Prove that a group of order $2^5 \cdot 7^5$ is not simple

Question

I can see that there are 8 Sylow 7-subgroups. I'm told I should consider the normalizer of one of the Sylow 7's, but I'm not sure what to do exactly.

Answer 1

The normalizer $H$ in question has index $8$ in the group $G$. Let $G$ act by right multiplication on the set of right cosets of $H$ in $G$. This yields a homomorphism $\varphi : G \to S_{8}$. Since $7$ is the highest power of $7$ that divides the order of $S_{8}$, the kernel $\bigcap_{g \in G} H^{g}$ of $\varphi$ is a normal subgroup of $G$ that is non-trivial (it has order divisible by $7^4$), and proper, as it is contained in $H$.

Answer 2

The Index Theorem:

If $G$ is a finite group and $H$ is a proper subgroup of $G$ such that $|G|$ does not divide $|G:H|!$, then $H$ contains a nontrivial normal subgroup of G. In particular, $G$ is not simple.

Let $H$ be proper subgroup of order $7^5$(Sylow's first theorem),

$|G|=2^5.7^5$ does not divide $\displaystyle|G:H|!=\bigg(\frac{|G|}{|H|}\bigg)!=\bigg(\frac{2^5.7^5}{7^5}\bigg)!=32!$

Therefore, $H$ contains non-trivial normal proper subgoup of $G$.

Hence, $G$ is not simple group.