Question about normal subgroups in finite groups

Question

I want to show that if $G$ is a finite group and $H$ is normal in G, and $K$ is a subgroup of $G$, and $H\cap K = \{e\}$ then $|HK|=|H||K|$

Answer 1

Let $h,h'\in H$ and $k,k'\in K$ s.t. $hk=h'k'$. Then $h'^{-1}h=k'k^{-1}$. Then $k'k^{-1}$ belongs to $K$ because it's a product of elements of $K$, and to $H$ because it's equal to $h'^{-1}h$. Therefore $k'k^{-1}\in H\cap K=\{e\}$, which implies $k=k'$ and $h=h'$. This prove that products of the form $hk$ are all distinct as $h$ runs over $H$ and $k$ runs over $K$. Thus $HK$ has $|H||K|$ elements.

Answer 2

Hint: Let $H$ act on $K$ by $h\cdot k=hk$. The orbits of this action are disjoint, and their union is $HK$.

Answer 3

Assume $hk=h'k'$ for some $h,h'\in H$, $k,k'\in K$. Then $k k'^{-1}=h^{-1}h'$. We have $$ k k'^{-1}\in K $$ $$ h^{-1}h'\in H $$ Thus both expressions are in $K\cap H=\{e\}$. Hence, $k=k'$, $h=h'$. This establishes bijection between $HK$ and $H\times K$.