Isomorphism between G and Z x Z_2 if G has a normal subgroup isomorphic to Z_2

1. Jan 14, 2010

3029298

1. The problem statement, all variables and given/known data
If G contains a normal subgroup H which is isomorphic to $$\mathbb{Z}_2$$, and if the corresponding quotient group is infinite cyclic, prove that G is isomorphic to $$\mathbb{Z}\times\mathbb{Z}_2$$

3. The attempt at a solution
$$G/H$$ is infinite cyclic, this means that any $$g\{h1,h2\}$$ is generated by some $$\gamma\{h1,h2\}$$ with $$\gamma\in G$$. $$\gamma=g^n$$ because H is normal. But now?

2. Jan 14, 2010

ystael

Do you know any general property of a group $$G$$ such that, when $$G$$ has this property and $$N$$ is a normal subgroup of $$G$$, you can conclude that $$G \cong N \times G/N$$?

(Hint: in the direct product $$H \times K$$ of two groups, what is the relationship between the subgroups $$H \times 1$$ and $$1 \times K$$?)

3. Jan 14, 2010

3029298

I do not know any general property of this kind... the subgroups H x 1 and 1 x K only have the identity in common and (H x 1)(1 x K)=H x K, but I do not see how this helps...

4. Jan 14, 2010

ystael

The "general property of $$G$$" I was referring to is "$$G$$ is abelian". One way to understand the thing that makes direct products special is that the factors commute with each other: in the product above, $$(h, 1)(1, k) = (1, k)(h, 1) = (h, k)$$.

In your original problem, what happens if $$G$$ is abelian? What happens if it's not?

5. Jan 16, 2010

3029298

I really do not understand... what is the use of the fact that the factors of the direct products commute with each other?