Does $G\cong H\times K$ imply $H\unlhd G?$

Question

I want to prove the following exercise.

If a group $G$ is the direct product of subgroups $H,K$, then $K$ is isomorphic to $G/H$.

To prove this, I think I need first to show $H$ is normal in $G$.

I can show that there is a normal subgroup $J$ in $G$ that is isomorphic to $H.$ But I don’t know how to show $H$ is a normal subgroup in $G.$

I’m not sure but my guessing is that $gHg^{-1} = J$ for some $g \in G$, so that $H = J.$ But . . . maybe there’s a counterexample.

Also, if I show $H$ is normal somehow, I still don’t know how the conclusion of the exercise follows from it. When $G = H\times K$ means the internal direct product of its normal subgroups, I can solve the exercise. If not, I know that, by using $\pi_k : H\times K \to K$ (the canonical projection), we can show $H\times K/\ker(\pi_k)$ is isomorphic to $K.$ But how can i show that $G/H$ is isomorphic to $H\times K/\ker(\pi_k)$?

Can somebody help?

Thank you!

Answer 1

By definition, if $G$ is the internal direct product of $H$ and $K$, then $H$ and $K$ are both normal, $H\cap K=\{e\}$, and $HK=G$. There is no need to prove this because this is what you are given. Then you can use the relevant isomorphism theorem to show that $HK/H\simeq K/(K\cap H) $.

Answer 2

The group $G\cong H\times K$ with $H\cong \langle S_H\mid R_H\rangle$ and $K\cong \langle S_K\mid R_K\rangle$ has as a presentation $$G\cong \langle S_H\cup S_K\mid R_H\cup R_K\cup X\rangle,$$ where $X=\{hk=kh\mid h\in S_H\land k\in S_K\}$, from which it is easy to see that $H\cong L$ such that $L\unlhd G$. (Why?)