An abelian group $G$ and onto group homomorphism $h: G \longrightarrow \mathbb{Z}$

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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 (=

Best Answer

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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