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# A Frobenius group analog for Camina triples

Shawn T. Burkett  Department of Mathematical Sciences, Kent State University,
Kent, Ohio 44240, U.S.A<EMAIL_ADDRESS>and  Mark L. Lewis  Department of
Mathematical Sciences, Kent State University, Kent, Ohio 44240, U.S.A.
<EMAIL_ADDRESS>
###### Abstract.

Frobenius groups are an object of fundamental importance in finite group
theory. As such, several generalizations of these groups have been considered.
Some examples include: A Frobenius–Wielandt group is a triple $(G,H,L)$ where
$H/L$ is almost a Frobenius complement for $G$; A Camina pair is a pair
$(G,N)$ where $N$ is almost a Frobenius kernel for $G$; A Camina triple is a
triple $(G,N,M)$ where $(G,N)$ and $(G,M)$ are almost Camina pairs. In this
paper we study triples $(G,N,M)$ where $(G,N)$ and $(G,M)$ are almost
Frobenius groups.

##  1\. Introduction

Frobenius groups are ubiquitous in finite group theory as they arise very
naturally in a variety of situations. In addition, Frobenius groups have been
generalized in a number of different ways. In this paper, we present a new
generalization of Frobenius groups.

Recall that if $G$ is a Frobenius group, then there is a normal subgroup $N$
that has complement $H$ in $G$ so that $(|N|,|H|)=1$ and $C_{G}(x)\leq N$ for
all $x\in N\setminus\\{1\\}$. In particular, every element of $G\setminus N$
acts fixed-point-freely on $N$. One key property of Frobenius groups is that
the conjugacy class of any element $g\in G\setminus N$ is a union of cosets of
$N$. This is the basis of the generalization of Camina pairs that have been
studied widely in the literature.

Recall that the pair $(G,N)$ is called a Camina pair if $N$ is a proper,
nontrivial normal subgroup of $G$ where every element $g\in G\setminus N$ has
the property that its conjugacy class is a union of cosets of $N$. It is not
difficult to see that every Frobenius group with its Frobenius kernel yields a
Camina pair. In fact, it is possible to characterize the Camina pairs that are
Frobenius groups. If $(G,N)$ is a Camina pair, then $G$ is a Frobenius group
with Frobenius kernel $N$ if and only if $(|G:N|,|N|)=1$ and if and only if
there is a complement $H$ for $N$ in $G$.

Camina pairs have been generalized further to Camina triples. A Camina triple
is a triple $(G,N,M)$ where $M\leq N$ are nontrivial proper normal subgroups
of $G$ so that every element $g\in G\setminus N$ satisfies the property that
its conjugacy class is a union of cosets of $M$. Camina triples motivate our
generalization of Frobenius groups.

To be precise, we study groups $G$ with two nontrivial proper normal subgroups
$M$ and $N$, with $M\leq N$, such that every element of $G$ lying outside of
$N$ acts fixed-point-freely on $M$. Observe that if $M=N$ then $G$ is
Frobenius group with Frobenius kernel $N$. We will call such a triple
$(G,N,M)$ a Frobenius triple. In our first theorem, we show that Frobenius
triples play the same role for Camina triples that Frobenius groups play for
Camina pairs.

######  Theorem A.

Let $(G,N,M)$ be a Camina triple. The following are equivalent:

  1. (1)

$(G,N,M)$ is a Frobenius triple;

  2. (2)

$(\lvert G:N\rvert,\lvert M\rvert)=1$;

  3. (3)

There exists a subgroup $H\leq G$ so that $G=HN$ and $H\cap M=1$.

Although Frobenius triples and Frobenius groups are defined similarly, we will
see that the structure of Frobenius triples is far less restricted. For
example, we will see that there exist Frobenius triples $(G,N,M)$ where $M$
has arbitrarily large Fitting height, whereas Frobenius kernels are always
nilpotent. We will also see that one has no control over the structure of the
quotient $N/M$. In fact, any group can be a direct factor of $N/M$ when
$(G,N,M)$ is a Frobenius triple.

In [18], Wieldandt studies finite groups $G$ possessing a subgroup $H$ for
which every conjugate of $H$ intersects $H$ within some fixed normal subgroup
$L$ of $H$. These triples $(G,H,L)$ have come to be known as
Frobenius–Wielandt triples. We will show that Frobenius–Wielandt groups and
Frobenius triples are intimately connected.

######  Theorem B.

Let $M$ and $N$ be nontrivial, proper normal subgroups of $G$ satisfying
$M\leq N$. Let $H\leq G$ satisfy $G=HM$ and $H\cap M=1$. Then $(G,N,M)$ is a
Frobenius triple if and only if $(G,H,H\cap N)$ is a Frobenius–Wielandt
triple.

Frobenius groups have a particularly nice structure from a representation-
theoretic perspective as every irreducible character of a Frobenius group $G$
with Frobenius kernel $N$ either has $N$ in its kernel or is induced from a
nonprincipal (irreducible) character of $N$. In fact, this is a condition that
characterizes the Frobenius kernel. Camina pairs and Camina triples have
similar definitions in terms of certain characters inducing homogeneously.

######  Theorem C.

Let $M$ and $N$ be nontrivial, proper normal subgroups of $G$ satisfying
$M\leq N$. If $(\lvert G:N\rvert,\lvert M\rvert)=1$, then $(G,N,M)$ is a
Frobenius triple if and only if $\psi^{G}$ is irreducible for every character
$\psi\in\mathrm{Irr}(N\mid M)$.

As a straightforward application of Theorems B and C, we can describe the
character theory of a Frobenius–Wielandt group $(G,H,L)$ in the situation that
$H$ has a normal complement.

######  Theorem D.

Let $(G,H,L)$ be a Frobenius–Wielandt group with kernel $N$. Suppose that
$M\lhd G$ satisfies $G=HM$ and $H\cap M=1$. Then $N=LM$, and $(G,N,M)$ is a
Frobenius triple. In particular, $C_{M}(g)=1$ for every $g\in G\setminus N$
and $\psi^{G}\in\mathrm{Irr}(G)$ for every $\psi\in\mathrm{Irr}(N\mid M)$.

As far as we are aware, this is the only paper to address the character theory
of Frobenius–Wielandt groups, albeit in a somewhat limited case. However, in
the situation of Theorem D we see that the Frobenius–Wieldandt kernel enjoys
similar properties held by Frobenius kernels. In some sense, the quotient
$N/M$ can be considered a measure of how close the Frobenius–Wielandt kernel
is to being a Frobenius kernel in this situation, if $L$ is chosen as small as
possible.

The Frobenius kernel of a Frobenius group is uniquely determined. It is
unfortunately not the case that $N$ or $M$ is uniquely determined if $(G,N,M)$
is a Frobenius triple. We will see however that for a fixed $N$ there is a
maximal choice of $M$ and for a fixed $M$ there is a minimal choice of $N$. A
similar result is true of Camina triples. In [1], the present authors
construct for any normal subgroup of $G$ a pair of normal subgroups that can
be used to characterize Camina triples. These constructions satisfy a
compatibility condition on the lattice of normal subgroups of $G$ known as a
(monotone) Galois connection. In Section 7 we discuss similar constructions
for Frobenius triples. Specifically, we define for any normal subgroup $N$ of
$G$ two subgroups $C(G\mid N)$ and $I(G\mid N)$ that also characterize
Frobenius triples: $(G,N,M)$ is a Frobenius triple if and only if $C(G\mid
M)\leq N$, which happens if and only if $M\leq I(G\mid N)$. In particular,
these constructions give another Galois connection on the lattice of normal
subgroups of $G$.

##  2\. Frobenius and Camina triples

Frobenius groups have many interesting properties. One such property is that
the conjugacy class of any element lying outside of the Frobenius kernel $N$
is a union of cosets of $N$. This property has been generalized to obtain
Camina pairs. Let $G$ be a group and let $N$ be a proper, nontrivial normal
subgroup of $G$. We say $(G,N)$ is a Camina pair if every the conjugacy class
of every element of $G\setminus N$ is a union of $N$-cosets and the normal
subgroup $N$ is said to be a Camina kernel.

The following result surveys some of the known results about Frobenius groups
and their connection to Camina pairs.

######  Theorem 2.1.

Let $N$ be a normal subgroup of $G$. The following are equivalent:

  1. (1)

The group $G$ is a Frobenius group with Frobenius kernel $N$.

  2. (2)

Every element $1\neq x\in N$ satisfies $C_{G}(x)\leq N$.

  3. (3)

([2, Lemma 1]) The pair $(G,N)$ is a Camina pair and $(\lvert G:N\rvert,\lvert
N\rvert)=1$.

  4. (4)

([3, Proposition 3.2]) The pair $(G,N)$ is a Camina pair and $G$ splits over
$N$.

There are many conditions that are equivalent to the definition of Camina
pairs. We now give one.

######  Lemma 2.2 ([2, Lemma 1]).

Let $N$ be a normal subgroup of $G$. Then $(G,N)$ is a Camina pair if and only
if $\lvert C_{G}(g)\rvert=\lvert C_{G/N}(gN)\rvert$ for every element $g\in
G\setminus N$.

For proofs of the above results, and other information about Camina pairs, we
refer the reader to [2, 3, 4, 17].

The idea of Camina pairs has been generalized even further. Let $M\leq N$ be
proper, nontrivial normal subgroups of a group $G$. We say $(G,N,M)$ is a
Camina triple if the conjugacy class of every element of $G\setminus N$ is a
union of $M$-cosets. It is not difficult to see that $(G,N)$ is a Camina pair
if and only if $(G,N,N)$ is a Camina triple. Mattarei first considered these
in his Ph. D. dissertation [16], although the term “Camina triple” never
appears in [16]. As far as we can tell, the term “Camina triple” first appears
in [14]. Camina triples are studied extensively by Mlaiki in [17], where he
proves several results about Camina triples that are analogous to known
results about Camina pairs. In fact, some of these are generalizations of
results of the second author appearing in [15]. One such result is the
following.

######  Lemma 2.3 (cf. [17, Theorem 4]).

Let $M$ and $N$ be normal subgroups of $G$ satisfying $M\leq N$. Then
$(G,N,M)$ is a Camina triple if and only if $\lvert C_{G}(g)\rvert=\lvert
C_{G/M}(gM)\rvert$ for every element $g\in G\setminus N$.

In some ways, both $N$ and $M$ are trying to be a Camina kernel when $(G,N,M)$
is a Camina triple. Unfortunately, one cannot expect to say anything about the
structure of the quotient group $N/M$ in general. In fact, it is not difficult
to see that if $(G,N)$ is a Camina pair and $A$ is any group, then $(G\times
A,N\times A,N)$ is a Camina triple.

As mentioned above, a Frobenius group $G$ is a group possessing a nontrivial
normal subgroup $N$ so that $C_{G}(x)\leq N$ for each $1\neq x\in N$. When
$(G,N)$ is a Camina pair, we already have $\lvert C_{G}(g)\rvert=\lvert
C_{G/N}(gN)\rvert$ for every $g\in G\setminus N$. Thus a Camina pair $(G,N)$
is a Frobenius group exactly when the canonical homomorphism $C_{G}(g)\to
C_{G/N}(gN)$ is an isomorphism. Applying this idea in the situation of Lemma
2.3 leads to the following generalization of a Frobenius group.

Let $M$ and $N$ be nontrivial proper normal subgroups of $G$ with $M\leq N$.
If $C_{G}(x)\leq N$ for every element $1\neq x\in M$, we will call $(G,N,M)$ a
Frobenius triple.

Using the idea preceding the definition, we show that Frobenius triples are,
in fact, examples of Camina triples.

######  Lemma 2.4.

A Frobenius triple is a Camina triple.

###### Proof.

Let $(G,N,M)$ be a Frobenius triple and let $g\in G\setminus N$. Then
$C_{M}(g)=1$, when means that the natural homomorphism $C_{G}(g)\to
C_{G/M}(gM)$ is injective. Since $\lvert C_{G}(g)\rvert\geq\lvert
C_{G/M}(gM)\rvert$ (e.g., see [12, Corollary 2.24]), we have $\lvert
C_{G}(g)\rvert=\lvert C_{G/M}(gM)\rvert$. Thus $(G,N,M)$ is a Camina triple by
Theorem 2.3. ∎

We know from Theorem 2.1 that a Camina pair $(G,N)$ is a Frobenius group if
and only if $(\lvert G:N\rvert,\lvert N\rvert)=1$. We now prove the
equivalence of the first two statements of Theorem A, which shows that a
similar relationship exists between Camina pairs and Frobenius triples.

######  Lemma 2.5.

A Camina triple $(G,N,M)$ is a Frobenius triple if and only if $(\lvert
G:N\rvert,\lvert M\rvert)=1$.

###### Proof.

First assume that $(\lvert G:N\rvert,\lvert M\rvert)=1$. Write $\pi=\pi(M)$
and let $1\neq m\in M$. Let $g\in G\setminus N$. We may write $g=xy$ where
$o(x)$ is a $\pi$-number and $o(y)$ is a $\pi^{\prime}$-number and $xy=yx$.
Since $o(xN)$ divides $o(x)$ and $(\lvert G:N\rvert,\lvert M\rvert)=1$, it
follows that $x\in N$ and $y\notin N$. Note that $M$ is solvable (see [17,
Lemma 2.7]). Since $(G,N,M)$ is a Camina triple and $y\notin N$, we have
$\lvert C_{G}(y)\rvert=\lvert C_{G/M}(yM)\rvert$. Since $(o(y),\lvert
M\rvert)=1$ and $M$ is solvable, we see that $C_{G/M}(yM)=C_{G}(y)M/M$ [13,
Corollary 3.28]. These two facts imply that $C_{M}(y)=1$. Since $y\in\langle
g\rangle$, we deduce that $g\notin C_{G}(m)$ and so $C_{G}(m)\leq N$.

Now let $(G,N,M)$ be a Frobenius triple. Let $p$ be a prime dividing $\lvert
M\rvert$, and let $P\in\mathrm{Syl}_{p}(G)$. Then $P\cap
M\in\mathrm{Syl}_{p}(M)$ and $P\cap M\lhd P$. Let $1\neq y\in M\cap Z(P)$.
Then $P\leq C_{G}(y)\leq N$, so $p$ does not divide $\lvert G:N\rvert$. ∎

##  3\. Connection to Frobenius–Wielandt triples

In this section, we consider another generalization of Frobenius groups due to
Wielandt. First, we show that if we have a Frobenius triple, then we can find
a “pseudo-complement.” That is if $(G,N,M)$ is a Frobenius triple, then there
exists a subgroup $H$ satisfying $G=HN$ and $H\cap M=1$. If $N=M$, then $H$
would be a complement. Note that we only need the index of $N$ to be coprime
to the order of $M$ for $H$ to exist. Hence, this is really a generalization
of the existence part of the Schur–Zassenhaus theorem.

######  Lemma 3.1.

Let $M$ and $N$ be normal subgroups of $G$ so that $(|G:N|,|M|)=1$. Then
$M\leq N$ and there exists a subgroup $H\leq G$ so that $G=HN$ and $H\cap
M=1$. Moreover, if $(G,N,M)$ is a Frobenius triple, then $(HM,(H\cap N)M,M)$
is also a Frobenius triple.

###### Proof.

The first statement is clear since $NM/N\cong M/(M\cap N)$ and $(\lvert
G:N\rvert,\lvert M\rvert)=1$. We prove the second statement by induction on
$\lvert G\rvert$. First we claim that there exists a maximal subgroup of $G$
not containing $N$. To see this, suppose on the contrary that $N\leq\Phi(G)$.
Then every prime divisor of $\lvert M\rvert$ divides $\lvert\Phi(G)\rvert$ and
does not divide $\lvert G:\Phi(G)\rvert$, which is impossible. Thus we may
find a maximal subgroup $K$ not containing $N$, as claimed. If $M\cap K=1$,
then we are done. So we assume that $M\cap K>1$. Then $\lvert G:N\rvert=\lvert
K:K\cap N\rvert$ is relatively prime to $\lvert K\cap M\rvert$, so by the
inductive hypothesis there exists $H\leq K$ so that $K=H(K\cap N)$ and $H\cap
M=1$. Thus $G=NK=NH$ and $H\cap M=1$, as desired.

Suppose $(G,N,M)$ is a Frobenius triple; so we have $C_{HM}(x)\leq HM\cap
N=(H\cap N)M$ for every element $1\neq x\in M$. Thus $(HM,(H\cap N)M,M)$ is a
Frobenius triple. ∎

When $M$ and $N$ are normal subgroups of $G$ so that $(\lvert G:N\rvert,\lvert
M\rvert)=1$, we call the subgroup $H$ of Lemma 3.1 a psuedo-complement for
$(G,N,M)$.

In [18], Wielandt studies groups $G$ that have a nontrivial, proper subgroup
$H$ and a normal subgroup $L$ of $H$ that is proper in $H$ for which
$H^{g}\cap H\leq L$ for every element $g\in G\setminus H$. In this situation,
Espuelas [6] says that $(G,H,L)$ is a Frobenius-Weilandt group. To be
consistent with our notation, we will instead say that $(G,H,L)$ is a
Frobenius-Wielandt triple. If $(G,H,L)$ is a Frobenius–Wieldandt triple, there
is a unique normal subgroup $N$ of $G$ satisfying $G=NH$ and $N\cap H=L$. This
subgroup is given by $N=G\setminus\bigcup_{g\in G}(H\setminus L)^{g}$ and is
called the Frobenius–Wielandt kernel. Note that the uniqueness of $N$ depends
on $L$. On the other hand, observe that $L$ is not uniquely defined; one can
easily replace $L$ by any normal subgroup of $H$ that contains $L$. For this
reason Espuelas defines the a Frobenius-Wielandt complement to be the quotient
$H/L$.

Our next result shows that Frobenius triples and Frobenius–Wielandt groups are
intimately connected.

######  Theorem 3.2.

Let $M\leq N$ be proper, nontrivial normal subgroups of $G$ and assume that
$H$ is a psuedo-complement for $(G,N,M)$. Then $(HM,(H\cap N)M,M)$ is a
Frobenius triple if and only if $(HM,H,H\cap N)$ is a Frobenius–Wielandt
triple.

###### Proof.

First, suppose that $(HM,H,H\cap N)$ is a Frobenius–Wielandt triple; so that
$H^{x}\cap H\leq H\cap N$ for every element $1\neq x\in M$. Observe that this
implies that $C_{H}(x)\leq H\cap N$ for every $1\neq x\in M$. Let $K$ be the
Frobenius–Wielandt kernel. Then $K=HM\setminus\bigcup_{x\in M}(H\setminus
N)^{x}$ and $K$ satisfies $HM=HK$ and $H\cap K=H\cap N$. Since $M\leq N$, it
follows that $M\leq K$. Also, $\lvert HM:K\rvert=\lvert HK:K\rvert=\lvert
H:H\cap K\rvert=\lvert H:H\cap N\rvert=\lvert HM:(H\cap N)M\rvert$, which
gives $K=(H\cap N)M$. Thus, we see that $HM\setminus N=HM\setminus(H\cap N)M$
is partitioned by the sets $(H\setminus N)^{x}$ for $x\in M$. Let $g\in
HM\setminus N$, let $1\neq x\in M$ and assume that $[g,x]=1$. Then there
exists $y\in M$ so that $g\in(H\setminus N)^{y}$. So $g=h^{y}$ for some $h\in
H\setminus N$ and it follows that $(x^{y^{-1}})^{h}=x^{y^{-1}}$. Thus $h\in
C_{H}(x^{y^{-1}})\leq H\cap N$. But $h\in H\setminus N$, a contradiction.
Thus, no such element $g$ exists and so $C_{HM}(x)\leq(H\cap N)M$. It follows
that $(HM,(H\cap N)M,M)$ is a Frobenius triple.

Now, assume that $(HM,(H\cap N)M,M)$ is a Frobenius triple. This implies that
$H\cap N$ will be a proper, normal subgroup of $H$. Consider elements $g\in
HM$ and $h\in H^{g}\cap H$. We see that $H^{g}\cap H=H^{x}\cap H$ for some
element $x\in M$ and so $h=k^{x}$ for some $k\in H$. Then
$[k,x]=k^{-1}k^{x}=k^{-1}h\in H\cap M=1$, so $k=h=k^{x}$. This implies that
$x\in C_{M}(k)$, and so, $h=k\in H\cap N$. Hence $HM$ is a Frobenius–Wielandt
group with Frobenius-Wielandt complement $H/(H\cap N)$, as desired. ∎

As a consequence of Theorem 3.2, we see that every Frobenius triple contains a
subgroup that is a Frobenius–Wielandt triple. Let $(G,H,L)$ be a
Frobenius–Wielandt triple and suppose that $M$ is a normal subgroup of $G$
satisfying $H\cap M=1$. Then $(HM,H,L)$ is also a Frobenius–Wielandt triple.
Let $N\leq HM$ be the Frobenius–Wielandt kernel. Then $H\cap N=L$ and so
$(HM,H,H\cap N)$ is Frobenius triple by Theorem 3.2. So at least in this
situation, a Frobenius–Wielandt triple contains a subgroup that is a Frobenius
triple. This is not always possible however. As an example, let $G$ be the
general linear group $\mathrm{GL}_{2}(\mathbb{F}_{3})$, let
$H\in\mathrm{Syl}_{2}(G)$, and let $N$ be the special linear group
$\mathrm{SL}_{2}(\mathbb{F}_{3})$. Then $H\cap H^{x}\leq H\cap N$ for every
$x\in G\setminus N$ and so $(G,H,H\cap N)$ is a Frobenius–Wieldandt triple
with kernel $N$. However, $Z(G)$ is the unique minimal normal subgroup of $G$
and $Z(G)\leq H$. Hence, there is no normal subgroup $M$ of $G$ with $H\cap
M=1$.

Under the added assumption that $(G,N,M)$ is a Camina triple, we obtain a
result slightly stronger than Theorem 3.2.

######  Lemma 3.3.

Let $(G,N,M)$ be a Camina triple. If $H$ is a psuedo-complement for $(G,N,M)$,
then $(G,N,M)$ is a Frobenius triple if and only if $(HM,(H\cap N)M,M)$ is a
Frobenius triple.

###### Proof.

When $(G,N,M)$ is a Frobenius triple, the fact that $(HM,(H\cap N)M,M)$ is a
Frobenius triple follows from Lemma 3.1.

Conversely, suppose $(HM,(H\cap N)M,M)$ is a Frobenius triple. It follows from
Lemma 2.5 that $(|HM:(H\cap N)M|,|M|)=1$. Since $\lvert HM:(H\cap
N)M\rvert=\lvert H:H\cap N\rvert=\lvert G:N\rvert$, we have $(|G:N|,|M|)=1$,
and in light of Lemma 2.5 again, we conclude that $(G,N,M)$ is a Frobenius
triple. ∎

Proposition 3.2 of [3] shows that $G$ is a Frobenius group with Frobenius
kernel $N$ if and only if $(G,N)$ is a Camina pair and $G$ splits over $N$.
Our next result is a generalization of this fact for Camina triples, and its
proof completes the proof of Theorem A.

######  Theorem 3.4.

Let $(G,N,M)$ be a Camina triple. Then $(G,N,M)$ is a Frobenius triple if and
only if there exists a subgroup $H\leq G$ so that $G=HN$ and $H\cap M=1$.

###### Proof.

If $(G,N,M)$ is a Frobenius triple, then there exists a subgroup $H\leq G$ so
that $G=NH$ and $H\cap M=1$ by Lemma 3.1.

Conversely, assume that there exists a subgroup $H\leq G$ so that $G=HN$ and
$H\cap M=1$. We may replace the triple $(G,N,M)$ by the triple $(HM,(H\cap
N)M.M)$ by Lemma 3.3, and thus, we may assume that $G=HM$. Fix the elements
$g\in H\setminus N$ and $z\in M$. Since $(G,N,M)$ is a Camina triple, there
exists an element $y\in G$ so that $[g,y]=z$ (see [17, Theorem 2.1 (v)]. Write
$y=hm$, where $h\in H$ and $m\in M$. Then $[g,y]=[g,m][g,h]^{m}$ and so it
follows that $[g,h]\in H\cap M=1$. Thus $[g,m]=z$. This means that
$\\{[g,m]\mid m\in M\\}=M$, and so, $g$ acts fixed-point-freely on $M$.
Consider elements $1\neq x\in M$ and $h\in H^{x}\cap H$. Then there exists an
element $k\in H$ so that $h=k^{x}$, and so, $k^{-1}h=k^{-1}k^{x}=[k,x]\in
H\cap M=1$. Hence $k\in C_{H}(x)\leq H\cap N$, from which it follows that
$H^{x}\cap H\leq H\cap N$. Thus $(G,H,H\cap N)$ is a Frobenius–Wielandt
triple, and therefore, the conclusion that $(G,N,M)$ is a Frobenius triple
follows from Theorem 3.2. ∎

We now prove Theorem B. Notice that condition (1) is the hypothesis that
$(G,N,M)$ is a Frobenius triple. On the other hand, condition (4) is the
condition that $(G,H,H\cap N)$ is a Frobenius-Wielandt triple.

######  Theorem 3.5.

Let $M$ and $N$ be proper, nontrivial normal subgroups of $G$ that satisfy
$M\leq N$, and assume that $M$ has a complement $H$ in $G$. The following are
equivalent:

  1. (1)

$C_{G}(x)\leq N$ for every element $1\neq x\in M$;

  2. (2)

$C_{H}(x)\leq H\cap N$ for every element $1\neq x\in M$;

  3. (3)

$C_{G}(h)\leq H$ for every element $h\in H\setminus N$;

  4. (4)

$H\cap N<H$ and $H^{g}\cap H\leq H\cap N$ for every element $g\in G\setminus
H$.

  5. (5)

Every element $g\in G\setminus N$ is conjugate to an element of $H$;

  6. (6)

If $h\in H\setminus N$, then $h$ is conjugate to every element of $hM$.

###### Proof.

The observation that condition (1) implies condition (2) is obvious. Assume
condition (2) now. Consider an element $h\in H\setminus N$ and suppose that
$g=km\in C_{G}(h)$ where $k\in H$ and $m\in M$. Then $1=[km,h]=[k,h]^{m}[m,h]$
and so $[k,h]\in H\cap M=1$. Thus $[m,h]=1$ and so $m=1$ and $g\in H$. Hence
(3) holds.

Now assume (3). Since $G=HM$, we have $G=NH$ and $|G:N|=|H:H\cap N|$. Because
$N$ is proper in $G$, we have that $H\cap N$ is proper in $H$. Consider an
element $g\in G\setminus H$. There exists elements $y\in H$ and $1\neq x\in M$
so that $g=yx$. Observe that $H^{g}=H^{yx}=H^{x}$. Suppose $h\in H^{g}\cap
H=H^{x}\cap H$. Then there exists $k\in H$ so that $h=k^{x}$ and so
$k^{1}h=[k,x]\in H\cap M=1$. So $h=k\in C_{H}(x)\leq H\cap N$ and (4) follows.

Assume that (4) holds. Thus, $(HM,H,H\cap N)$ is a Frobenius–Wielandt group
with kernel $K=HM\setminus\left[\bigcap_{x\in M}(H^{x}\setminus N)\right]$.
Since $M\leq N$, we see that $M\leq K$. Since $HM=HK$ and $H\cap K=H\cap N$,
we deduce that $\lvert HM:K\rvert=\lvert HK:K\rvert=\lvert H:H\cap
N\rvert=\lvert HM:(H\cap N)M\rvert$ and so $K=(H\cap N)M=HM\cap N=N$. By the
description of $K=HM\cap N$ given above, it is clear that every element of
$HM$ lying outside of $N$ is contained in one of the sets $(H\setminus N)^{x}$
for some element $x\in M$. Thus (5) is proved.

Given (5), let $h\in H\setminus N$ and $m\in M$. We wish to show that $h$ is
conjugate to $hm$. By (5), some conjugate of $hm$ lies in $H$, say
$(hm)^{g}\in H$. Observe that $(hm)^{g}\in h^{g}M=h^{k}M$, where $k\in H$ and
$k^{-1}g\in M$, since $(h^{k})^{-1}h^{g}=[(h^{k})^{-1},k^{-1}g]$. So
$(h^{k})^{-1}(hm)^{g}\in H\cap M=1$ and hence $(hm)^{gk^{-1}}=h$, which proves
(6).

Finally assume (6). We will show that $(G,N,M)$ is a Camina triple. To do this
we will show that every element $g\in G\setminus N$ is conjugate to every
element in $gM$. Write $g=hm$, where $h\in H\setminus N$ and $m\in M$. Let
$x\in M$. By (6), we can find $y\in G$ so that $g^{y}=(hm)^{y}=h$ and $z\in G$
so that $h=(hx)^{z}$. So $(hm)^{yz^{-1}}=hx$, as required. So $(G,N,M)$ is a
Camina triple. Since there exists a psuedo-complement $H$, we may use Theorem
3.4 to see that $(G,N,M)$ is a Frobenius triple, and thus, condition (1)
follows. ∎

Note that the assumption $G=HM$ is necessary in the statement of Theorem B. To
see this, let $G=C_{3}\times S_{3}$. Let $M$ be a Sylow 3-subgroup of $S_{3}$,
let $H$ be a Sylow $2$-subgroup of $S_{3}$ and let $N$ be a Sylow $3$-subgroup
of $G$. Then $(G,N,M)$ is a Frobenius triple, $G=HN$ and $H\cap M=1$. However,
conditions (3)–(6) of Theorem B do not hold.

##  4\. Properties of Frobenius triples

We now study Frobenius triples and we determine some of their properties. We
first show that if we have a Frobenius triple where the “psuedo complement”
actually complements the lower term in the triple, then in fact, we have a
Frobenius group.

######  Lemma 4.1.

Let $(G,N,M)$ be a Frobenius triple. If there exists a subgroup $H\leq G$ that
satisfies $H\cap N=1$, then $MH$ is a Frobenius group with Frobenius kernel
$M$. In particular, $M$ is nilpotent.

###### Proof.

Suppose $1\neq m\in M$, then $C_{G}(m)\leq N$. It follows that
$C_{MH}(m)=C_{G}(m)\cap MH\leq N\cap MH=M$. By Theorem 2.1, this implies $HM$
is a Frobenius group. ∎

In the situation of Corollary 4.1, we can find an element of prime order in
$G\setminus N$ and, of course, $M$ is nilpotent. Using a similar argument as
in Lemma 4 of [2], we see that having an element of prime order in $G\setminus
N$ is always enough to guarantee that $M$ is nilpotent. Note that if $(G,N,M)$
is a Frobenius triple and $G\setminus N$ contains an element of prime order,
then this next lemma is even easier to prove.

######  Lemma 4.2.

Let $(G,N,M)$ be a Camina triple. If there is an element $x\in G\setminus N$
of prime order, then $M$ is nilpotent.

###### Proof.

Let $g\in M$ and let $x\in G\setminus N$ with $o(x)=p$. Since $(G,N,M)$ is a
Camina triple, $x^{-1}$ is conjugate to $x^{-1}g$, which is conjugate to
$gx^{-1}$. In particular, $(gx^{-1})^{p}=1$. Rewriting this, we have
$gg^{x}g^{x^{2}}\dotsb g^{x^{p-1}}=x^{p}=1$. The result now follows from a
result of Hughes, Kegel and Thompson (see [10, V 8.13 Hauptsatz]). ∎

In general, $M$ is not nilpotent when $(G,N,M)$ is a Frobenius triple. In
fact, one may find examples of Frobenius triples $(G,N,M)$ where the Fitting
height $h(M)$ of $M$ is arbitrarily large. This next theorem contains most of
the work of proving these examples exist.

######  Theorem 4.3.

For every positive integer $n$ and every prime $p$, there exists a group $G$
that satisfies the following conditions:

  1. (i)

$G$ is a semidirect product of a cyclic group $H$ of order $p^{n}$ acting
faithfully on a group $M$.

  2. (ii)

$M$ is a solvable $p^{\prime}$-group whose Fitting height is $n$.

  3. (iii)

The Fitting subgroup of $M$ is minimal normal in $G$.

  4. (iv)

For every nonidentity element $x\in M$, $p$ divides $\lvert G:C_{G}(x)\rvert$.

###### Proof.

We work by induction on $n$. When $n=1$, we take $G$ be a Frobenius group
whose Frobenius complement has order $p$ and whose Frobenius kernel $M$ is
minimal normal in $G$. Note that $G$ will satisfy the stated conditions.

We now assume $n\geq 1$, and we assume that we have an example $G$ that
satisfies the given conditions, and we prove that there exists a group that
satisfies the conditions for $n+1$. Thus, $G$ is the semi-direct product of a
cyclic group $H$ of order $p^{n}$ acting faithfully on the solvable
$p^{\prime}$-group $M$ whose Fitting height is $n$ such that the Fitting
subgroup of $M$ is minimal normal in $G$ and every non identity element $x\in
M$ satisfies $\lvert H:C_{H}(x)\rvert$ is divisible by $p$. We can write
$H=\langle h\rangle$. Take $\tilde{H}$ to be a cyclic group of order $p^{n+1}$
and write $\tilde{H}=\langle\tilde{h}\rangle$. We can define an action of
$\tilde{H}$ on $M$ by $m^{\tilde{h}}=m^{h}$ for all $m\in M$, and note that
$Z=\langle\tilde{h}^{p^{n}}\rangle$ will be the kernel of this action. We now
define $\Gamma$ to be the semi-direct product of $\tilde{H}$ acting on $M$.
Observe that $Z$ will be the center of $\Gamma$ and $Z\times M$ will be the
Fitting subgroup of $\Gamma$.

Take $\mu\in\mathrm{Irr}(M)$ and $\lambda\in\mathrm{Irr}(Z)$ to be
nonprincipal characters, and take $\chi$ to be an irreducible constituent of
$(\mu\times\lambda)^{G}$. It is not difficult to see that $\chi$ will have to
be faithful. Let $q$ be a prime that does not divide $|G|$. Thus, we may view
$\chi$ as an irreducible $q$-Brauer character. Now, choose $V$ to be an
irreducible $Z_{q}$-module that affords $\chi$ as a $q$-Brauer character. We
now have an action of $\Gamma$ on $V$. It is not difficult to see that
$\Gamma$ will act faithfully on $V$. Observe that $C_{V}(Z)$ will be a proper
$G$-submodule of $V$, and so, $C_{V}(Z)=0$.

We now define $\tilde{G}$ to be the semi-direct product of $\Gamma$ acting on
$V$, and we define $\tilde{M}$ to be $MV$, so $\tilde{M}$ is the semi-direct
product of $M$ acting on $V$. It follows that $\tilde{M}$ is a solvable
$p^{\prime}$-subgroup. Note that $\tilde{G}$ will be the semi-direct product
of $\tilde{H}$ acting faithfully on $\tilde{M}$. Observe that $V$ will be the
Fitting subgroup of $\tilde{M}$, and since $M$ has Fitting height $n$, this
implies that $\tilde{M}$ has Fitting height $n+1$. Since $V$ is the Fitting
subgroup of $\tilde{M}$, it follows that the Fitting subgroup of $\tilde{M}$
is minimal normal in $G$. Finally, suppose $x$ is a nonidentity element of
$\tilde{M}$. Suppose first that $x\not\in V$. Then $C_{\tilde{G}}(x)ZV/ZV\leq
C_{\tilde{G}/VZ}(xVZ)$. Observe that $\tilde{G}/VZ$ is isomorphic to $G$ and
$\tilde{H}/Z$ is isomorphic to $H$, and so, we have $p$ divides
$|\tilde{G}/VZ:C_{\tilde{G}/VZ}(xVZ)|$. In this case, we will have $p$ divides
$\lvert\tilde{G}:C_{\tilde{G}}(x)\rvert$. Suppose $x\in V$. Since
$C_{V}(Z)=0$, we see that $Z$ is not contained in $C_{\tilde{G}}(x)$. On the
other hand, we know that the Sylow $p$-subgroups of $\tilde{G}$ are cyclic; so
this implies that a Sylow subgroup of $C_{\tilde{G}}(x)$ will be trivial. This
implies that $p^{n}$ divides $|\tilde{G}:C_{\tilde{G}}(x)|$. We have now
proven that $\tilde{G}$ with $\tilde{H}$ and $\tilde{M}$ satisfy the
conclusions of the theorem for $n+1$. ∎

We now show the existence of the stated examples.

######  Corollary 4.4.

Let $n$ be a positive integer. There is a Frobenius triple $(G,N,M)$ where
$h(M)=n$.

###### Proof.

Take $G$, $H$ and $M$ to be as in Theorem 4.3. Let $D$ be the subgroup of $H$
of index $p$, and let $N=DM$. Then $C_{H}(x)\leq D=H\cap N$ for every $1\neq
x\in M$ and so $(G,N,M)$ is a Frobenius triple by Theorem B. ∎

Although $M$ can have arbitrarily large Fitting height when $(G,N,M)$ is a
Frobenius triple, we can bound the Fitting height of $M$ in terms of the order
of elements in $G\setminus N$ by appealing to results of F. Gross regarding
fixed-point-free automorphisms.

######  Theorem 4.5.

Let $M$ be a normal subgroup of $G$ and suppose that $C=C(G\mid N)<G$. For
each prime $p\in\pi(G/C)$, write $p^{n_{p}}$ for the minimum order of a
$p$-element in $G\setminus C$. Let $\mathcal{N}=\\{n_{p}:p\in\pi(G\setminus
C),\ \text{$p$ odd}\\}$, let $m_{o}$ be the minimum value in $\mathcal{N}$
when $\mathcal{N}\neq\varnothing$; otherwise let $m_{o}=1$, and let
$m_{e}=\max\\{2n_{2}-2,n_{2}\\}$ when $\lvert G/C\rvert$ is even; otherwise
let $m_{e}=1$. Then $h(M)\leq\min\\{m_{o},m_{e}\\}$.

###### Proof.

Since every $g\in G\setminus C$ acts fixed-point-freely on $M$, this follows
from Theorem 1 of [8] and Theorem 1.1 of [9]. ∎

##  5\. Character theory of Frobenius triples

It is well-known that $G$ is a Frobenius group with Frobenius kernel $N$ if
and only if the induced character $\psi^{G}$ is irreducible for every
nonprincipal character $\psi\in\mathrm{Irr}(N)$. A Camina triple can also be
characterized by a condition on its irreducible characters. Since these
conditions will be used repeatedly throughout this section, we list them here.
For a normal subgroup $M$ of $G$, we use $\mathrm{Irr}(G\mid M)$ to denote the
set of all irreducible characters of $G$ that do not contain $M$ in their
kernel.

######  Lemma 5.1 (cf. [17, Theorem 4]).

Let $M$ and $N$ be proper, nontrivial normal subgroups of $G$ that satisfy
$M\leq N$. The following are equivalent:

  1. (1)

$(G,N,M)$ is a Camina triple.

  2. (2)

Every character $\psi\in\mathrm{Irr}(N\mid M)$ induces homogeneously to $G$.

  3. (3)

Every character $\psi\in\mathrm{Irr}(G\mid M)$ vanishes on $G\setminus N$.

The following result a known consequence of Clifford’s Theorem. This result
(essentially) appears as Lemma 2.2 of [7]. We include a short proof here.

######  Lemma 5.2.

Let $N$ be a normal subgroup of $G$, and suppose that the character
$\chi\in\mathrm{Irr}(G)$ vanishes on $G\setminus N$. Let $\psi$ be an
irreducible constituent of $\chi_{N}$. Then $\psi^{G}=\lvert
T:N\rvert^{1/2}\chi$, where $T=I_{G}(\psi)$ is the stabilizer of $\psi$ in
$G$. In particular, $\psi$ is fully-ramified with respect to $T/N$.

###### Proof.

Write $e=\langle\psi^{G},\chi\rangle$. Consider the character
$\xi\in\mathrm{Irr}(G)$. Then

$\displaystyle\langle\psi^{G},\xi\rangle$ $\displaystyle=\frac{1}{\lvert
G\rvert}\sum_{g\in G}\psi^{G}(g)\overline{\xi(g)}=\frac{1}{\lvert
G\rvert}\sum_{g\in N}\psi^{G}(g)\overline{\xi(g)}$
$\displaystyle=\frac{1}{\lvert G\rvert}\sum_{g\in N}\frac{1}{\lvert
N\rvert}\sum_{x\in G}\psi^{x}(g)\overline{\xi(g)}=\frac{1}{\lvert
G\rvert}\sum_{g\in N}\frac{\lvert T\rvert}{e\lvert
N\rvert}\chi(g)\overline{\xi(g)}$ $\displaystyle=\frac{\lvert
T:N\rvert}{e}\cdot\frac{1}{\lvert G\rvert}\sum_{g\in
G}\chi(g)\overline{\xi(g)}=\frac{\lvert
T:N\rvert}{e}\,\langle\chi,\xi\rangle=\frac{\lvert
T:N\rvert}{e}\,\delta_{\chi,\xi}.$

It follows that $\psi^{G}=e\chi=\frac{\lvert T:N\rvert}{e}\chi$ and so also
that $\psi^{G}=\lvert T:N\rvert^{1/2}\chi$. ∎

We now show that the irreducible characters of Frobenius triples satisfy a
condition that is similar to the one satisfied by Frobenius groups.

######  Theorem 5.3.

If $(G,N,M)$ is a Frobenius triple, then $\psi^{G}$ is irreducible for every
character $\psi\in\mathrm{Irr}(N\mid M)$.

###### Proof.

We work by induction on $\lvert G\rvert$. Consider a character
$\psi\in\mathrm{Irr}(N\mid M)$. It is not difficult to see from Lemma 5.2 and
[12, Theorem 6.11] that $\psi^{G}\in\mathrm{Irr}(G)$ if and only if $N$ is the
stabilizer of $\psi$. Using the inductive hypothesis, we have $I_{G}(\psi)\cap
K=I_{K}(\psi)=N$ for any nontrivial proper subgroup $K/N$ of $G/N$ and so it
follows that $I_{G}(\psi)=N$ if $G/N$ has composite order. Thus it suffices to
assume that $G/N$ has prime order. Let $\chi\in\mathrm{Irr}(G\mid\psi)$. Since
$\chi$ vanishes on $G\setminus N$ by Lemma 5.1, we have $\psi^{G}=\lvert
I_{G}(\psi):N\rvert^{1/2}\chi$ by Lemma 5.2. Since $G/N$ has prime order, we
must have $I_{G}(\psi)=N$ and so $\psi^{G}$ is irreducible. ∎

When $M=N$, we recover the well-known fact that every nonprincipal irreducible
character of a Frobenius kernel $N$ induces irreducibly to $G$. A standard
proof of this fact is a straightforward application of Brauer’s Permutation
Lemma (see [12, Theorem 6.32], for example). We can actually prove Theorem 5.3
in a similar manner by using a variation of this result, which will also be
used to obtain another character-theoretic characterization of Frobenius
triples. Our proof is essentially the same argument used in the proof [12,
Theorem 6.32].

######  Lemma 5.4.

Let $N$ be a normal subgroup of $G$. Set
$\mathrm{Cl}_{G}(N)=\\{\mathrm{cl}_{G}(n)\mid n\in N\\}$ and
$\mathrm{Irr}_{G}(N)=\\{\mathrm{orb}_{G}(\psi)\mid\psi\in\mathrm{Irr}(N)\\}$.
Suppose that the group $A$ acts on $G$ via automorphisms. If $N$ is
$A$-invariant, then $A$ acts on $\mathrm{Cl}_{G}(N)$ and
$\mathrm{Irr}_{G}(N)$, and each $a\in A$ fixes the same number of elements in
$\mathrm{Cl}_{G}(N)$ and $\mathrm{Irr}_{G}(N)$.

###### Proof.

Since $A$ acts on $G$ by automorphisms, it is clear that $A$ acts on
$\mathrm{Cl}_{G}(N)$. Also for every $\psi\in\mathrm{Irr}(N)$, $a\in A$ and
$g\in G$, we have $(\psi^{g})^{a}=(\psi^{a})^{g^{a}}$ and so $a$ maps
$\mathrm{orb}_{G}(\psi)$ to $\mathrm{orb}_{G}(\psi^{a})$. Note that for each
$\kappa\in\mathrm{Cl}_{G}(N)$ and $\lambda\in\mathrm{Irr}_{G}(N)$, the
character $\chi_{\lambda}=\sum_{\xi\in\lambda}\xi$ is constant on the elements
of $\kappa$. For each $\kappa\in\mathrm{Cl}_{G}(N)$, let $g_{\kappa}\in\kappa$
and let $X$ be the matrix $(\chi_{\lambda}(g_{\kappa}))$. For each
$\lambda\in\mathrm{Irr}_{G}(N)$ and $a\in A$, define
$\chi_{\lambda}^{a}=\chi_{\lambda^{a}}$ and observe that
$\chi_{\lambda}^{a}(n^{a})=\chi_{\lambda}(n)$ for every
$\lambda\in\mathrm{Irr}_{G}(N)$, $a\in A$ and $n\in N$. Let $P$ and $Q$ denote
the permutation representations associated to the actions of $A$ on
$\mathrm{Irr}_{G}(N)$ and $\mathrm{Cl}_{G}(N)$, respectively. Then the
$(\lambda,\kappa)$-entry of $P(a)X$ is
$\sum_{\mu}P(a)_{\lambda\mu}\chi_{\mu}(g_{\kappa})=\chi_{\lambda}^{a}(g_{\kappa}),$
and the $(\lambda,\kappa)$-entry of $XQ(a)$ is
$\sum_{\gamma}\chi_{\lambda}(g_{\gamma})Q(a)_{\gamma\kappa}=\chi_{\lambda}(g_{\kappa}^{a^{-1}})=\chi_{\lambda}^{a}(g_{\kappa}).$
Thus, we see that $P(a)X=XQ(a)$.

To complete the proof, it suffices to show that $X$ is invertible since then
$P(a)$ and $Q(a)$ would have the same trace. To see this, first note that $X$
must be square by Brauer’s Permutation Lemma. Write
$\mathrm{Cl}_{G}(N)=\\{\kappa_{1},\dotsc,\kappa_{n}\\}$ and
$\mathrm{Irr}_{G}(N)=\\{\lambda_{1},\dotsc,\lambda_{n}\\}$. Define the
matrices $T=(\delta_{ij}\lvert\lambda_{i}\rvert)$ and
$D=(\delta_{ij}\lvert\kappa_{i}\rvert)$, where (as usual) $\delta_{ij}$
denotes the Kronecker $\delta$ function. Let $X^{\dagger}$ denote the
conjugate transpose of $X$. Then $XDX^{\dagger}=\lvert G\rvert T$ since
$\langle\chi_{\lambda_{i}},\chi_{\lambda_{j}}\rangle=\delta_{ij}\lvert\lambda_{i}\rvert$.
Therefore, we conclude that $\det(X)\neq 0$ and so $X$ is invertible, as
desired. ∎

We can also characterize Frobenius triples in terms of the stabilizers of
nonprincipal characters.

######  Lemma 5.5.

Let $M\leq N$ be proper, nontrivial normal subgroups of $G$ satisfying $M\leq
N$. Then $(G,N,M)$ is a Frobenius triple if and only if the stabilizer of
every nonprincipal character in $\mathrm{Irr}(M)$ is contained in $N$.

###### Proof.

First assume that $(G,N,M)$ is a Frobenius triple. Fix $\theta$ to be a
nonprincipal irreducible character of $M$, and consider an element $g\in G$
that stabilizes $\theta$. Note then that $g$ fixes the $N$-orbit in
$\mathrm{Irr}_{N}(M)$ containing $\theta$. From Lemma 5.4, it follows that $g$
fixes a nonidentity $N$-class in $\mathrm{Cl}_{N}(M)$, say the $N$-conjugacy
class of $x\in M$. Then there exists an element $n\in N$ so that
$x^{g}=x^{n}$, which yields $gn^{-1}\in C_{G}(x)\leq N$. It follows that $g\in
N$. The proof of the reverse implication is similar. ∎

We now present an alternate proof of Theorem 5.3 using Lemma 5.4.

###### Alternate proof of Theorem 5.3.

Let $g\in G\setminus N$. Note that $g$ cannot fix the $N$-conjugacy class of
any nonidentity element of $M$. For if it did, say the $N$-class of $x\in M$,
then there would exist $n\in N$ so that $x^{g}=x^{n}$ and so $gn^{-1}\in
C_{G}(x)$. But this cannot happen since $C_{G}(x)\leq N$ and $g\notin N$.
Since $g$ fixes no $N$-classes of nonidentity elements of $M$, $g$ fixes no
$N$-orbits of nonprincipal irreducible characters of $M$ by Lemma 5.4. By
Clifford’s Theorem, we deduce that $\psi^{g}_{M}\neq\psi_{M}$ for any
$\psi\in\mathrm{Irr}(N\mid M)$, and hence $\psi^{g}\neq\psi$ for any
$\psi\in\mathrm{Irr}(N\mid M)$. It follows that $\psi^{G}\in\mathrm{Irr}(G)$
for each character $\psi\in\mathrm{Irr}(N\mid M)$. ∎

Theorem 5.3 and Lemma 5.4 yield the following result about the the action of a
group on the set of conjugacy classes of a normal subgroup.

######  Corollary 5.6.

Let $(G,N,M)$ be a Frobenius triple. Then every element $g\in G\setminus N$
fixes exactly as many $N$-conjugacy classes in $N\setminus M$ as $gM$ fixes
nonidentity conjugacy classes of $N/M$.

###### Proof.

By Brauer’s Permutation Lemma [12, Theorem 6.32], the number of $N$-conjugacy
classes fixed by $g$ is equal to the number of irreducible characters of $N$
fixed by $g$. Since $g$ fixes no nonidentity classes of $M$ and no elements of
$\mathrm{Irr}(N\mid M)$ by Theorem 5.3, the result follows. ∎

Note that if $G$ is an extraspecial $p$-group, $N$ is a maximal subgroup of
$G$, and $M=Z(G)$ then $\psi^{G}\in\mathrm{Irr}(G)$ for every character
$\psi\in\mathrm{Irr}(N\mid M)$. However $(G,N,M)$ is not a Frobenius triple
since $G$ is a $p$-group. In particular, the converse of Theorem 5.3 does not
hold in general. However, as Theorem C indicates, it does hold if we include
an extra arithmetic condition.

###### Proof of Theorem C.

Assume that $(\lvert G:N\rvert,\lvert M\rvert)=1$. If $(G,N,M)$ is a Frobenius
triple, then $\psi^{G}\in\mathrm{Irr}(G)$ for every $\psi\in\mathrm{Irr}(N\mid
M)$ by Theorem 5.3. Conversely, assume that $\psi^{G}\in\mathrm{Irr}(G)$ for
every $\psi\in\mathrm{Irr}(N\mid M)$. Then $(G,N,M)$ is a Camina triple by
Lemma 5.1. Hence the result follows from Theorem A. ∎

We now prove Theorem D, which provides character-theoretic information about
Frobenius–Wielandt triples $(G,H,L)$ in the special situation that $H$ has a
normal complement, $M$. In the situation of Theorem D, suppose that $L$ is
chosen to be as small as possible. Then the Frobenius–Wielandt kernel $N$
satisfies conditions satisfied by Frobenius kernels. Of course, if $L=1$ then
$N=M$ and $N$ is a Frobenius kernel. Outside of the situation $L=1$, since
every character $\psi\in\mathrm{Irr}(N\mid M)$ induces irreducibly to $G$
(alternatively since every $g\in G\setminus N$ acts fixed-point-freely on
$M$), the size of the quotient group $N/M$ is in some sense a measure of how
close $N$ is to being a Frobenius kernel.

###### Proof of Theorem D.

Since $H\cap M=1$, we see that $H^{g}\cap M=1$ for all $g\in G$. Since
$N=G\setminus\bigcup_{g\in G}(H\setminus L)^{g}$, we deduce that $M\leq N$. So
$N=HM\cap N=(H\cap N)M=LM$. Since $G=HN=HM$ and $H\cap M=1$, we see that
$(G,(H\cap N)M,M)=(G,N,M)$ is a Frobenius triple by Theorem 3.2. The remaining
statement follows from Theorem C. ∎

We conclude this section with an application. We use the above results to
provide an analog for Frobenius–Wielandt triples of a well-known
characterization of Frobenius complements. Fix a subgroup $H\leq G$. Then $H$
is a Frobenius complement if and only if $N_{G}(X)\leq H$ for every subgroup
$1<X\leq H$. If $G$ is a group and $G=H\ltimes M$, then a similar necessary
and sufficient condition exists for $(G,H,L)$ to be a Frobenius–Wielandt
triple.

######  Lemma 5.7.

Let $M$ be a normal subgroup of $G$ and suppose that $H\leq G$ satisfies
$G=HM$ and $H\cap M=1$. Let $L\lhd H$ and assume that $N_{G}(X)\leq H$ for
every subgroup $X\leq H$ satisfying $X\nleq L$. Then $(G,H,L)$ is a
Frobenius–Wielandt triple, and hence $(G,LM,M)$ is a Frobenius triple.

###### Proof.

Let $g\in G$ and assume that $h\in(H\cap H^{g})\setminus L$. Then there exists
$k\in H$ and $x\in M$ such that $h=k^{x}$. Then $k^{-1}h=[k,x]\leq H\cap M=1$
and so $h=k\in C_{G}(x)$. In particular, $x$ normalizes $\langle h\rangle$
Since $h\notin L$, it follows that $N_{G}(\langle h\rangle)\leq H$. Hence
$x=1$ and $g\in H$. The last statement follows from Theorem D. ∎

We now prove the theorem.

######  Theorem 5.8.

Let $M$ and $N$ be proper, nontrivial normal subgroups of $G$ with $M\leq N$.
Let $H\leq G$ and suppose that $G=HM$ and $H\cap M=1$. Then $G$ satisfies the
equivalent conditions of Theorem 3.5 if and only if $N_{G}(X)\leq H$ for every
subgroup $X\leq H$ satisfying $X\nleq N$.

###### Proof.

The one direction is Lemma 5.7. To prove the converse, let $(G,N,M)$ be a
Frobenius triple. Let $X\leq H$ with $X\nleq H\cap N$ and let $g\in N_{G}(X)$.
Then $X=X\cap X^{g}\leq H\cap H^{g}$. Thus $g\in H$, since $X$ cannot be
contained in $H\cap N$, and since $H^{g}\cap H\leq H\cap N$ if $g\notin H$, by
Theorem 3.5. ∎

The example $G=C_{3}\times S_{3}$ from earlier shows that is necessary to
assume that $G=HM$ in Theorem 5.8.

##  6\. Multiple Frobenius triples

We now consider ways of constructing new Frobenius triples from existing ones.

######  Lemma 6.1.

Let $(G,N,M)$ be a Frobenius triple, and let $1<K$ be a normal subgroup of $G$
contained in $N$ satisfying $M\cap K<M$. Then $(G/K,N/K,MK/K)$ is a Frobenius
triple. In particular, if $N=M\times K$ then $G/K$ is a Frobenius group with
Frobenius kernel $N/K$.

###### Proof.

Let $\psi\in\mathrm{Irr}(G/K\mid MK/K)$. Then,considering $\psi$ as a
character of $G$, observe that $M\nleq\ker(\psi)$ and so
$\psi^{G}\in\mathrm{Irr}(G)$. Thus $\psi^{G/K}\in\mathrm{Irr}(G/K)$. Also
$\lvert MK/K\rvert=\lvert M/(M\cap K)\rvert$ divides $\lvert M\rvert$, which
is relatively prime to $\lvert G:N\rvert=\lvert G/K:N/K\rvert$. Thus
$(G/K,N/K,MK/K)$ is a Frobenius triple by Theorem C. ∎

We have the following immediate consequence when $K<M$.

######  Corollary 6.2.

Let $(G,N,M)$ be a Frobenius triple. If $1<K<M$ is normal in $G$, then
$(G/K,N/K,M/K)$ is a Frobenius triple.

In the next several results, we consider what happens when a group contains
two different Frobenius triples. We first consider products of the bottom
factors.

######  Lemma 6.3.

Let $M_{1},M_{2}$ and $N$ be normal subgroups of $G$ satisfying
$M_{1}M_{2}\leq N$. If $(G,N,M_{1})$ and $(G,N,M_{2})$ are Frobenius triples,
then so is $(G,N,M_{1}M_{2})$.

###### Proof.

First observe that $\mathrm{Irr}(N\mid M_{1}M_{2})=\mathrm{Irr}(N\mid
M_{1})\cup\mathrm{Irr}(N\mid M_{2})$. Thus every $\psi\in\mathrm{Irr}(N\mid
M_{1}M_{2})$ induces irreducibly to $G$. Since $(\lvert G:N\rvert,\lvert
M_{i}\rvert)=1$ for $1,2$, we see that $(\lvert G:N\rvert,\lvert
M_{1}M_{2}\rvert)=1$. Hence $(G,N,M_{1}M_{2})$ is a Frobenius triple. ∎

We obtain the following interesting consequence of Lemma 6.3.

######  Corollary 6.4.

Let $L_{1}$ and $L_{2}$ be normal subgroups of $G$. Then $\langle
C_{G}(x):1\neq x\in L_{1}L_{2}\rangle=\langle C_{G}(x):1\neq x\in L_{1}\cup
L_{2}\rangle$.

###### Proof.

Write $C_{i}=\langle C_{G}(x):1\neq x\in L_{i}\rangle$ for $1=1,2$ and
$C=\langle C_{G}(x):1\neq x\in L_{1}\cup L_{2}\rangle=C_{1}C_{2}$. Write
$N=\langle C_{G}(x):1\neq x\in L_{1}L_{2}\rangle$. It is easy to see that
$C\leq N$. If $C=G$, then we are done so assume that $C<G$. Then $(G,C,L_{1})$
and $(G,C,L_{2})$ are Frobenius triples. By Lemma 6.3 $(G,C,L_{1}L_{2})$ is
also a Frobenius triple; thus $C_{G}(x)\leq C$ for every $1\neq x\in
L_{1}L_{2}$. Hence $N\leq C$, as required. ∎

We next consider the case when two different quotients are Frobenius triples.
The most important case is when $K_{1}\cap K_{2}=1$.

######  Lemma 6.5.

Let $M$ and $N$ be normal subgroups of $G$ satisfying $M\leq N$. Let $K_{1}$
and $K_{2}$ be normal subgroups of $G$ properly contained in $M$. If
$(G/K_{i},N/K_{i},M/K_{i})$ is a Frobenius triple for $i=1,2$, then so is
$(G/(K_{1}\cap K_{2}),N/(K_{1}\cap K_{2}),M/(K_{1}\cap K_{2}))$.

###### Proof.

Write $K_{12}=K_{1}\cap K_{2}$. Let $g\in G\setminus N$. Let $D_{i}=\\{x\in
G:[g,x]\in K_{i}\\}$ for $i=1,2$, let $D_{12}=\\{x\in G:[g,x]\in K_{12}\\}$,
and observe that $D_{12}=D_{1}\cap D_{2}$. Since $(G/K_{i},N/K_{i},M/K_{i})$
is a Frobenius triple for $i=1,2$, it follows that $D_{i}\cap M\leq K_{i}$ for
$i=1,2$. Hence, $D_{12}\cap M=(D_{1}\cap D_{2})\cap M\leq K_{1}\cap
K_{2}=K_{12}$, and we deduce that $C_{M/K_{12}}(gK_{12})=D_{12}/K_{12}\cap
M/K_{12}$ is trivial. It follows that $(G/K_{12},N/K_{12},M/K_{12})$ is a
Frobenius triple, as desired. ∎

We now return to the case where we have two Frobenius triples, and we consider
the intersection of the bottom subgroups.

######  Lemma 6.6.

If $(G,N_{1},M_{2})$ and $(G,N_{2},M_{2})$ are Frobenius triples and
$M_{1}\cap M_{2}>1$, then $(G,N_{1}\cap N_{2},M_{1}\cap M_{2})$ is a Frobenius
triple.

###### Proof.

For each element $1\neq x\in M_{1}\cap M_{2}$, we have $C_{G}(x)\leq N_{1}$
since $x\in M_{1}$ and $(G,N_{1},M_{1})$ is a Frobenius triple; similarly
$C_{G}(x)\leq N_{2}$. ∎

##  7\. A Galois connection

Let $G$ be a group. For a character $\chi$ of $G$, the vanishing-off subgroup
of $\chi$—denoted $V(\chi)$—is the smallest subgroup $V$ of $G$ such that
$\chi(g)=0$ for all $g\in G\setminus V$. For a nontrivial, proper normal
subgroup $M$ of $G$ Mlaiki [17] defines the subgroup $V(G\mid M)$ to be the
product of the subgroups $V(\chi)$ for $\chi\in\mathrm{Irr}(G\mid M)$. This
subgroup gives a convenient way of characterizing Camina triples. Mlaiki shows
that $(G,N,M)$ is a Camina triple if and only if $V(G\mid M)\leq N$.

In [1] the authors define another subgroup $U(G\mid N)$ for any normal
subgroup $N$ of $G$ that can also be used to characterize Camina triples. For
$M,N\lhd G$, we show that $V(G\mid M)\leq N$ if and only if $M\leq U(G\mid
N)$. This not only gives the alternate characterization, but also shows that
the two maps $V(G\mid\rule{7.11317pt}{0.5pt})$ and
$U(G\mid\rule{7.11317pt}{0.5pt})$ provide a (monotone) Galois connection on
the lattice of normal subgroups of the group $G$. In the section we explore a
similar Galois connection coming from Frobenius triples.

We define $C(G\mid M)=\langle C_{G}(x)\mid 1\neq x\in M\rangle$. It is not
difficult to see that if $N$ is a proper normal subgroup of $G$ containing
$M$, then $(G,N,M)$ is a Frobenius triple if and only if $C(G\mid M)\leq N$.
Furthermore, note that $M$ will be a Frobenius kernel if and only if $C(G\mid
M)=M$.

For $N\lhd G$, define $I(G\mid N)=\prod_{M\in\mathcal{F}}M$, where
$\mathcal{F}=\\{M\lhd G:C(G\mid M)\leq N\\}$. By Lemma 6.3, it is clear that
$(G,N,M)$ is a Frobenius triple if and only if $M\leq I(G\mid N)$. It is also
clear that $N$ will be a Frobenius kernel if and only if $N=I(G\mid N)$.

We summarize these results below, as well as some other basic properties that
hold in general for Galois connections.

######  Lemma 7.1.

Let $1<M,N<G$ be normal subgroups of $G$. The following hold:

  1. (1)

$C(G\mid M)\leq N$ if and only if $M\leq I(G\mid N)$;

  2. (2)

$C(G\mid M_{1}M_{1})=C(G\mid M_{1})C(G\mid M_{2})$;

  3. (3)

$I(G\mid N_{1}\cap N_{2})=I(G\mid N_{1})\cap I(G\mid N_{2})$;

  4. (4)

$M\leq I(G\mid C(G\mid M))$ and $C(G\mid I(G\mid N))\leq N$;

  5. (5)

$Z(G)\cap I(G\mid N)=1$;

  6. (6)

$(G,N,M)$ is a Frobenius triple if and only if $C(G\mid M)\leq N$;

  7. (7)

$G$ is a Frobenius group if and only if $C(G\mid F(G))=F(G)$, which happens if
and only if $F(G)=I(G\mid F(G))$;

  8. (8)

$Z(G)M\leq V(G\mid M)\leq C(G\mid M)$ and $I(G\mid N)\leq U(G\mid N)\leq N\cap
G^{\prime}$.

###### Proof.

Statement (1) is clear from the definitions. Since $C(G\mid M_{1})C(G\mid
M_{2})=\langle C_{G}(x):1\neq x\in M_{1}\cup M_{2}\rangle$, statement (2) is
just Corollary 6.4. One direction of statement (3) is clear from the
definition; the other follows immediately from Lemma 6.6. Statement (4)
follows immediately from statement (1). Since every $g\in G\setminus N$ acts
fixed-point-freely on $I(G\mid N)$, statements (5), (6) and (7) follow.

We now prove (8). Since $(G,C(G\mid M),M)$ is a Frobenius triple, it is also a
Camina triple by Lemma 2.4. Thus $V(G\mid M)\leq C(G\mid M)$ follows from [17,
Theorem 2.1]. Similarly $I(G\mid N)\leq U(G\mid N)$ [1, Lemma 5.8]. The
remaining two containments follow from Lemmas 5.2, 5.3 and 5.4 of [1] and
Lemma 2.4 of [17]. ∎

The next result gives two alternate descriptions of $I(G\mid N$). As is
standard, when $\psi$ is an irreducible character of a normal subgroup $N$ of
$G$, we let $I_{G}(\psi)$ denote the stabilizer of $\psi$ under the natural
action of $G$ on $\mathrm{Irr}(N)$ induced by conjugation.

######  Lemma 7.2.

Let $1<M,N<G$ be normal subgroups of $G$. Write $\pi=\pi(G/N)$. The following
hold:

  1. (1)

A normal subgroup $M$ of $G$ satisfies $M\leq I(G\mid N)$ if and only if
$M\leq O_{\pi^{\prime}}(G)$ and $\psi^{G}\in\mathrm{Irr}(G)$ for every
$\psi\in\mathrm{Irr}(N\mid M)$;

  2. (2)

$I(G\mid
N)=O_{\pi^{\prime}}(G)\cap\bigcap\limits_{\begin{subarray}{c}\psi\in\mathrm{Irr}(N)\\\
I_{G}(\psi)>N\end{subarray}}\ker(\psi)$.

###### Proof.

Statement (1) is a restatement of Theorem C.

Since $(G,N,I(G\mid N))$ is a Frobenius triple, $I(G\mid N)\leq
O_{\pi^{\prime}}(G)$ by Lemma 2.5. Let $K$ be the intersection of the kernels
of all irreducible characters $\psi$ of $N$ satisfying $I_{G}(\psi)>N$.
Observe that every $\chi\in\mathrm{Irr}(N\mid I(G\mid N))$ induces irreducibly
to $G$ by Theorem C. It is easy to see that this is equivalent to
$I_{G}(\psi)=N$ for all $\psi\in\mathrm{Irr}(N\mid I(G\mid N))$. Since
$I_{G}(\psi)=N$ if and only if $\psi\in\mathrm{Irr}(G\mid K)$, we see that
$\mathrm{Irr}(G\mid I(G\mid N))\subseteq\mathrm{Irr}(G\mid K)$, which gives
$I(G\mid N)\leq K$. Now let $L=O_{\pi^{\prime}(G)}\cap K$. Then
$\mathrm{Irr}(G\mid L)\subseteq\mathrm{Irr}(G\mid K)$, so every
$\psi\in\mathrm{Irr}(G\mid L)$ induces irreducibly to $G$. Since $L\leq
O_{\pi^{\prime}}(G)$, it follows from Theorem C that $(G,N,L)$ is a Frobenius
triple. Thus $L\leq I(G\mid N)$ by Lemma 7.1. ∎

##  8\. Examples

We now present a number of examples to see what can occur in Frobenius
triples. The first result tells us that we cannot expect to say anything about
the structure of $N/M$ when $(G,N,M)$ is a Frobenius triple.

######  Lemma 8.1.

Let $(G,N,M)$ be a Frobenius triple and let $A$ be any group. Then $(G\times
A,N\times A,M)$ is a Frobenius triple.

###### Proof.

Consider any element $1\neq x\in M$. Then $C_{G\times A}(x)=C_{G}(x)\times
A\leq N\times A$. ∎

In particular, if $H$ is any group, then there is a Frobenius triple $(G,N,M)$
so that $N/M\cong H$ by taking $G$ to be a Frobenius group with Frobenius
kernel $A$. This lemma shows that $(G\times A,N\times A,N)$ is a Frobenius
triple.

We next show that if we have a subgroup of prime order acting coprimely, then
the resulting semidirect product yields a Camina triple.

######  Example 8.2.

Let $N$ be an abelian group and $H$ be a cyclic group of prime order not
dividing $\lvert N\rvert$. Suppose that $H$ acts on $N$ and let $G=H\ltimes
N$. We claim that $(G,N,G^{\prime})$ is a Frobenius triple. To see this, first
note that $(\lvert G:N\rvert,\lvert G^{\prime}\rvert)=1$. Let
$\psi\in\mathrm{Irr}(N\mid G^{\prime})$. Since $G^{\prime}\nleq\ker(\psi)$,
$\psi$ does not extend to $G$. Thus $\psi^{G}$ is irreducible and
$(G,N,G^{\prime})$ is a Frobenius triple, as claimed.

We next present an example of a Frobenius triple which is not split over the
bottom subgroup.

######  Example 8.3.

Let $\mathbb{F}$ be the finite field of cardinality 64, and let
$K\leq\mathrm{GL}_{3}(\mathbb{F})$ be the subgroup of all upper triangular
unipotent matrices. Let $D\leq\mathrm{GL}_{3}(\mathbb{F})$ be the subgroup of
all diagonal matrices of the form $\mathrm{diag}(1,a,a^{2})$, where
$a\in\mathbb{F}^{\times}$. Let $H$ be a Sylow $3$-subgroup of $D$. Then $G=HK$
is a Frobenius group with Frobenius kernel $K$. Let $M=K^{\prime}$, and let
$N\leq G$ satisfy $\lvert N:K\rvert=3$. Then $(G,N,M)$ is a Frobenius triple
that does not split over $M$, since $M\leq\Phi(G)$.

Next, we consider a class of examples that has arisen in a number of places in
the literature.

######  Example 8.4.

Let $(G,N)$ be a Camina pair with $G/N$ a $p$-group for some prime $p$. Assume
that $G$ is not a $p$-group and that $(G,N)$ is not a Frobenius group. Chillag
and Macdonald conjectured in [3] that $G$ is a Frobenius group with Frobenius
complement isomorphic to the quaternion group $Q_{8}$ and $\lvert
G:N\rvert=4$. This conjecture is verified by Dark and Scoppola in [5] in the
case that $G/N$ is abelian. Isaacs [11] showed that $G$ must have a solvable
normal $p$-complement $M$ and that $C_{G}(x)\leq N$ for every nonidentity
element $x\in M$. Around the same time, Chillag, Mann and Scoppola
independently discovered similar results in [4] and showed that $(G,P,P\cap
N)$ is a Frobenius–Wielandt group with Frobenius–Wielandt kernel $N$, where
$P$ is a Sylow $p$-subgroup of $G$. In particular we see that $(G,N,M)$ is a
Frobenius triple.

We know that when $G$ is a Frobenius group, the set of Frobenius complements
are conjugate. It is natural to ask what can be said about the psuedo-
complements for a Frobenius triple. We present two examples to show that very
little can be said. The first shows that there is not a unique conjugacy class
of such psuedo-complements. In fact, it also established that there is not
even a unqiue conjugacy class of maximal psuedo-complements.

######  Example 8.5.

Take $G=A\times B$, where $A=B=S_{3}$, $M=A^{\prime}$, and $N=A^{\prime}\times
B$. It is not difficult to see that if $x\in M\setminus\\{1\\}$, then
$C_{G}(x)=N$. It follows that $(G,N,M)$ is a Frobenius triple. Take $D$ to be
the diagonal subgroup of $G$; observe that the only maximal subgroup
containing $D$ is $DM$. (This can be verified this with MAGMA). So $D$ is
itself maximal subject to the condition that $G=DN$ and $D\cap M=1$. Let
$H=P\times B$, where $P$ is a Sylow $2$-subgroup of $G$. It is not difficult
to see that $H$ is a pseudo-complement of $G$. Then $\lvert{H}\rvert=12$ is
the maximum order of a psuedo-complement. It turns out that every psuedo-
complement of order $12$ is conjugate to $H$. (This is also checked with
MAGMA). So there is not a unique conjugacy class of maximal psuedo-
complements, and there can exist conjugacy classes of psuedo-complements of
different orders.

The next example shows that there can even be two different conjugacy classes
of psuedo-complements of maximal possible order.

######  Example 8.6.

Take $G$ to be the Frobenius group $C_{2}\ltimes(C_{3}\times C_{3})$. Let $P$
be a Sylow $2$-subgroup and let $M$ be minimal normal. Let $L_{1},L_{2},L_{3}$
be the remaining minimal normal subgroups of $G$. Let $H_{i}=PL_{i}$ for
$i=1,2,3$. Let $N=F(G)$. Then $\lvert{M}\rvert=3$ and $\lvert{G:N}\rvert=2$.
Also, $G=H_{i}N$ and $H_{i}\cap M=1$ for $i=1,2,3$. Note that $H_{i}$ and
$H_{j}$ are not conjugate if $i\neq j$. Thus there is not a unique conjugacy
class of psuedo-complements of maximal order. Also, note that $(G,N,M)$ is a
Frobenius triple, so this doesn’t even hold for them.

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