Suppose that $H,N,M$ are subgroups of $G$, $M$ is a normal subgroup of $N$. Show that $(N\cap H)M$ is a subgroup of $N$.

Question

Suppose that $H,N,M$ are subgroups of $G$, $M$ is a normal subgroup of $N$. Show that $(N\cap H)M$ is a subgroup of $N$.

I know all elements in $(N\cap H)M$ should be also in $N$. I just need to test the criteria of subgroups. However, I really don't know how to prove.

Answer 1

It is a direct application of the more general result:

Let $G$ be a group, $H,K$ subgroups of $G$ such that $K$ is normal. Then $HK=\{hk\mid h\in H,\: k\in K\}$ is a subgroup of $G$.

All that has to be proved is that, if $\:hk$ and $h_1k_1\in HK$, then $(hk)(h_1k_1)\in HK$.

Indeed, $(hk)(h_1k_1)=h(kh_1)k_1$. Now, as $K$ is normal, $kh_1=h_1 k'$ for some $k'\in K$, so $\;(hk)(h_1k_1)=h(h_1k')k_1=(hh_1)(k'k_1)$, which belongs to $HK$.

Answer 2

Just another perspective:

If $G$ is a group with $H,K$ subgroups where $K$ is normal, let $\pi:G\to G/K$ be the quotient map. Then

$$HK = \pi^{-1}(\pi(H)).$$

Since the image and preimage of groups are also groups, we see that $HK$ must be a group as well.