$. It is tempting to ask whether $\eta^{*}(N)+\eta(H)$ is the lower
bound for $\eta(G)$. It turns out the answer is no. Let $G$ be the semidirect
product of $D=D_{8}$ acting faithfully on $N=Z_{3}\times Z_{3}$. (In the
SmallGroups library in Magma, this is SmallGroup (72,50).) It is not difficult
to see that $D$ has two orbits on cyclic subgroups of order $3$ in $N$ and so,
$\eta^{*}(N)=2$. We know $\eta(D)=3$; so $\eta^{*}(N)+\eta(D)=5$. On the other
hand, one can see that the maximal cylic subgroups of $G$ have orders $4$ and
$6$, and that there is one conjugacy class of cyclic subgroups of order $4$
and two conjugacy classes of cyclic subgroups of order $6$; so $\eta(G)=3$. We
now provide a lower bound; however, we do not have any examples where this
lower bound is met.

######  Proposition 2.3.

Let $N$ be a normal subgroup of a group $G$ and let $\eta^{*}(N)$ be the
number of $G$-orbits on the $N$-conjugacy classes of maximal cyclic subgroups
of $N$. Then $\eta(G)\geq\eta^{*}(N)$. In particular,

  1. (i)

if $N$ is central in $G$ then $\eta(G)\geq\eta(N)$.

  2. (ii)

if $|G:N|=k$ then $\eta(G)\geq\eta(N)/k$.

Proof. Let $C_{1},\ldots,C_{m}$ be a set of representatives for the $G$-orbits
of the maximal cyclic subgroups of $N$. We know that each $C_{i}$ is contained
in some maximal cyclic subgroup of $G$. It suffices to show that if $i\neq j$,
then $C_{i}$ and $C_{j}$ are not $G$-conjugate to subgroups of the same
maximal cyclic subgroup of $G$. Without loss of generality, we may assume
$i=1$ and $j=2$. Suppose that $C_{1}$ and $C_{2}$ are $G$-conjugate to
subgroups of the same maximal cyclic subgroup $D$ of $G$. Replacing $C_{1}$
and $C_{2}$ by conjugates if necessary, we may assume that $C_{1}$ and $C_{2}$
are subgroups of $D$. Since $C_{1}$ and $C_{2}$ are contained in $N$, we see
that $C_{1}$ and $C_{2}$ are subgroups of $D\cap N$. Now, $D\cap N$ is a
cyclic subgroup of $N$ and $C_{1}$ and $C_{2}$ are maximal cyclic subgroups of
$N$; so we have $C_{1}=D\cap N=C_{2}$, but this contradicts the fact that
$C_{1}$ and $C_{2}$ are representatives of distinct orbits. $\Box$

We close by mentioning that the situation for subgroups is more complicated.
Consider the wreath product $C_{3}\wr C_{3}$ and its base group $C_{3}\times
C_{3}\times C_{3}$. Then $\eta(C_{3}\wr C_{3})=7$ but $\eta(C_{3}\times
C_{3}\times C_{3})=13$.

##  3 The set $G^{-}$

Let $G$ be a finite group. We define the set

$G^{-}=\\{g\in G:\langle g\rangle\;{\rm
is\;not\;a\;maximal\;cyclic\;group}\\}.$

For a prime $p$, we set $G^{\\{p\\}}=\\{g^{p}:g\in G\\}$.

We now give a characterization of this set.

######  Lemma 3.1.

Let $G$ be a group. Then

$G^{-}=\\{g^{q}\mid g\in G,q{\rm\;a\;prime\;dividing\;the\;order\;of\;}g\\}.$

In particular, if $G$ is a $p$-group then $G^{-}=G^{\\{p\\}}$.

Proof: Suppose $x\in\\{g^{q}\mid g\in G,q{\rm~{}a~{}prime~{}and~{}}q|o(g)\\}$.
There exists $y\in G$ so that $y^{q}=x$ for some prime $q$ with $q$ dividing
$o(y)$. We see that $\langle x\rangle<\langle y\rangle$, and thus, $x\in
G^{-}$. This yields $\\{g^{q}\mid g\in
G,q{\rm~{}a~{}prime~{}and~{}}q|o(g)\\}\subseteq G^{-}$. Now, suppose $x$ is in
$G^{-}$, so that $\langle x\rangle$ is a proper subgroup of $\langle h\rangle$
for some $h\in G$, so some prime $q$ must divide $|\langle h\rangle:\langle
x\rangle|$. We can then choose an element $y$ of $\langle h\rangle$ such that
$\langle y\rangle$ contains $\langle x\rangle$ and $|\langle y\rangle:\langle
x\rangle|=q$, so $\langle x\rangle=\langle y^{q}\rangle$ with $q$ dividing
$o(y)$. In particular, $o(x)=o(y)/q$. We know that $x=(y^{q})^{l}$ for some
integer $l$ that is relatively prime to $o(y^{q})=o(y)/q$. If $q$ does not
divide $l$, then $l$ is relatively prime to $o(y)$. We have that
$x=y^{ql}=(y^{l})^{q}$ and $q$ divides $o(y^{l})=o(y)$. We now suppose that
$q$ does divide $l$. This implies that $q$ does not divide
$m=o(y)/q=o(y^{q})$. Set $l^{*}=l+m$. Hence,
$(y^{q})^{l^{*}}=(y^{q})^{l+m}=y^{ql}=x$. We prove that $l^{*}$ is coprime to
$o(y)$. Since $q$ divides $l$ and not $m$, we see that $q$ does not divide
$l^{*}$. Suppose $p$ is a prime divisor of $m=o(y)/q$. Thus, $p$ does not
divide $l$; so again $p$ does not divide $l^{*}$. It follows that $l^{*}$ is
coprime to $o(y)$. We have $x=(y^{l^{*}})^{q}$ and $q$ divides $o(y^{l^{*}})$.
This proves the claim. In particular, we have $G^{-}\leq\\{g^{q}\mid g\in
G,q{\rm~{}a~{}prime~{}and~{}}q|o(g)\\}$, and equality follows. $\Box$

In this next result, we show that when $G$ is a $p$-group, $G^{-}$ is
contained in every normal subgroup whose quotient has exponent $p$. We also
show that it is rare when $G$ is not a $p$-group for $G^{-}$ to be contained
in a proper subgroup. Notice that Theorem 1.1 follows from conclusion (1).

######  Theorem 3.2.

Let $G$ be a group. Then the following statements are true:

  1. 1.

If $N$ is a normal subgroup of $G$, then $G^{-}\subseteq N$ if and only if
every element in $G\setminus N$ has prime power order and every element in
$G/N$ has prime order. In particular, $G/N$ is either (1) a $p$-group with
exponent $p$, (2) a Frobenius group whose Frobenius kernel is a $p$-group with
exponent $p$ and whose Frobenius complement has order $q$ for some prime
$q\neq p$, or (3) $A_{5}$.

  2. 2.

If $M$ is a normal subgroup of $G$ and $p$ is a prime so that $|G:M|=p$, then
$G^{-}\subseteq M$ if and only if every element in $G\setminus M$ has
$p$-power order.

  3. 3.

If $G$ is a $p$-group and $N$ is a normal subgroup so that $G/N$ has exponent
$p$, then $G^{-}\subseteq N$.

Proof: We begin by considering an element $g\in G$ whose order is divisible by
at least two primes, say $p$ and $q$. It follows that $g^{p}$ and $g^{q}$ lie
in $G^{-}$. It is not difficult to see that $\langle g\rangle=\langle
g^{p},g^{q}\rangle$, so $g\in\langle G^{-}\rangle$. It follows that if
$\langle G^{-}\rangle\leq N$ where $N$ is a normal subgroup of $G$, then every
element in $G\setminus N$ must have prime power order. Furthermore, if $p$ is
the prime that divides $o(g)$, then $g^{p}\in G^{-}\subseteq N$ and so,
$(gN)^{p}=N$. Hence, every element of $G/N$ has prime order. Conversely,
suppose $N$ is a normal subgroup of $G$ so that every element in $G\setminus
N$ has prime power order and every element in $G/N$ has prime order. Let $h\in
G^{-}$. We know that $h=g^{p}$ where $g\in G$ and $p$ is a prime that divides
$o(g)$. If $o(g)$ is also divisible by some prime other than $p$, then we
cannot have $g\in G\setminus N$, so $g$ and hence $h$ must lie in $N$. We may
now suppose that $o(g)$ is a power of $p$. If $g\in N$, then we are done, so
we assume $g\not\in N$. We see that $o(gN)$ divides $o(g)$ and is prime, we
have $o(gN)=p$ and so, $h=g^{p}\in N$. This proves that $G^{-}\subseteq N$.

The study of groups where all the elements have prime order has a long
history. It appears that they were first studied by Higman in [12]. This work
continued in [3], [9], [17], [18], and [19]. The study of such groups
culminated by being completely classified by Cheng, et. al in [7]. In
particular, they prove that if $G$ is a group where all elements have prime
order, then $G$ is one of the following: (i) a $p$-group of exponent $p$, (ii)
a Frobenius group whose Frobenius kernel is a $p$-group of exponent $p$ and
whose Frobenius complement has order $q$ for distinct primes $p$ and $q$, or
(iii) $A_{5}$. Applying this result to $G/N$ completes the proof of (1).

We have the condition that $M$ is normal in $G$ and $|G:M|=p$. Note that if
every element in $G\setminus M$ has $p$-power order, then (1) implies that
$G^{-}\subseteq M$. On the other hand, if $G^{-}\subseteq M$, then every
element $g\in G\setminus M$ has prime power order by (1) and $o(gM)=p$ which
implies that $p$ divides $o(g)$. It follows that every element in $G\setminus
M$ has $p$-power order. Suppose $G$ is a $p$-group and $N$ is a normal
subgroup so that $G/N$ has exponent $p$. Then we may apply (1) to see that
$G^{-}\subseteq N$. $\Box$

We note that there are $p$-groups $G$ where $G^{\\{p\\}}$ is a subgroup of $G$
and $p$-groups where $G^{\\{p\\}}$ is not a subgroup of $G$. We next show that
for $G^{-}$ to be a subgroup, it must be the case that all of the elements of
$G$ have prime power order.

######  Lemma 3.3.

Let $G$ be a group. If $G^{-}$ is a subgroup of $G$, then every element of $G$
must have prime power order.

Proof: Suppose $h$ is an element of $G$ that does not have prime power order.
Then there is an element $g$ so that $\langle g\rangle$ is maximal cyclic and
$h\in\langle g\rangle$. It follows that $o(g)$ is not a prime power. Hence,
there exist nonidentity elements $a,b$ that are powers of $g$ so that $g=ab$
and $(o(a),o(b))=1$. We see that $g\not\in G^{-}$, but $a,b\in G^{-}$, and
this implies that all elements of $G$ have prime power order. $\Box$

Higman in [12] classifies solvable groups where all elements have prime power
order. However, we present a different viewpoint to describe these groups. In
particular, if $G$ is a solvable group, then all the elements have prime power
order if and only if (1) $G$ is a $p$-group or (2) $G$ is either a Frobenius
group or a $2$-Frobenius group that is also a $\\{p,q\\}$-group for primes $p$
and $q$. Recall that a group $G$ is a $2$-Frobenius group if there exist
normal subgroups $L\leq K\leq G$ so that $G/L$ and $K$ are Frobenius groups
with Frobenius kernels $K/L$ and $L$, respectively. This can be seen using the
Grueneberg-Kegel graph (GK-graph) (sometimes called the prime graph). This
graph has the primes dividing $|G|$ as its vertices, and there is an edge
between $p$ and $q$ if there is some element $g\in G$ so that $pq$ divides
$o(g)$. The GK-graph of a group where all the elements have prime power order
will be an empty graph (i.e., a graph with no edges). Note that the GK-graph
of $G$ consists of a single vertex if and only if $G$ is a $p$-group. When $G$
is solvable, it is known that the GK-graph is disconnected if and only if $G$
is a Frobenius group or a $2$-Frobenius group and that it has two connected
components (see [22]). Assuming all elements of $G$ have prime power order,
this implies that $|G|$ is divisible by two primes $p$ and $q$.

Let $p$ and $q$ be primes so that $p$ divides $q-1$. It is not difficult to
see that there will exist a Frobenius group $G$ whose Frobenius kernel is
cyclic of order $q^{2}$ and a Frobenius complement has order $p$. It is easy
to see that $G^{-}$ is going to be the unique subgroup of order $q$. This
gives an example of a group $G$ where $G^{-}$ is a subgroup.

Now let $p$ and $q$ be primes so that $p^{2}$ divides $q-1$. In this case, we
take $G$ to be the Frobenius group whose Frobenius kernel has order $q$ and
whose Frobenius complement has order $p^{2}$. It is not difficult to see in
this case that $G^{-}$ will be the elements of order $p$ in $G$ along with
$1$, and that they do not form a subgroup. This gives an example where $G^{-}$
is not a subgroup.

Let $l(G)$ be the number of conjugacy classes of cyclic subgroups of $G$.
Notice that $\eta(G)\leq l(G)-1$ when $G$ is a nontrivial group. If every
element of $G$ has prime order, then every cyclic subgroup will be maximal
cyclic and so $\eta(G)=l(G)-1$. Conversely, if $l(G)-1=\eta(G)$, then every
nontrivial cyclic subgroup will be maximal cyclic and so will have prime
order. It will follow that every element of $G$ will have prime order. Hence,
$\eta(G)=l(G)-1$ if and only every element of $G$ has prime order. Note that
we have seen that every element of $G$ has prime order if and only if (1) $G$
is a $p$-group of exponent $p$, (2) $G$ is a Frobenius group where the
Frobenius kernel is a $p$-group of exponent $p$ and a Frobenius complement has
order $q$ for primes $p$ and $q$, or (3) $G\cong A_{5}$.

##  4 $\eta$ and quotients

In this next result we show that $\eta(G/N)\leq\eta(G)$ when $N$ is a normal
subgroup. We then give necessary and sufficient conditions for
$\eta(G/N)=\eta(G)$. We also refine the necessary and sufficient conditions
when $G$ is a $p$-group. Let $S$ and $T$ be subsets of $G$, then we set
$ST=\\{st\mid s\in S,t\in T\\}$. When $N$ is a normal subgroup of $G$, we
write $SN/N=\\{sN\mid s\in S\\}$ as a subset of $G/N$.

######  Theorem 4.1.

Let $N$ be a proper normal subgroup of $G$. Then

  1. 1.

$\eta(G/N)\leq\eta(G)$.

  2. 2.

$\eta(G/N)=\eta(G)$ if and only if the following conditions hold:

    1. (a)

$N\subseteq G^{-}$,

    2. (b)

$(G/N)^{-}=\\{gN\in G/N\mid gN\subseteq G^{-}\\}$, and

    3. (c)

Every element $x\in G\setminus G^{-}$ satisfies the condition that every
element of $xN\setminus G^{-}$ is conjugate to a generator of $\langle
x\rangle$.

  3. 3.

If $G$ is a $p$-group and $\eta(G/N)=\eta(G)$, then $G^{-}$ is a union of
$N$-cosets.

  4. 4.

$\eta(G/N)=\eta(G)$ and $G^{-}$ is a union of $N$-cosets if and only if for
every $x\in G\setminus G^{-}$ every element of $xN$ is conjugate to a
generator of $\langle x\rangle$.

  5. 5.

If $\eta(G/N)=\eta(G)$ and $G^{-}$ is a union of $N$-cosets, then the
following hold:

    1. (a)

$G^{-}N=G^{-}$, and

    2. (b)

$(G/N)^{-}=G^{-}N/N$.

Proof. Fix an element $x\in G$. First note $\langle x\rangle$ is maximal
cyclic if and only if $x\in G\setminus G^{-}$. Similarly, $\langle xN\rangle$
is maximal cyclic if and only if $xN\in G/N\setminus(G/N)^{-}$.

Suppose $\langle xN\rangle$ is maximal cyclic in $G/N$. Let $\langle g\rangle$
be a maximal cyclic subgroup containing $\langle x\rangle$. It follows that
$\langle xN\rangle\leq\langle gN\rangle$. By the maximality of $\langle
xN\rangle$, we have that $\langle xN\rangle=\langle gN\rangle$. Let
$x_{1},\dots,x_{n}\in G$ be chosen so that $\langle
x_{1}N\rangle,\dots,\langle x_{n}N\rangle$ are representatives of the
conjugacy classes of maximal cyclic subgroups of $G/N$. For each $i$, choose
$g_{i}\in G$ so that $\langle g_{i}\rangle$ is a maximal cyclic subgroup of
$G$ containing $\langle x_{i}\rangle$. The work above shows that $\langle
x_{i}N\rangle=\langle g_{i}N\rangle$. Suppose $\langle g_{i}\rangle$ and
$\langle g_{j}\rangle$ are conjugate for $1\leq i,~{}j\leq n$. Hence, there
exists $y\in G$ so that $\langle g_{j}\rangle=\langle g_{i}\rangle^{y}$. This
implies that $\langle x_{j}N\rangle=\langle g_{j}N\rangle=\langle
g_{i}N\rangle^{yN}=\langle x_{i}N\rangle^{yN}$. Since the $x_{i}$’s were
chosen to be representatives, we see that $i=j$. Because the $\langle
g_{i}\rangle$’s form a subset of a set of representatives of the conjugacy
classes of maximal cyclic subgroups of $G$, we have $\eta(G/N)\leq\eta(G)$.

Now assume $\eta(G/N)=\eta(G)$. I.e., we may assume that $\langle
g_{1}\rangle,\dots,\langle g_{n}\rangle$ are representatives of the maximal
cyclic subgroups of $G$. Suppose there exists $x\in N\setminus G^{-}$. Then
$\langle x\rangle$ is a maximal cyclic subgroup of $G$. Notice that
$g_{1},\dots,g_{n}$ all lie outside of $N$. Thus, $\langle x\rangle$ will not
be conjugate to any $\langle g_{i}\rangle$, a contradiction. Hence, we must
have that $N\subseteq G^{-}$.

We now work to show $(G/N)^{-}=\\{gN\mid gN\subseteq G^{-}\\}$. Suppose that
$xN\in(G/N)^{-}$ and suppose there exists $y\in xN\setminus G^{-}$. It follows
that $\langle y\rangle$ is maximal cyclic in $G$ and $\langle yN\rangle$ is
not maximal cyclic in $G/N$. In particular, $\langle y\rangle$ cannot be
conjugate to one of the $\langle g_{i}\rangle$’s, contradiction. Thus, we have
$xN\subseteq G^{-}$. Conversely, suppose now that $xN\subseteq G^{-}$. We know
that if $xN\not\in(G/N)^{-}$, then $\langle xN\rangle$ is maximal cyclic in
$G/N$. We have seen $\langle xN\rangle$ is conjugate to $\langle
g_{i}N\rangle$ for some $i$, and so, there exists $g\in G\setminus G^{-}$ so
that $xN=gN$, and we do not have $xN\subseteq G^{-}$, a contradiction. Thus,
we have that $(G/N)^{-}=\\{gN\mid gN\subseteq G^{-}\\}$.

We now prove that if $x\in G\setminus G^{-}$ and $y\in xN\setminus G^{-}$,
then $y$ is conjugate to a generator of $\langle x\rangle$. We have that
$\langle x\rangle$ and $\langle y\rangle$ are maximal cyclic subgroups of $G$
and $\langle xN\rangle=\langle yN\rangle$ is a maximal cyclic subgroup of
$G/N$. Using the fact that $\eta(G)=\eta(G/N)$, this implies that $\langle
x\rangle$ and $\langle y\rangle$ must be conjugate to the same $\langle
g_{i}\rangle$ and so, they are conjugate to each other. We conclude that $y$
must be conjugate to a generator of $\langle x\rangle$.

Conversely, assume that $N\subseteq G^{-}$, $(G/N)^{-}=\\{gN\mid gN\subseteq
G^{-}\\}$ and every element $x\in G\setminus G^{-}$ satisfies the condition
that every element of $xN\setminus G^{-}$ is conjugate to a generator of
$\langle x\rangle$. Define a map from the maximal cyclic subgroups of $G$ to
the maximal cyclic subgroups of $G/N$ by $C\mapsto CN/N$. Suppose $C$ is a
maximal cyclic subgroup of $G$. Then $C=\langle x\rangle$ for some $x\in
G\setminus G^{-}$. It follows that $xN\in G/N\setminus(G/N)^{-}$. We see that
$CN/N=\langle xN\rangle$ is a maximal subgroup of $G/N$. Also, if $C=\langle
x\rangle=\langle y\rangle$, then $\langle xN\rangle=\langle yN\rangle$, so the
map is well-defined. Suppose $D$ is a maximal cyclic subgroup of $G/N$. Then
$D=\langle xN\rangle$ where $xN\in G/N\setminus(G/N)^{-}$. As we have seen,
this implies that $xN\setminus G^{-}$ contains some element $g$, and thus,
$\langle g\rangle$ is a maximal cyclic subgroup of $G$. Also, $\langle
g\rangle N/N=\langle gN\rangle=\langle xN\rangle=D$. This implies that the map
is onto. Finally, suppose that $x,y\in G\setminus G^{-}$ yields $\langle
xN\rangle=\langle yN\rangle$. This implies that some generator of $\langle
y\rangle$ lies in $xN$. By hypothesis, this generator will be conjugate to a
generator of $\langle x\rangle$. This implies that $\langle x\rangle$ and
$\langle y\rangle$ are conjugate. Hence, our map yields a bijection of
conjugacy classes, and we conclude that $\eta(G)=\eta(G/N)$.

We now suppose that $G$ is a $p$-group and assume $N$ is a normal subgroup so
that $\eta(G/N)=\eta(G)$. Suppose $x\in G\setminus G^{\\{p\\}}=G\setminus
G^{-}$. The above work shows that $xN\in G/N\setminus(G/N)^{\\{p\\}}$. Let
$y\in xN$ and suppose $y\in G^{\\{p\\}}$, so $y=g^{p}$ for some element $g\in
G$. This implies that $xN=yN=g^{p}N=(gN)^{p}$ and this contradicts $xN\in
G/N\setminus(G/N)^{\\{p\\}}$. We conclude that $xN\cap G^{-}$ is empty. This
implies that $G\setminus G^{-}$ and $G^{-}$ are both unions of cosets of $G$.

If $G^{-}$ is a union of $N$-cosets and $\eta(G/N)=\eta(G)$, then when $x\in
G\setminus G^{-}$, we must have that $xN\cap G^{-}$ is empty, and so, every
element $y\in xN$ lies in $G\setminus G^{-}$ and so applying conclusion
(2)(c), we have that $y$ is conjugate to a generator of $\langle x\rangle$.
Conversely, suppose for every $x\in G\setminus G^{-}$ that every element of
$xN$ is conjugate to a generator of $\langle x\rangle$. Consider the element
$n\in N\setminus\\{1\\}$. Since $\langle n\rangle$ cannot be conjugate to
$\langle 1\rangle$, we must have that $n\in G^{-}$, and so, $N\subseteq
G^{-}$. If $x\in G\setminus G^{-}$, then every generator of $\langle x\rangle$
will lie in $G\setminus G^{-}$. Since every element of $xN$ is conjugate to a
generator of $\langle x\rangle$, we have that $xN\subseteq G\setminus G^{-}$,
and we deduce that $G\setminus G^{-}$ and hence, $G^{-}$ is a union of
$N$-cosets. It follows that $G^{-}N\subseteq G^{-}$, and we deduce that
$G^{-}N=G^{-}$. This implies that $(G/N)^{-}=\\{gN\mid gN\subseteq
G^{-}\\}=\\{gN\mid g\in G^{-}\\}=G^{-}N/N$. It follows that (a), (b), and (c)
of (2) are met, so $\eta(G)=\eta(G/N)$. Notice that we have proved both (4)
and (5). $\Box$

We note that we cannot completely drop the hypothesis that $G$ is a $p$-group
in conclusion (3). Let $G$ be the dihedral group of order $30$. Let $M$ be the
normal subgroup of order $3$ and $N$ the normal subgroup of order $5$. It
follows that $G/M$ is a dihedral group of order $10$ and $G/N$ is a dihedral
group of order $6$. We see that $\eta(G/M)=\eta(G/N)=\eta(G)=2$. On the other
hand, it is not difficult to see that $G^{-}=N\cup M$, and $G^{-}$ is not a
union of $N$-cosets or $M$-cosets. If $x\in NM\setminus G^{-}$, then it is not
difficult to see that neither $xN\cap G^{-}$ nor $xM\cap G^{-}$ is empty.
Also, $G^{-}N=NM=G^{-}M$.

In Section 2, we considered the case where $G$ is the semi-direct product of
$H$ acting on $N$. Using Lemma 2.3, we saw that $\eta(G)\geq\eta^{*}(N)$.
Since $H\cong G/N$, we can use Theorem 4.1 (1) to see that
$\eta(G)\geq\eta(G/N)=\eta(H)$. Notice that the examples of $D_{8}$ acting on
$Z_{3}\times Z_{3}$ is an example where $\eta(G)=\eta(H)$.

For $p$-groups, we can find a largest normal subgroup such that $\eta$ of the
quotient is equal to $\eta$ of the group. In this situation, we also cannot
completely drop the hypothesis that $G$ is a $p$-group from Corollary 4.2.
Again, let $G$ be the dihedral group of order $30$, let $M$ be the normal
subgroup of order $3$, and let $N$ the normal subgroup of order $5$. We have
seen that $\eta(G/M)=\eta(G/N)=\eta(G)=2$. Notice that $G/NM$ is cyclic of
order $2$, and so, $\eta(G/MN)=1$.

######  Corollary 4.2.

Let $G$ be a noncyclic finite $p$-group for some prime $p$.

  1. (i)

Let $N$ and $M$ be normal subgroups of $G$. If $\eta(G)=\eta(G/N)=\eta(G/M)$,
then $\eta(G)=\eta(G/NM)$.

  2. (ii)

There exists a characteristic subgroup $X$ of $G$ so that $\eta(G)=\eta(G/X)$
and if $N$ is a normal subgroup $G$ so that $\eta(G/N)=\eta(G)$, then $N\leq
X$.

Proof. (i) As $N$ and $M$ both lie inside $G^{-}$ by Theorem 4.1, it follows
that $NM\subseteq G^{-}$ again by Theorem 4.1. Suppose now that $g\in
G\setminus G^{-}$. We need to show that every element in $gMN$ is conjugate to
a generator of $\langle g\rangle$. An arbitrary element of $gMN$ has the form
$gmn$ where $m\in M$ and $n\in N$. Applying Theorem 4.1 to $M$ in $G$, we see
that every element of $gM$ is conjugate to a generator of $\langle g\rangle$.
In particular, $\langle gm\rangle$ is conjugate to $\langle g\rangle$. This
implies that $gm\in G\setminus G^{-}$. We then apply Theorem 4.1 to $N$ in $G$
to see that every element in $gmN$ is conjugate to a generator of $\langle
gm\rangle$. We obtain $\langle gmn\rangle$ is conjugate to $\langle
gm\rangle$. Using the transitivity of conjugate subgroups, we see that
$\langle g\rangle$ and $\langle gmn\rangle$ are conjugate. We conclude that
$gnm$ is conjugate to a generator of $\langle g\rangle$ as desired. We have
now shown that $NM$ satisfies the conditions of Theorem 4.1, and so,
$\eta(G/MN)=\eta(G)$.

(ii) Let $\mathcal{M}=\\{M\unlhd G\mid\eta(G/M)=\eta(G)\\}$. Let
$X=\prod_{M\in\mathcal{M}}M$. It is not difficult to see that $X$ is
characteristic in $G$ and if $N$ is normal with $\eta(G/N)=\eta(G)$, then
$N\leq X$. Since $|G|$ is finite, we can use (i) inductively to see that
$\eta(G/X)=\eta(G)$. $\Box$

Our next corollary states that $N\leq G^{\prime}$ when $\eta(G)=\eta(G/N)$ and
every Sylow subgroup of $G/G^{\prime}$ is noncyclic. Note that if $G$ is a
noncyclic $p$-group, then $G/G^{\prime}$ is noncyclic. It follows that this
next corollary applies to $p$-groups.

We now show that the hypothesis that $G/G^{\prime}$ is not cyclic is
necessary. Let $M$ be a cyclic group of order $3$ and let $H$ be a cyclic
group of order $4$. Note that $H$ has a nontrivial action on $M$ and let $G$
be the resulting semidirect product. Take $N$ to be the subgroup of order $2$
in $H$ and note that $N=Z(G)$. It is easy to see that $M=G^{\prime}$. Note
that $G/N\cong S_{3}$, and so, $\eta(G/N)=2$. We claim that $\eta(G)=2$, also.
To see this, observe that $MN$ is cyclic of order $6$. Suppose $g\in
G\setminus MN$. It follows that $gN\in G/N\setminus MN/N$. Since $G/N$ is
isomorphic to $S_{3}$, we see that $gN$ is conjugate to $hN$ where $h$ is the
generator of $H$. Since $N\leq H$, this implies that $g$ is conjugate to
either $h$ or $h^{3}$. We conclude that $\eta(G)=2$. Finally, note that $N$ is
not contained in $G^{\prime}=M$. We now present a second example to see that
we need to assume that all Sylow subgroups of $G/G^{\prime}$ are noncyclic,
take $K$ to be a group so that $K/K^{\prime}$ is noncyclic, and take $C$ to be
a cyclic $p$-group for a prime $p$ so that $p$ does not divide $|K|$. Let
$G=K\times C$. We see that $\eta(G/C)=\eta(K)=\eta(K\times C)=\eta(G)$ and $C$
is not contained in $G^{\prime}$. This is Theorem 1.2.

######  Corollary 4.3.

Suppose $G$ is a group such that every nontrivial Sylow subgroup of
$G/G^{\prime}$ is not cyclic and $N$ a normal subgroup such that
$\eta(G)=\eta(G/N)$. Then $N\leq G^{\prime}$.

Proof. Let $\Phi/G^{\prime}=\Phi(G/G^{\prime})$, so $\Phi$ is the preimage of
the Frattini subgroup of $G/G^{\prime}$. We claim that $N\leq\Phi$. To obtain
a contradiction, we suppose that $N$ is not contained in $\Phi$. Let
$M=N\cap\Phi$, and we have $M<N$. We know that $G/\Phi$ is the direct product
of elementary abelian $p$-groups for various primes $p$. In particular, there
is a subgroup $H$ so that $G=H(N\Phi)$ and $H\cap N\Phi=\Phi$. This implies
that $\Phi\leq H$ and so, $G=NH$. Also, $H\cap N=H\cap(N\Phi\cap N)=(H\cap
N\Phi)\cap N=\Phi\cap N=M$. It follows that $G/M=H/M\times N/M$. Observe that
$G/N\cong H/M$ and $G/H\cong N/M$. We see that $N/M$ is nontrivial. If $N/M$
is not cyclic, then by Lemma 2.1 (i), we have
$\eta(G/M)\geq(\eta(N/M))(\eta(H/M))>\eta(H/M)$. On the other hand, suppose
$N/M$ is cyclic, and let $p$ be a prime that divides $|N:M|$. Observe that
$N/M\cong N\Phi/\Phi$ and $G/\Phi=N\Phi/\Phi\times H/\Phi$. Since we are
assuming no Sylow subgroup of $G/G^{\prime}$ is cyclic, we see that no Sylow
subgroup of $G/\Phi$ is cyclic. It follows that $p$ must divide $|H:\Phi|$,
and so, $p$ divides $|H:M|$. We now apply Lemma 2.1 (iv) to see that
$\eta(G/M)>\eta(H/M)$. In both cases, we have proved that
$\eta(G/M)>\eta(H/M)$. By Theorem 4.1, we have
$\eta(G)\geq\eta(G/M)>\eta(H/M)=\eta(G/N)$. This however contradicts
$\eta(G)=\eta(G/N)$. Thus, we must have that $N\leq\Phi$.

Suppose $N$ is not contained in $G^{\prime}$. Then we can find $n\in
N\setminus G^{\prime}$ so that $nG^{\prime}$ has order $p$ in $G/G^{\prime}$
for some prime $p$. Since $G/G^{\prime}$ is not cyclic, we know that $G/\Phi$
is not cyclic. Hence, we can find $g\in G\setminus\Phi$, so that $\langle
nG^{\prime}\rangle$ is not contained in $\langle gG^{\prime}\rangle$. In
particular, $\langle gG^{\prime}\rangle$ and $\langle gnG^{\prime}\rangle$ are
distinct subgroups of $G/G^{\prime}$. Notice that $g,gn\in
G\setminus\Phi\subseteq G\setminus G^{-}$. By Theorem 4.1, we know that
$\eta(G)=\eta(G/N)$ implies that every element in $gN\setminus G^{-}$ is
conjugate to a generator of $\langle g\rangle$. This implies that $gn$ is
conjugate to a generator of $\langle g\rangle$. Hence, $gnG^{\prime}$ is
conjugate to an element of $\langle gG^{\prime}\rangle$. Since $G/G^{\prime}$
is abelian and $\langle gnG^{\prime}\rangle$ and $\langle gG^{\prime}\rangle$
are distinct subgroups of $G/G^{\prime}$, this is a contradiction. $\Box$

Since it is well known when $G$ is a noncyclic $p$-group that $G/G^{\prime}$
is not cyclic, Corollary 4.3 yields:

######  Corollary 4.4.

If $G$ is a noncyclic $p$-group and $N$ is a normal subgroup so that
$\eta(G/N)=\eta(G)$, then $N\leq G^{\prime}$.

Applying Theorem 4.1 to $p$-groups of exponent $p$, we see that $\eta(G)$
grows proportionally to $\log_{p}(|G|)$.

######  Corollary 4.5.

Suppose $n\geq 2$ and $G$ is a $p$-group of order $p^{n}$ and exponent $p$.
Then $\eta(G)\geq n+p-1$.

Proof. If $|G|=p^{2}$ then $\eta(G)=p+1$. We proceed by induction on the order
of $G$ and assume $|G|\geq p^{3}$. Suppose $z$ is a central element of $G$,
then $\eta(G/\langle z\rangle)<\eta(G)$ by Theorem 4.1. Furthermore,
$G/\langle z\rangle$ is a noncyclic $p$-group of exponent $p$. The result
follows.$\Box$

A covering $\\{H_{i}\\}$ of a group $G$ is called a partition if $H_{i}\cap
H_{j}$ is trivial for all $i\neq j$. (I.e., $H_{i}\cap H_{j}=1$ for $i\neq
j$.) Also, a group $G$ is tidy if for all $x\in G$ the set ${\rm
Cyc}_{G}(x)=\\{y\in G:\langle x,y\rangle\;\;{\rm is}\;{\rm cyclic}\\}$ is a
subgroup of $G$. Suppose $G$ is a finite $p$-group and $G$ has a covering that
consists of cyclic groups that is also a partition. Clearly, a cyclic group is
tidy, so it follows from [10, Corollary 2.5] that $G$ is tidy. The tidy
$p$-groups have been classified in [14]; they are either (i) cyclic, (ii) have
exponent $p$ or (iii) $p=2$ and $G$ is dihedral or generalized quaternion.
Hence, if $G$ is a partitioned $p$-group, then either $G$ is cyclic and
$\eta(G)=1$, $G$ is dihedral or generalized quaternion and $\eta(G)=3$, or $G$
has exponent $p$ and $\eta(G)\geq\log_{p}(|G|)+p-1$.

The following result gives a dichotomy of normal subgroups. Our proof requires
that $G$ be a $p$-group, but we do not have any counterexamples when $G$ is
not a $p$-group, so it may be possible to weaken the hypothesis on this
corollary. Finally, we have Theorem 1.3.

######  Corollary 4.6.

Let $G$ be a noncyclic $p$-group and $N$ a nontrivial normal subgroup of $G$
such that $\eta(G)=\eta(G/N)$.

  1. (i)

Suppose $M\unlhd G$ then either $N\leq M$ or $M\leq G^{-}$.

  2. (ii)

If there exists a maximal cyclic group $\langle x\rangle$ which is normal,
then $N\leq\langle x\rangle$; so $N$ is cyclic.

Proof. (i) Suppose that $N$ is not contained in $M$ and $M$ is not contained
in $G^{-}$. Then we can find $n\in N\setminus M$ and $m\in M\setminus G^{-}$.
It follows that $m\in M$ and $mn$ is not in $M$. Since $M$ is normal, $mn$
cannot be conjugate to any generator of $\langle m\rangle$. Since $m$ is not
in $G^{-}$, we should have that every element of $mN$ is conjugate to a
generator of $\langle m\rangle$, and $mn$ is not; so we have a contradiction
to Theorem 4.1.

(ii) We are assuming that $\langle x\rangle$ is normal in $G$ and $x\in
G\setminus G^{-}$. Thus, $\langle x\rangle$ is not contained in $G^{-}$. So by
(i) we have $N\leq\langle x\rangle$. $\Box$

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Mariagrazia Bianchi: Dipartimento di Matematica F. Enriques, Università degli
Studi di Milano, via Saldini 50, 20133 Milano, Italy.
<EMAIL_ADDRESS>
Rachel D. Camina: Fitzwilliam College, Cambridge, CB3 0DG, UK.
<EMAIL_ADDRESS>
Mark L. Lewis: Department of Mathematical Sciences, Kent State University,
Kent, Ohio, 44242 USA.
<EMAIL_ADDRESS>
Emanuele Pacifici: Dipartimento di Matematica e Informatica ”U. Dini” (DIMAI),
Universitá degli Studi di Firenze, Viale Morgagni 67/A, 50134 Firenze, Italy.
<EMAIL_ADDRESS>