$H,K$ be normal subgroups of $G$ such that $G$ is isomorphic with $H \times K$ , then is $G=HK$?

Question

Let $H,K$ be normal subgroups of $G$ such that $G$ is isomorphic with $H \times K$ , then is it necessary that $G=HK$ ? What if we also assume that every element of $H$ commutes with every element of $K$ ? Please help . Thanks in advance

Answer 1

If we assume that $G$ is finite and that every element of $H$ commutes with every element of $K$ and that $H\cap K=\{e\}$ then the answer is yes. And in fact we don't need to assume that $G$ is isomorphic to $H\times K$, just that $|G|=|H|\,|K|$:

Define $\phi:H\times K\to G$ by $\phi((h,k))=hk$. The fact that every element of $H$ commutes with every element of $K$ shows that $\phi$ is a homomorphism. If $\phi(h,k)=e$ then $k$ is the inverse of $h$; hence $k\in H$ so $k=e$ and hene $h=e$. So $\phi$ is injective, and hence $\phi$ is a bijection by our assumption on the cardinalities.

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If, as in the original version (and the current version as I write this) we don't assume that $G$ is finite then the answer is no. And it remains no if $H\cap K=\{e\}$ and everything commutes:

Let $G=\Bbb Z^2$. Let $H=\{(2n,0):n\in\Bbb Z\}$ and $K=\{(0,2n):n\in\Bbb Z\}$.