group of order 36 is solvable I need to prove that group G of order 36 is solvable.
This is what I came with so far:
There is either one 3-sylow subgroup or 4 3-sylow subgroups.
-first case: if there is only one 3-sylow subgroup than P- a 3-sylow subgroup of G is normal. it is also abelian and therefor solvable. [G:P] = 4 and therefor G/P is abelian so it is solvable and we are done.
-second case: if there are four 3-sylow subgroups it means by sylow theory that there is a subgroup N of index 4 so G/N is abelian and therefor solvable and N is of order 9 so it's also abelian and solvable. 
the only thing I'm missing is that in the second case I don't know how to prove that N is also normal in G.
 A: In the second case, the trick is as follows: Let $S$ denote the set of all 3-Sylow subgroups, so that $|S|=4$. Let $G$ act on $S$ by conjugation (note that this is a well-defined action). This gives a homomorphism $\varphi : G\to S_4$. Since $|G|>|S_4|$, this map cannot be injective. 
So let $N$ be the kernel of this homomorphism. Then $N$ is normal in $G$ and $G/N$ is isomorphic to a subgroup of $S_4$. Hence, $|N|\geq 3$, so I claim that both $N$ and $G/N$ are solvable. Let us consider the cases:


*

*$|N| = 3$, so $|G/N| = 12$. Then $N$ is cyclic and hence solvable. I claim that $G/N$ is also solvable: Consider the number $n_3$ of $3$-Sylow subgroups in $G/N$. If $n_3=1$, then we are done, but if $n_3=4$, then all the 3-Sylow subgroups intersect trivially, so there must be exactly $8=4\times 2$ elements of order 3 in $G/N$. Hence, all the remaining 4 elements must be in a single 2-Sylow subgroup, which must be consequently normal. Hence, $G/N$ is solvable.

*$|N|=6$, then $|G/N|=6$. Once again, one of the Sylow subgroups must be normal, so both groups are solvable.

*$|N|=9$, then $|G/N|=4$, so both are abelian - hence solvable.

*$|N|=12$, then $|G/N| = 3$. Again apply the argument from $1$ to see that both are solvable.
In all cases, both $N$ and $G/N$ are solvable, so $G$ is solvable.
