Neccesary and sufficient condition for a direct product of proper subgroups to be isomorphic to the whole group

Question

It is a known result that if $G$ is a group and $H,K \triangleleft G$ are normal subgroups such that $HK=G$ and $H \cap K= \left\{{1}\right\}$ then $G \cong H \times K$. Is it true that if $H,K \subset G$ are subgroups such that $G \cong H \times K$ then $H \cap K= \left\{{1}\right\}$? Apparently the whole converse of the theorem is true but I don't see why, specially the intersection condition.

Any help?

In advance thank you

Answer 1

This is not true in the way you wrote it. For example, consider $G = (\mathbb{Z}/2) \times (\mathbb{Z}/2)$. Let $H = K = (\mathbb{Z}/2) \times \{0\}$. Then $G$ is abstractly isomorphic to $H \times K$, but of course $H \cap K$ is not trivial.

Here's the correct version of the claim:

Proposition Let $H$ and $K$ be normal subgroups of a group $G$. The following are equivalent:

1. $HK = G$ and $H \cap K = \{1\}$
2. $G$ is the internal direct product of $H$ and $K$, meaning that the function $(h,k) \mapsto hk : H \times K \to G$ is a group isomorphism.