Question on simple subgroup $H$ and a normal subgroup $N$, of $G$

Question

This one is a bit strange to me, mainly the third hypothesis. It goes as follows:

Given a group (finite) group $G$, and $N, H \leq G$ such that

• $N$ is normal in $G$, and $H$ is simple
• $NH = G$
• $|H|$ has some prime divisor greater than $|N|$

Prove that

1. Every element of $N$ commutes with every element of $H$
2. $G\simeq N\times H$

I've done a similar exercise, which involved proving several equivalences of a group being a direct product of other groups. The hypotheses were a bit different though, and I can't see how I can work with these ones to find the same result. Any hints?

Answer 1

Observe that $$H \cap N \unlhd H$$ i.e. $H \cap N$ is a normal subgroup in $H$; but $H$ is simple, $H \cap N$ can't be $H$ due to the hypothesis about the prime divisor and thus $$H \cap N = 1 $$ and $|G| = |H| \cdot |N|$.

Consider the action of $G$ on the lateral classes of $H$ $$g \cdot xH = gxH$$ The kernel $K$ of this action is the largest normal subgroup of $G$ contained in $H$ (as you can check).

$H$ is simple and so there are only $2$ cases: $K= H $ or $K = 1 $, because clearly $K$ is normal in $H$ . The lateral classes of $H$ are $|N|$ and thus $$\frac{|G|}{|K|} \mid |N|! $$ By the hypothesis about the prime divisor and the result we found about $|G|$ this implies $K = H$, and thus $H$ is normal in $G$.

Thus $G \cong H \times N$, and this implies that every element of $N $ commutes with every element of $H$.