$G/F(G)$ is isomorphic to $X_1\times\cdots\times X_t$

Question

Since I still don’t know the answer, I’ve also asked it on math.overflow.

I saw a remark in an old post that

$G/F(G)$ is isomorphic to a group of the form $X_1 \times \ldots \times X_t,$ where each $X_i$ is isomorphic to a subgroup of a completely reducible solvable subgroup of ${\rm GL}(n_i,p_i)$ for some integer $n_i$ and prime $p_i$.

Before I ask my question, I need to fill in some details.

It is well-known that for a finite solvable group $G$, $G/F(G)$ is isomorphic to a subgroup of ${\rm Out}(F(G))$. Let $p_i(i=1,\cdots, t)$ be all the prime divisors of $|G|$. It follows that $G/F(G)$ is isomorphic to a subgroup of ${\rm Out}(O_{p_1}(G))\times\cdots\times{\rm Out}(O_{p_t}(G))$.

If $\Phi(G)=1$, then $\Phi(O_{p_i}(G))\le\Phi(G)=1$ for each $p_i$ dividing $|G|$. It follows that each $O_{p_i}(G)$ is elementary abelian, namely there exists a positive integer $n_i$ such that $O_{p_i}(G)=C_{p_i}^{n_i}$. Since we can see elementary abelian groups as vector spaces, ${\rm Aut}(O_{p_i}(G))\cong{\rm GL}_{n_i}(\mathbb{F}_{p_i})$. In addition, since $O_{p_i}(G)$ is abelian, ${\rm Inn}(O_{p_i}(G))$ is trivial and hence ${\rm Out}(O_{p_i}(G))\cong {\rm Aut}(O_{p_i}(G))\cong{\rm GL}_{n_i}(\mathbb{F}_{p_i})$. Then we know that $G/F(G)$ is isomorphic to a subgroup of ${\rm GL}_{n_1}(\mathbb{F}_{p_1})\times\cdots\times{\rm GL}_{n_t}(\mathbb{F}_{p_t})$.

If $\Phi(G)\neq 1$, we still can know that $\Phi(G/\Phi(G))$ is trivial. Since $G$ is solvable, we have $\Phi(G)\subseteq F(G)$ and $F(G)/\Phi(G)=F(G/\Phi(G))$. By the third isomorphism theorem, $$G/F(G)\cong G/\Phi(G)\big/F(G)/\Phi(G)= G/F(G)\cong G/\Phi(G)\big/F(G/\Phi(G)) .$$Hence we can get a similar conclusion.

My question is: By the reasoning above, I can only know that $G/F(G)$ is isomorphic to a subgroup of ${\rm GL}_{n_1}(\mathbb{F}_{p_1})\times\cdots\times{\rm GL}_{n_t}(\mathbb{F}_{p_t})$. In general, if $A$ is a subgroup of $A_1\times\cdots\times A_t$, we cannot say $A$ is of the form $X_1\times\cdots\times X_t$, where $X_i$ is isomorphic to a subgroup of $A_i$. But here each $X_i$ is isomorphic to a subgroup of ${\rm GL}(n_i,p_i)$. I want to know what was different in this specific case.

Thanks!

Answer 1

As I have said in comments, it is not hard to prove the remark as stated.

But here is a stronger statement that is definitely not true.

Let $G$ be solvable and suppose that $F(G)$ is a direct product of elementary abelian groups $C_{p_i}^{n_i}$ for fprimes $p_1,\ldots,p_t$. Then $G/F(G)$ is isomorphic to a direct product $X_1 \times \cdots \times X_t$, where $X_i$ is isomorphic to a solvable completely reducible subgroup of ${\rm GL}(n_i,p_i)$.

That is not true. I can construct an example in which $F(G) = C_2^2 \times C_5^2$. and $G/F(G)$ is isomorphic to a subgroup of index $2$ in$X_1 \times X_2$ with $X_1 \cong X_2 \cong S_3$, but $G/F(G)$ is indecomposable and is not itself isomorphic to $X_1 \times X_2$ for any $X_1 \le {\rm GL}(2,2)$ and $X_2 \le {\rm GL}(2,5)$.