Prove a group of order 351 is not simple without Sylow's Theorems 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?
 A: 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.
A: You can also get away just with Cauchy, Lagrange and group actions:
Take a subgroup $H$ of order $13$ and look at the set $\{H^g\mid g\in G\}$ of its conjugates. As $G$ acts on this set (by conjugation), its order divides $|G|=351$. Restrict this action to $H$, and look at the orbits of $H$. $H$ fixing $H^g$ is equivalent to $H$ being contained in the normalizer of $H^g$. If $H\ne H^g$ for such an $H^g$, then $H\cdot H^g$ is a subgroup of order $13^2$ leading to a contradiction. Hence $H$ is the only fixed point of the action of $H$ on $\{H^g\mid g\in G\}$ and all other orbits of $H$ have order $|H|=13$. As $H$ is not normal in $G$, there are other orbits and the set $\{H^g\mid g\in G\}$ has order $1\bmod 13$. As the order divides 351 and is $>1$, we get that $H$ has 27 conjugates. In particular, $H$ is self-normalizing (i.e., its own normalizer) and therefore (as it's abelian) also self-centralizing (i.e., its own centralizer).
Now take a subgroup $T$ of $G$ of order $3$, and look at the set $\{T^g\mid g\in G\}$ of its conjugates, on which $G$ acts by conjugation. As $H$ is self-centralizing (and therefore does not commute with any element of order $3$), this action restricted to $H$ is fixed point free, so there are at least $13$ conjugates that intersect trivially (i.e., in $1$).
As the conjugates of $H$ intersect in $1$, the set $G\setminus\bigcup_{g\in G} (H^g\setminus 1)$ has order $351-27\cdot 12 = 27$ and contains $\bigcup_{g\in G} T^g$, so by the last paragraph $T$ has exactly $13$ conjugates. So $G$ acts on a set of 13 elements transitively, and the stabilizer $S$ of any point is therefore a subgroup of $G$ of order $27$. Hence $S = \bigcup_{g\in G} T^g$, and $S$ is normal in $G$.
