Prove that there are no simple groups of order 224.
Let $G$ be a finite group such that $\vert G \vert = 224 = 2^5 \cdot 7$. We know that $n_2 \mid 7$ and $n_2 \equiv 1 \pmod 2$ and we know that $n_7 \mid 2^5$ and $n_7 \equiv 1 \pmod 7$. So we can say $n_2 = 1$ or $7$ and $n_7 = 1$ or $8$. Suppose, to the contrary that $G$ is a simple group. Then $n_7 = 8$ and $n_2 = 7$. So we can say there are $8 \cdot 6 = 48$ elements of order 7 and $7 \cdot 31 = 217$ elements of order 2, which gives us 265 elements, 266 including the identity, which contradicts the cardinality of the group. Hence, $G$ is not a simple group since we must have at least one of $n_7$ or $n_2$ being 1.
Is this approach correct? specifically how i said there were $7 \cdot 31$ elements of order 2? I saw a few similar approaches online and thought it would work nicely for this problem. Is it correct?
Thank you!
Best Answer
The following proves there must exist a normal Sylow $\,2-$subgroup of order $\,32\,$,
Suppose there are $\,n_2=7\,$ Sylow $\,2-$subgroups in $\,G\,$ . Making $\,G\,$ act on the set of these Sylow subgroups by conjugation (Mitt wrote about this but on the set of the other Sylow subgroups, which gives no contradiction), we get a homomorphism $\,G\to S_7\,$ which must be injective if $\,G\,$ is simple (why?).
But this cannot be since then we would embed $\,G\,$ into $\,S_7\,$ , which is impossible since $\,|G|\nmid 7!=|S_7|\,$ (why?)