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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 https://link.springer.com/article/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 by E.I. Khukhro https://www.sciencedirect.com/science/article/pii/S0021869312002700. 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" 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.

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 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 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" 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.

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 https://link.springer.com/article/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 by E.I. Khukhro https://www.sciencedirect.com/science/article/pii/S0021869312002700. 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 which is $FH$-indecomposable, then $G$ has a basis which is permuted by $H$ by Theorem 15.16 in "Character theory of finite groups" 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.

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 https://link.springer.com/article/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 by E.I. Khukhro https://www.sciencedirect.com/science/article/pii/S0021869312002700. 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" 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.

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 https://link.springer.com/article/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 by E.I. Khukhro https://www.sciencedirect.com/science/article/pii/S0021869312002700. 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 which is $FH$-indecomposable, then $G$ has a basis which is permuted by $H$ by Theorem 15.16 in "Character theory of finite groups" 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 also define isomorphism of $FH$-chief factors of $G$.

Any explanation, references suggestion and examples are appreciated.

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 https://link.springer.com/article/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 by E.I. Khukhro https://www.sciencedirect.com/science/article/pii/S0021869312002700. 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 which is $FH$-indecomposable, then $G$ has a basis which is permuted by $H$ by Theorem 15.16 in "Character theory of finite groups" 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.