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

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!

• This is actually a lot easier than what you are doing. Just let $G=H\times K$. Then, clearly $H,K$ are both normal in $G$, and are the quotients of the group modulo the other subgroup. The homomorphisms that induce these quotients are simply the forgetful morphisms. – Don Thousand Oct 25 at 14:26
• I think we need to interpret $G = H$ x $K$ as $G$ is isomorphic to $H$ x $K$.. – anadad Oct 25 at 14:29
• I don’t understand why H is normal in G clearly. – anadad Oct 25 at 14:30
• Because of the forgetful morphism from $H\times K\to K$. – Don Thousand Oct 25 at 14:32
• – Shaun Oct 25 at 14:36

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 $$HK/H\simeq K/(K\cap H)$$.
• @anadad If it is the direct product of subgroups, then it is the internal direct product. Even if it's external you can identify the subgroups with $H$ and $K$. – Matt Samuel Oct 25 at 14:39
• Thanks. So you mean that whenever I see $G = H \times K$ and $H,K$ are subgroups of G, I can assume $HK=G$, $H \cap K= e$ and they are normal in $G$ ? – anadad Oct 25 at 14:51
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?)