I recently posted a question regarding this same problem, and am looking for further assistance with my proof. The problem is as follows:
Let $G= H \times K$ be the direct product of two groups $H$ and $K$. Let $\tilde{H}=\{(h, e_{k});h \in H\}$ and $\tilde{K}=\{(e_{H},k); k \in K\}$. Show that $\tilde{H}$ and $\tilde{K}$ are normal subgroups of $G$
Here was my previous question.
My start to the proof:
Let $G= H \times K$ be the direct product of two groups $H$ and $K$. Let $\tilde{H}=\{(h, e_{k});h \in H\}$ and $\tilde{K}=\{(e_{H},k); k \in K\}$. $\tilde{H}$ and $\tilde{K}$ are subgroups of $G$ because these sets are subgroups of $H$ and $K$, whose direct product forms the group $G$. We want to see that they are normal in $G$, which is to say we want to see that they are closed under conjugation in $G$. Let $h_{1},h_{2} \in H$ so $(h_{1},e_{K}),(h_{2},e_{K}) \in \tilde{H}$. and $g_{1},g_{2} \in G$.
First, is my statement regarding $\tilde{H}$ and $\tilde{K}$ being subgroups correct, or should I write out the proof that these are subgroups? I thought there was sufficient reasoning to assume without the full proof, but I wanted to be sure. Additionally, to prove they are normal, can I assume that $H$ and $K$ are closed under conjugation, and apply this to the elements of $\tilde{H}$ and $\tilde{K}$ by showing that they are closed as well?
$\tilde H,\tilde K$ are easily seen to be groups, isomorphic to $H,K$ respectively.
Here's an idea: define $\varphi: H×K\to K$ by $\varphi (h,k)=k.$
It's easy to see that $\varphi $ is a homomorphism with kernel $\tilde H$
But the kernel of a homomorphism is always a normal subgroup.