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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 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 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.

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 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 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.

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's idea 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's idea 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.

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 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 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.

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's idea 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's idea is trying to show that: under the assumption of this question, every $\mathbb{F}_p[FH]$-module is completely reducible.

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's idea 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's idea 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.