If $G=HK$, $H,K$ are normal in $G$, and $H\cap K=\{e\}$ then $G$ is abelian

Question

If $G=HK$, $\,\,H,K$ are both normal in $G$, and $H\cap K=\{e\}$, then show that $G$ is abelian

I concluded that $H$ and $K$ commute with each other but I'm not sure how to proceed. Any hints?

Answer 1

This is false. For example, let $H$ and $K$ be non-abelian groups; then their product $H \times K$ satisfies all of the hypotheses but is not abelian.

Answer 2

It's trivial that $H$ and $K$ commute ($hkh^{-1}k^{-1}$ is both in $H$ and $K$ for all $h \in H, k \in K$) but you can't go anywhere from there.

Answer 3

The statement is false.

It comprises of the definition of $G$ being the internal direct product of $H$ and $K$, denoted $G=H\times K$.

If $H\cong\langle G_H\mid R_H\rangle$ and $K\cong\langle G_K\mid R_K\rangle$ are presentations of $H$ and $K$, respectively, then a presentation of $G$ is

$$H\times K\cong\langle G_H\cup G_K\mid R_H\cup R_K\cup\{ hk=kh \mid h\in G_H, k\in G_K\}\rangle,$$

from which one can see that $G$ is abelian if and only if both $H$ and $K$ are abelian.