Group of order $2^2 \cdot 3^n$, $n \geq 1$.

Question

Let $G$ be a group of order $2^2 \cdot 3^n$, with $n \geq 1$. The objective is to show that there is a normal subgroup of order $3^n$ or $3^{n-1}$; I can do the first part using the Sylow Theorems, but can't even tink about the second one. How to proceed?

Answer 1

We have that the number of $3$-sylow groups $n_3 =1 \mod 3$, and $n_3 = [ G: N_G(P)]\leq [G:P] = 4$, where $P$ is some $3$-sylow group. If $n_3 =1$ we are done as the single $3$-Sylow is fixed by automorphisms and therefore is normal.

Otherwise, if $n_3 =4$, $G$ acts (by conjugation) as permutations of the four $3$-Sylows, i.e. we have a map $\varphi: G \to S_4$, since conjugation acts transitively on the $3$-Sylows, its image must have order divisible by 4 (you cant act transitively on four elements otherwise), Since $\lvert S_4\rvert = 3 \cdot 8$, there are only two possibilities, $\lvert im(\varphi)\rvert = 4, 12$ and since $\lvert im(\varphi) \rvert \lvert ker(\varphi)\rvert = \lvert G \rvert = 4 \cdot 3^n$, we have that the kernel of $\varphi$ must have $3^n$ or $3^{n-1}$ elements, since we started by the asumption that we have four $3$-Sylows, we can't have a normal subgroup of order $3^n$, the kernel of that map must be a normal subgroup of order $3^{n-1}$.