Show that G is not simple where $|G| = 2^2 \cdot 5^3 \cdot 7^9$

Question

Show that if a group $G$ has order $|G|=2^2\cdot5^3\cdot7^9$, then $G$ is not simple.

$\textbf{Def:}$ A group $G$ is $\textbf{simple}$ if and only if $|G|>1$ and $\lbrace 1_G\rbrace$ as well as $G$ are its only normal subgroups.

I tried to use contradiction. First assuming that $G$ is simple. I was under the impression that we have to do this proof by cases with Sylow subgroups $n_2$, $n_5$, and $n_7$. I am stuck here since I thought we have to think about case 1 where $n_2 \equiv 1(\bmod2)$ and $n_2|5^3\cdot7^9 $. The trouble I'm having is wouldn't there be many factors where this case is true? Similarly for the other 2 Sylow subgroups.

Answer 1

Probably the intended solution is the argument in the comments using $n_7 = 50$, but it's worth knowing that we can completely avoid thinking about $n_7$. In fact we have the following.

Proposition: The order of a finite simple group $G$ is divisible by either $8$ or $12$ (assuming Feit-Thompson), or the smallest prime $p$ dividing the order is odd and divides it with multiplicity at least $3$ (not assuming Feit-Thompson, but Feit-Thompson implies this never happens).

Proof. By the Burnside normal $p$-complement theorem, if a Sylow $p$-subgroup $P$ is normal in its centralizer $N(P)$, then it has a normal complement (a subgroup splitting the short exact sequence $1 \to P \to G \to G/P \to 1$), and in particular $G$ is not simple. Let $p$ be the smallest prime dividing $|G|$ and write $\nu_p(G)$ for the exponent of $p$ dividing $|G|$. If $\nu_p(G) \le 2$ then the Sylow $p$-subgroup $P$ is abelian (it must be either $C_p, C_p \times C_p$, or $C_{p^2}$) and the order of its automorphism group is only divisible by primes $\le p$, unless $p = 2$. We now split into cases:

• $p$ odd (by Feit-Thompson this never happens): since $p$ is the smallest prime dividing $|G|$, the primes dividing $|N(P)|$ are all greater than $p$. It follows that $P$ is normal in $N(P)$ (since $N(P)/P$ can't act nontrivially on $P$), so $P$ has a normal complement. If $G$ is simple then this can't occur, so taking contrapositives, $\nu_p(G) \ge 3$.
• $p = 2$: $|\text{Aut}(P)|$ is only divisible by a prime $\ge 3$ when $P = C_2 \times C_2$, which has automorphism group $S_3$. If $G$ is simple then $P$ can't have a complement, so either $\nu_2(G) \ge 3$ or $P = C_2 \times C_2$ and the conjugation map $N(P)/P \to \text{Aut}(P)$ is surjective, which implies that $12 \mid |N(P)|$ and hence that $12 \mid |G|$. $\Box$