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!
Best Answer
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) $.