Let $|G|=2^n3$. Show that there’s a normal subgroup of order $2^n$ or $2^{n-1}$.

Question

Let $|G|=2^n3$. Show that there's a normal subgroup of order $2^n$ or $2^{n-1}$.

What I did so far:

I used the third Sylow theorem, which says that if $|G|=p^nb$ with $\operatorname{GCD}\{p,b\} = 1$ then the number of $p$-Sylow subgroups, $n_p$, divides $b$ and $n_p \equiv 1 \operatorname{mod}p$. From here, we have that $n_2 = 1$ or $n_2 = 3$. If $n_2 = 1$, then the $2$-Sylow subgroup of $G$ is normal and the order is $2^n$. If $n_2 = 3$ then I need to show somehow that there's a normal subgroup of order $2^{n-1}$.

Any hints on how to continue?

Answer 1

Hint: Assume $n_2=3$. Let $Syl_2(G)$ be the set of $2$-Sylow subgroups. Define an action of $G$ on $Syl_2(G)$ by conjugation, i.e $g.P=gPg^{-1}$. This action induces a homomorphism $\varphi:G\to S_{Syl_2(G)}\cong S_3$. Can you show $\varphi$ is surjective? If so, can you finish the exercise?