Suppose that a finite group $G$ admits a Frobenius group of
automorphisms $F H$ with kernel $F$ and complement $H$ such that $F$
acts without nontrivial fixed points (that is, such that $C\_G(F)=1$).
It is proved by Belyaev and Hartley in [Centralizers of finite nilpotent subgroups in locally finite groups](https://doi.org/10.1007/BF02367023) that $G$ is a solvable group.
So, there are some papers studied on the Fitting height of $G$, exponent of $G$, rank of $G$.
For example, a paper [Fitting height of a finite group with a Frobenius group of automorphisms](https://doi.org/10.1016/j.jalgebra.2012.05.011) by E.I. Khukhro.
I am curious about the following question.


**Question: If $G$ is an abelian group which is $FH$-indecomposable and $F$ acts fixed-point-freely on $G$, I am wondering whether it is true that $G$ is homocyclic $p$-group.**


Obviously, $G$ is a $p$-group for some prime $p$.
If the action of $FH$ on $G$ is coprime, then a result of M. Harris shows that $G$ is homocyclic without assuming that $FH$ being a Frobenius group.
So, we may assume that $p\mid |H|$.
If $G$ is an elementary abelian group, then $G$ has a basis which is permuted by $H$ by Theorem 15.16 in "[Character theory of finite groups](https://doi.org/10.1090/chel/359)" by I.M. Isaacs.


Here is an example of a very special case: $G$ is an abelian $2$-group which is $FH$-indecomposable and $FH\cong S\_3$.
Then $$G=C\_G(H)\times C\_G(H)^x$$ where $F=\langle x\rangle$.
Since $F$ acts fixed-point-freely on $G$, $C\_G(H)$ is a cyclic subgroup.
So, $G$ is a homocyclic 2-group. Also, power map defines isomorphism of $FH$-chief factors of $G$.


Any explanation, references, suggestion and examples are appreciated.



Finite abelian group admits a Frobenius group of automorphism
Let $V$ be an $\mathbb{F}\_p[X]$-module such that $p\mid |G|$.
Then $V$ is not necessarily completely reducible.
A classical example is: $X\cong C\_p$ and $V\cong C\_p \times C\_p$ and $XV$ is extraspecial $p$-group of order $p^3$.


Richard Lyons's [idea](https://mathoverflow.net/questions/444134/finite-abelian-group-admits-a-frobenius-group-of-automorphism#comment1147095_444134) is: when a group $X$ acting completely reducibly on every $X$-invariant elementary abelian subquotient of $A$, $A$ must decompose into a direct product of $X$-invariant homocyclic groups.
Here is a proof based on the idea of Richard Lyons.


**Proposition** Let $A$ be a finite abelian $p$-group on which a group $X$ acts.
Suppose that that for every $X$-invariant subquotient $B/C$ of $A$ $($i.e., both $B$ and $C$ are $X$-invariant)
such that $B/C$ is elementary abelian, $X$ acts completely reducibly on $B/C$.
Then $A$ is a direct product of $X$-invariant homocyclic subgroups.


**Proof** Let $A$ be a counterexample of minimal possible order.
Then $A$ is $X$-indecomposable.
Let $\overline{A}=A/\Phi(A)$, and observe that $X$ acts completely reducibly on $\overline{A}$.
Then
$$\overline{A}=\overline{B}\times \overline{C}$$
where $\overline{B}$ and $\overline{C}$ is completely reducible such that $\Phi(A)=B\cap C$, $\exp(B)=\exp (A)$.
By the minimality of $A$, $B$ and $C$ are both direct product of $X$-invariant homocyclic subgroups,
and, without loss of generality, we may assume that $B$ is homocyclic.
Also, $\Phi(B)=\Phi(A)$.
Let $F$ be a maximal homocyclic subgroup of $A$ containing $B$.
Then $\Phi(B)\leq \Phi(F)\leq \Phi(A)$.
However, since $\Phi(B)=\Phi(A)$, $\Phi(F)=\Phi(B)$, and hence $B=F$ (as $F$ is homocyclic).
By Krull-Remak-Schmidt's theorem, $A=B\times V$ where $V$ is elementary abelian (as $\Phi(A)=\Phi(B)$),
i.e. $A=B\Omega\_1(A)$, where $\Omega\_1(A)$ is a subgroup of $A$ generating by every element of order 2 in $A$.
By the assumption of this proposition, $\Omega\_1(A)$ is completely reducible.
Therefore, $\Omega\_1(A)=\Omega\_1(B)\times D$, where $\Omega\_1(B)$ and $D$ are $X$-invariant.
Observe that $B\cap D=\Omega\_1(B)\cap D=1$, and hence $A=B\times D$ is $X$-decomposible, a contradiction.


I guess Dave Benson's [idea](https://mathoverflow.net/questions/444134/finite-abelian-group-admits-a-frobenius-group-of-automorphism#comment1146669_444134) is trying to show that: under the assumption of this question, every $\mathbb{F}\_p[FH]$-module $M$ such that $C\_M(F)=1$ is completely reducible.


The proof of Theorem 15.16 in "Character theory of finite groups" (Isaacs) use the same idea suggested by Dave Benson.


