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I'm having trouble with the next statement:

Let $G$ be a group such that there exists a surjective group homomorphism $G \rightarrow \mathbb Z$, where $\mathbb Z$ denotes the group of the integers. Then for any subgroup of finite index $H$ of $G$, there also exists a surjective group homomorphism $H\rightarrow \mathbb Z$.

I really appreciate if someone gives me a hint or a proof of this fact.

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up vote 1 down vote accepted

Let $\phi \colon G \to \mathbb Z$ the an epimorphism. We consider the image $\phi[H]$ of $H$. $\phi[H]$ is a subgroup of $\mathbb Z$ and we have $\phi[H] \ne 0$, we would have an epimorphism $G/H \to \mathbb Z$, contradicting $\mathbb Z$ being infinite. So $\phi[H] = n\mathbb Z$ for some $n \ge 1$, but the latter group is isomorphic to $\mathbb Z$.

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If $\phi(H)=0$, then you can define a surjective function from $G/H$ to $\mathbb Z$ which is imposible. –  Antonio Dec 19 '12 at 7:19
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I think the word "otherwise" is missing just before "we would..." –  DonAntonio Dec 19 '12 at 8:10
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