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 $h_1k_1, h_2k_2 \in HK$ . Further, $$ h_1k_1h_2k_2 = h_1(h_2h_2^{-1})k_1h_2k_2 = h_1h_2(h_2^{-1}k_1h_2)k_2 $$ Since $K$ is a normal subgroup of $G$ and $h_2 \in G$ , then $h_2^{-1}k_1h_2 \in K$ . Therefore $h_1h_2(h_2^{-1}k_1h_2)k_2 \in HK$ , so $HK$ is closed under multiplication.

Also, $(hk)^{-1} \in HK$ for $h \in H$ , $k \in K$ , since $$ (hk)^{-1} = k^{-1}h^{-1}=h^{-1}hk^{-1}h^{-1} $$ 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 $h_0k_0K$ be some element of $HK/K$ ; since $k_0 \in K$ , then $h_0k_0K=h_0K$ , and $\phi(h_0)=h_0K$ . Now $$ \ker(\phi) = \{ h \in H \mid \phi(h)=K \} = \{ h \in H \mid hK=K \} $$ and if $hK=K$ , then we must have $h \in K$ . So $$ \ker(\phi) = \{ h \in H \mid h \in K \} = H \cap K $$

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 $HK/K$ .