Sylow theorems exercise

Question

I just learned the Sylow theorems and I would like someone to to help me use these theorems. I'm also a beginner in group theory in general.

This is the exercise:

Prove that if we have $G,H$ two groups with orders $56$ and $2^4 5^6$ respectively, then $G,H$ are not simple groups.

For $|G|=56=7* 2^3$ we have from Sylow theorems that $|Syl_7(G)|=1\pmod7$

Also $|Syl_7(G)|=7k+1\mid|G|$ thus $7k+1\mid8$ and we must have that $|Syl_7(G)|=1$

so $G$ has a unique normal 7-Sylow subgroup, thus $G$ is not simple.

Is this proof right?

How can I tackle the second part of the exercise?

Thank you in advance!

Answer 1

Let $H$ be a simple group of order $2^45^6$, by Sylow's Theorem, $n_5\equiv 1 \bmod 5$ and $n_5\mid16$. So the possible values of $n_5$ is $1$ or $16$. Since $H$ is simple, $n_5=16$.

So there is a homomorphism $\phi:H\rightarrow S_{16}$ where $\ker \phi\leq N_H(P)$ for some $P\in Syl_5(H)$.
Since $H$ is simple, $\ker \phi=1$.
Thus we have $H\cong\phi(H)\leq S_{16}$.
This means that $|H|$ divides $16!$ which is a contradiction.

Answer 2

You got a problem with your solution to $|G| = 56$.

Denote by $n_k$ the number of the $k$-sylow subgroup of $G$.

We have $n_7 | 8$ and $n_7=1 mod 7$ thus $n_7 \in \{1,8\}$.

If $n_7 = 1$ , as you said, we are done.

Otherwise, we have $8$ 7-sylow subgroups , each of order $7$ so each is cyclic and they all intersects trivially ({e} ) , so the $7$-sylow subgroups "contributes " $8*6 = 48$ different elements, thus $n_2$ must be equal to one and we are done.