Showing the product of two normal subgroups is normal [closed]

Prove that if $H$ or $K$ are normal subgroups then $HK=\{hk\mid h\in H,k\in K\}$ is a subgroup. Then if both are normal subgroups, prove that HK is normal.

Clearly $HK=\{hk:h\in H,k\in K\}$ is nonempty (as $H$ and $K$ are nonempty). We appeal to the subgroup test to show that $HK$ is indeed a subgroup of $G$.

Suppose that $h_1k_1$, $h_2k_2$ are two elements of $HK$ (so $h_1,h_2\in H$ and $k_1,k_2\in K$). We must show that $(h_1k_1)(h_2k_2)^{-1}\in HK$ also. Then by the subgroup test, we can conclude that $HK<G$ ("$<$" denotes subgroup here).

We have $(h_2k_2)^{-1}=k_2^{-1}h_2^{-1}$ (check this), and so: $$(h_1k_1)(h_2k_2)^{-1}= (h_1k_1)k_2^{-1}h_2^{-1}.$$

Writing $k_1k_2^{-1}=k\in K$, the above becomes $h_1 kh_2^{-1}\enspace (*)$.

Case 1: $K$ normal

If $K$ is normal, then $h_2 kh_2^{-1}=\overline{k}$ for some $\overline{k}\in K$. Left multiplying $h_2 kh_2^{-1}=\overline{k}$ by $h_2^{-1}$ gives : $kh_2^{-1}=h_2^{-1}\overline{k}$. With this in mind $(*)$ becomes:

$$h_1 kh_2^{-1}= h_1h_2^{-1}\overline{k}$$ Since $H$ is a subgroup $h_1h_2^{-1}\in H$ and so the above element belongs to $HK$ as desired.

Case 2: $H$ normal

The argument here is incredibly similar to the one above. $kh_2^{-1}k^{-1}=h$ for some $h\in H$ by normality of $H$ (since $h_2^{-1}\in H$) and so $kh_2^{-1}=hk$. Substituting this into $(*)$ gives the desired result.

Thus $HK$ is a subgroup of $G$ provided at least one of $H,K$ is normal.

$HK$ is normal if $H$ and $K$ are:

Let $hk$ be an arbitrary element of $HK$ and let $g\in G$. we must show $g(hk)g^{-1}\in HK$. Since $g^{-1}g=e$, where $e$ is the group identity of $G$, we can rewrite the above as: $$g(hk)g^{-1}=(gh)g^{-1}g(kg)=(ghg^{-1})(gkg^{-1}).$$ Since $H$ is normal $ghg^{-1}\in H$. Since $K$ is normal $gkg^{-1}\in K$. Thus the above belongs to $HK$, proving $HK$ is normal.

Hint:

$$(hk)^g:=g^{-1}hkg=(g^{-1}hg)(g^{-1}kg)$$