First, we shall prove that $HK$ is a subgroup of $G$: Since $e\in H$ and $e\in K$, clearly $e=e^{2}\in HK$. Take $h_{1},h_{2}\in H,k_{1},k_{2}\in K$. Clearly $h_{1}k_{1},h_{2}k_{2}\in HK$. Further,

Since $K$ is a normal subgroup of $G$ and $h_{2}\in G$, then $h_{2}^{{-1}}k_{1}h_{2}\in K$. Therefore $h_{1}h_{2}(h_{2}^{{-1}}k_{1}h_{2})k_{2}\in HK$, so $HK$ is closed under multiplication.

Also, $(hk)^{{-1}}\in HK$ for $h\in H$, $k\in K$, since

and $hk^{{-1}}h^{{-1}}\in K$ since $K$ is a normal subgroup of $G$. So $HK$ is closed under inverses, and is thus a subgroup of $G$.

Since $HK$ is a subgroup of $G$, the normality of $K$ in $HK$ follows immediately from the normality of $K$ in $G$.

Clearly $H\cap K$ is a subgroup of $G$, since it is the intersection of two subgroups of $G$.

Finally, define $\phi\colon H\rightarrow HK/K$ by $\phi(h)=hK$. We claim that $\phi$ is a surjective homomorphism from $H$ to $HK/K$. Let $h_{0}k_{0}K$ be some element of $HK/K$; since $k_{0}\in K$, then $h_{0}k_{0}K=h_{0}K$, and $\phi(h_{0})=h_{0}K$. Now

and if $hK=K$, then we must have $h\in K$. So

Thus, since $\phi(H)=HK/K$ and $\ker{\phi}=H\cap K$, by the First Isomorphism Theorem we see that $H\cap K$ is normal in $H$ and that there is a canonical isomorphism between $H/(H\cap K)$ and $HK/K$.