I am trying to prove a group of order $351$ is not simple. I know this has many answers that involve Sylow's Theorems, but I was trying to do this using an alternative method.
My idea is the following:
Let $G$ be a group such that $|G| = 351$.
If I can prove the existence of a subgroup $H$ of order $117$, then, by Lagrange's Theorem $[G: N] = 3$. Since $3$ is the smallest prime diving the order of $G$, then this would show that $H$ is normal in $G$. This of course shows that $G$ is not simple.
The problem is that I do not know if a subgroup of order $117$ necessarily exists. If it does exist, I do not know how to show it. Are Sylow's Theorem's the only way to solve this problem or will my method work?
Best Answer
There are fourteen groups of order $351$. One has no subgroup of order $117$. It has exactly six conjugacy classes of subgroups, of orders $1$, $3$, $9$, $13$, $27$ and $351$. Thus you cannot do it in such a way, and Sylow's theorems will be needed I expect.