# Internal and external direct products of finite groups are isomorphic.

Let $G$ be a group with identity element $e$. Let $H$ and $K$ be normal subgroup of $G$ such that $H∩K={e}$. Prove that
$(H×K)≅(H⊕K)$

$(H×K)$ is the internal direct product between $H$ and $K$
$(H⊕K)$ is the external direct product between $H$ and $K$

I know that since H and K be normal subgroup of G such that $H∩K={e}$, $hk=kh$ for all $h∈H$ and $k∈K$.
I also know that $(h,k)∈(H⊕K)$ . But I’m not sure I know how to link them together and show that $(H×K)≅(H⊕K)$

Hint. Show that every element of the internal direct product may be uniquely represented as a product $hk$ for some $h\in H$, $k\in K$. Then use this to show that $\varphi(hk)=(h,k)$ is an isomorphism.
• I have a theorem that say $H$ and $K$ be normal subgroup of $G$ such that $H∩K=e$, then $hk$ is unique for all $h∈H$ and $k∈K$ so I don't have to prove that. <br> Are you suggesting me to make a fucntion $φ:(H×K)→(H⊕K)$ defined by $φ(hk)=(h,k)$ then show this function is bijective and preserving structure? Apr 18, 2013 at 1:26
• ok, I proved that theorem, again. Now I'm going to prove it's injective. Just want to double check, if $(h,k)=(x,y)$ then $h=x$ and $k=y$. Is this true? Apr 18, 2013 at 3:22