Prove that a group of order 150 is not simple.

Question

Prove that a group of order 150 is not simple.

Answer 1

If we have the following:

For any finite group $G$ and subgroup $H$ we have $|G:H|=n$ then there exists a normal subgroup $K$ of $G$ contained in $H$ and $|G:K|$ divides $n!$.

So suppose that we have that $G$ is not simple and so by Sylow theory we have $6$ subgroups of order $25$, denote one $X$. We also have that $K=\{e\}$ above.

Then consider $N_{G}(X)$

Now we have that $|G:N_{G}(X)|=6$

Then as $|K|=1$ we have $|G|$ divides $6!$ but $150$ does not so this is a contradiction and so $G$ is simple.

Although looking at it I think we need an argument using group actions to prove the proposition that I used so it probably doesn't really help. (It's basically the argument in the other answer but using my proposition to cover up the group action I think.)

Answer 2

A little different approach: if the 5-Sylow subgroup is not normal (and thus not unique), then there are 6 such subgroups, which means that the normalizer of any of them has index 6.

But if the group were simple, then making it act on the left cosets of that normalizer we'd get an embedding of the group into $\,S_6\,$ , something that is absurd as $\,150\nmid6!$