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Let $G$ be an abelian group and $h: G \longrightarrow \mathbb{Z}$ a group homomorphism which is onto.

(a) Prove that there exists a group homomorphism $f: \mathbb{Z} \longrightarrow G$ such that $hf$ is the identity map on $\mathbb{Z}$.

(b) Prove that $G$ is isomorphic to $\mathbb{Z} \times$ (ker $h$).

I'm not sure how to begin for part (a) -- from what I understand, group homomorphisms are not equivalence relations (and, in particular, not necessarily symmetric), so how can I know that the group homomorphism $f$ even exists ? Does it have to do with the fact that both $G$ and $\mathbb{Z}$ are given to be abelian groups?

For part (b), we can use the First Isomorphism Theorem. In particular, $G/ker(h) \cong im(h)$ $\Rightarrow$ $G/ker(h) \cong \mathbb{Z}$, since $h$ is an onto group homomorphism, and thus, $im(h)$ coincides with $\mathbb{Z}$.

Now, after reading When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold? , I'm convinced that this gives us the desired isomorphism $G \cong \mathbb{Z} \times ker(h)$, because the composition $hf$ is the identity map based on what we show in part (a).

Is this correct ?

Thanks for all your help (=

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Here are some hints:

For (a), notice a group homomorphism $h : \mathbb{Z} \to G$ is completely determined by $h(1)$, and moreover, we can send $1$ wherever we want.

Since $f : G \to \mathbb{Z}$ is surjective, we know $f(x) = 1$ for some $x$. Can you use this, and the discussion above, to find a group homomorphism with the desired properties?

For (b), you are on the right track. Once you've constructed $h$ as above, use the fact that an abelian group $G$ is isomorphic to $X \times Y$ if and only if both

  1. every element of $G$ can be written as $xy$ for $x \in X$ and $y \in Y$
  2. $X \cap Y = \{ 0 \}$

I hope this helps ^_^

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    $\begingroup$ Very cool. I knew (but hadn't recalled) $h$ is determined by $h(1)$ -- that made it click for me. (= $\endgroup$ – michiganbiker898 Jan 20 '20 at 23:31
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    $\begingroup$ Glad to hear it! Good luck with everything ^_^ $\endgroup$ – HallaSurvivor Jan 20 '20 at 23:32
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Hint: I'll use additve notation because the groups are abelian. (a) group homomorphisms are functions. You need a function $f$ from $\Bbb{Z}$ to $G$ such that $h(f(y)) = y$ for every $y \in \Bbb{Z}$. In particular you need $h(f(1)) = 1$, and because $h$ is onto you know you can pick some $x \in G$ such that $h(x) = 1$ and let $f(1) = x$. But if $h$ and $f$ are homomorphisms, what can you now say about $h(f(2)) = h(f(1+1)))$ and $h(f(-1))$ and $h(f(-3)) = h(f(-1 + -1 +-1))$ etc.

(b) Given $f$ as in part (a) and any $x \in G$, $h(x - f(h(x))) = 0$ (do you see why)? Use this to construct a function $i$ from $G$ to $\Bbb{Z} \times \mathrm{ker}(h)$ and show that $i$ is an isomorphism. (It's in this part that you need to know that $G$ is abelian.)

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    $\begingroup$ (b) seems very clear now. Thank you (= $\endgroup$ – michiganbiker898 Jan 20 '20 at 23:32
  • $\begingroup$ There is also a missing word between "you" and "to" in the last sentence that needs to be added, namely the word "need". $\endgroup$ – Geoffrey Trang Jan 21 '20 at 2:03
  • $\begingroup$ @GeoffreyTrang: thanks! I was one right bracket short too. Fixed. $\endgroup$ – Rob Arthan Jan 21 '20 at 20:13

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