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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.

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up vote 2 down vote accepted

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.

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I am unsure of the cosets of N. Are they just N and $\{1_G\} X H$? –  Jim_CS Oct 26 '12 at 9:02
    
@Jim_CS: Take any $\langle g,h\rangle\in D$; the coset $N\langle g,h\rangle=\{\langle g',1_H\rangle\langle g,h\rangle:g'\in G\}=\{\langle g'g,h\rangle:g'\in G\}=G\times\{h\}$, so you get one coset for each $h\in H$. –  Brian M. Scott Oct 26 '12 at 11:59
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