Proving the direct product D of two groups G & H has a normal subgroup N such that N isomorphic to G and D/N isomorphic to H Let $D = G \times H$ be the direct product of groups $G$ and $H$. Prove that $D$ has a normal subgroup $N$, such that $N$ is isomorphic to $G$ and $D/N$ is isomorphic to $H$.
Here's where I stand...I know what a direct product is and I know what a normal subgroup is but I have no idea how to prove that the direct product of two such arbitrary groups has a normal subgroup.
And then with the isomorphisms. I know how to show basic isomorphisms, show that a mapping is homomorphic and then show that it is onto and one to one, but I don't know how you would show isomorphisms from completely arbitrary subgroups to groups.

HINT: Let $N=G\times\{1_H\}$, where $1_H$ is the identity in $H$. For any $\langle g,h\rangle\in G\times H$,
$$\langle g,h\rangle N\langle g,h\rangle^{-1}=\langle g,h\rangle N\langle g^{-1},h^{-1}\rangle=\dots\;?\tag{1}$$
If you make use of the fact that $N=G\times\{1_H\}$, it shouldn’t be too hard to explain why $(1)$ eventually simplifies to $N$, showing that $N$ is normal in $G\times H$.
Then you just have to figure out what the cosets of $N$ are in $D=G\times H$; once you’ve done that, an isomorphism between $D/N$ and $H$ should be pretty apparent.
