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

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 (=

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 ^_^

• Very cool. I knew (but hadn't recalled) $h$ is determined by $h(1)$ -- that made it click for me. (= – michiganbiker898 Jan 20 '20 at 23:31
• Glad to hear it! Good luck with everything ^_^ – HallaSurvivor Jan 20 '20 at 23:32

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.)

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