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Let $G$ be an abelian group, and let $f:G\rightarrow \mathbb{Z}$ be an epimorphism (surjective homomorphism).
Prove that there exist a subgroup $H\subseteq G$ such that $H\cong \mathbb{Z}$, and $G\cong H\times (\ker f)$ .

My attempt so far:
Let $h\in G$ be the element for which $f(h)=1_{\mathbb{Z}}$, and consider the subgroup $H=\langle h\rangle$ of $G$.
Now, I want to argue that $H\cong \mathbb{Z}$. we know that two cyclic groups of the same order are isomorphic. can I use this fact somehow? Although $H$ and $\mathbb{Z}$ are both cyclic, obviously $\mathbb{Z}$ is of infinite order, so can I say the same about $H$ and conclude that these two are isomorphic because of that?
I'm not even sure $H$ must be infinite...

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    $\begingroup$ $f(nh)=n$, so $nh\not={mh}$ when $n\not={m}$. Then $H$ is infinite. $\endgroup$ – gaoxinge Jan 25 '14 at 13:17
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    $\begingroup$ Does the theorem "cyclic groups of same order are isomorphic" apply here? $\endgroup$ – so.very.tired Jan 25 '14 at 13:24
  • $\begingroup$ I think you are right.@so.very.tired $\endgroup$ – gaoxinge Jan 25 '14 at 13:32
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    $\begingroup$ OK, thank you very much for the help. :) $\endgroup$ – so.very.tired Jan 25 '14 at 13:35
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    $\begingroup$ This is just to point out that "Let $h \in G$ be the element for which $f(h) = 1_{\mathbb Z}$" is a mistake, because in general $h$ is not unique. You need to say "an element" not "the element". $\endgroup$ – Derek Holt Jan 25 '14 at 13:49
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As gaoxinge mentioned, the "right" way to look at this problem is a short exact sequence. Of course we can do it elementary as well:

If $H=\langle h\rangle\le G$ is finite, $n\, h=0_G$ for some $n\ge 1$. Thus $$ n = n\cdot 1_{\mathbb Z} = n\cdot f(h) = f(n\, h) = f(0_G) = 0_{\mathbb Z}, $$ which is a contradiction. So $H$ is in fact infinite and isomorphic to $\mathbb Z$.

Can you take it from here?

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  • $\begingroup$ Yes, thank you... :) $\endgroup$ – so.very.tired Jan 25 '14 at 13:37
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This problem is talking about the split property of a short exact sequence.

Hint: $0\rightarrow\mathrm{ker}(f)\rightarrow{G}\rightarrow\mathbb{Z}\rightarrow{0}$ is must spilit because $\mathbb{Z}$ is free abelian group.

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    $\begingroup$ This is not part of our course material (Exact sequences), so unfortunately the hint isn't helping too much, but thanks anyway! $\endgroup$ – so.very.tired Jan 25 '14 at 13:26
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    $\begingroup$ @so.very.tired: So you dont know the semi-direct product?? $\endgroup$ – mrs Jan 25 '14 at 13:28
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    $\begingroup$ @B.S. I know what direct-product is... is that the same? $\endgroup$ – so.very.tired Jan 25 '14 at 13:34
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    $\begingroup$ @so.very.tired: In some cases they are the same but in a general case the answer is no. $\endgroup$ – mrs Jan 25 '14 at 13:38

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