Question: Show that if $G$ is a group with two normal subgroups $H$ and $K$ such that $G=HK$ and $H\cap K=\{e\}$, then the map $(h,k)\mapsto hk$ is an isomorphism of groups from $H\times K$ to $G$
Comments: I would like to know where I can find a proof to this statement as it seems like a standard result.
To me the statement seems intuitively obvious since $H$ will look like $\{e,h_1,...,h_n\}$ and $K=\{e,k_1,...,k_m\}$ and $G=\{e,h_1,...,h_n,k_1,h_1k_1,...,h_nk_1,...,k_m,h_1k_m,...,h_nk_m\}$
Therefore $K$ will have $m+1$ elements, $H$ will have $n+1$ elements and $G$ will have $(m+1)(n+1)$ elements and there would be a 1-1 and onto mapping -but how do I prove this?
The homomorphism property comes from taking $a=(h_1,k_1)$ and $b=(h_2,k_2)$
Then $\varphi(ab)=\varphi(h_1h_2,k_1k_2)=h_1h_2k_1k_2=h_1k_1h_2k_2=\varphi(h_1,k_1)\varphi(h_2,k_2)=\varphi(a)\varphi(b)$
This is my understanding of this question, how should I prove the bijective part of this map, also if possible what is the name of this proof or where can I find one.