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

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Homework Statement


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

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?
 
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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?)
 
ystael said:
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?)

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...
 
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?
 
I really do not understand... what is the use of the fact that the factors of the direct products commute with each other?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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