proof of second isomorphism theorem for groups

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,

$$h_1{k}_1{h}_2{k}_2=h_1(h_2{h}_2^{-1})k_1{h}_2{k}_2=h_1{h}_2(h_2^{-1}{k}_1{h}_2)k_2$$

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

$${(hk)}^{-1}=k^{-1}{h}^{-1}=h^{-1}h{k}^{-1}{h}^{-1}$$

and $h{k}^{-1}{h}^{-1}\in K$ 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 fromthe 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 $\varphi :H\to HK/K$ by $\varphi (h)=hK$.We claim that $\varphi$ 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 $\varphi (h_0)=h_{0}K$.Now

$$\ker (\varphi )=\{h\in H\mid \varphi (h)=K\}=\{h\in H\mid hK=K\}$$

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

$$\ker (\varphi )=\{h\in H\mid h\in K\}=H\cap K$$

Thus, since $\varphi (H)=HK/K$ and $\ker \varphi =H\cap K$,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 $HK/K$.