Any group of order $5^2q^m$ is not simple if both $q$ is a prime and $m$ a positive integer such that $5^2\nmid q^m!$

Question

Let $G$ be a group of order $5^2 q^m$ where $q$ is a prime and $m$ a positive integer such that $5^2\nmid q^m!$. Show that $G$ is not simple.

Attempt: Assume that $G$ is simple. The fact that $5^2\nmid q^m!$ implies that $$q^m < 5^2=25\tag{1}$$ Now if $n_5$ is the number of Sylow $5$-subgroups then $$n_5 = q^k \text{ for some } k=1,2,\dots, m \text{ such that } q^k \equiv 1 \pmod{5}\tag{2}$$ Doing the same for $n_q$ we get that the possible values for $n_q$ are $2,5$ and $10$. From the condition $n_q \equiv 1 \pmod{q}$ we get that the possible values for $q$ are $2$ and $3$

Now consider the cases:

• $q=2$: Relation $(2)$ tells us that $2^k \equiv 1 \pmod{5}$ so $m\geq k \geq 4$. On the other hand, the inequality at $(1)$ gives us $m \leq 4$ so $|G|=5^2\cdot 2^4$ but in that case $5^2 \mid 2^4!$ so this can't be the case.
• $q=3$: Doing the same procedures as above we get that $m\geq 4$ but inequality $(1)$ gives $m < 3$ so again that can't be the case.

I have two questions: Is the above solution correct? If not, could you point out the mistake? If yes, is there a more elegant solution that I am missing (at first sight the factorial on the divisibility condition screams for some group action)?

Any comment is appreciated.

EDIT: When calculating $n_q$ I wrote $n_q \mid 10$ instead of $n_q\mid 25$. Now the possible values for $n_q$ are $5$ and $25$ which again give $q=2$ or $q=3$ as the only possibilities (thanks Robert Shore for pointing out my mistake).

Answer 1

If $5^2 \nmid q^m!$, then $q^m \leq 9$. But then $q, q^m \neq 1 \pmod 5$ (because $6$ is a product of distinct primes) so the $5$-Sylow subgroup of $G$ is normal.

Answer 2

Though it is not an answer to your question, you can choose to apply this approach instead.

If $q=5$, this statement is trivial, thus we assume $q\neq 5$ then.

Let $Q$ be a Sylow $5$-subgroup of $G$, and let $G/Q$ be the collection of all cosets of $Q$. (Here $G/Q$ is not necessarily a group. We simply abuse this notation for convenience.)

Then $|Q|=25$ and $|G/Q|=q^m$.

We may let $G$ act on $G/Q$ by left multiplication, i.e., $$g\cdot aQ:=(ga)Q.$$ This action has a group homomorphism representation $\varphi:G\to\operatorname{Sym}(G/Q)\cong S_{q^m}$.

Let us consider the kernel $K$ of $\varphi$, which is clearly a normal subgroup of $G$. We claim that $K$ is also non-trivial: Suppose not. Then by the first isomorphism theorem, we have $$|\operatorname{im}(\varphi)|=|G/K|=|G|=5^2q^m.$$ Then by Lagrange theorem, since $\operatorname{im}(\varphi)\leq \operatorname{Sym}(G/Q)\cong S_{q^m}$, the order of $\operatorname{im}(\varphi)$ divides $q^m!$. Together we can see that $5^2q^m\mid q^m!$, which implies $5^2\mid q^m!$ as well, a contradiction.

Since $G$ has a non-trivial normal subgroup, it is not simple.