I am working on the following problem:

Let $G$ an Abelian group and $f: G \to \Bbb Z$ a surjective homomorphism. Prove that $G \cong \ker(f) \times \Bbb Z$

By means of the First Isomorphism Theorem, we can obtain that $G / \ker(f) \cong \Bbb Z$. Is there some natural way to proceed from this point, invoking probably some other theorem, or would one need to define a map and prove that is it an isomorphism?

More generally, is it true that given an Abelian group $A$ and a subgroup $B$, $$A/B \cong C \Rightarrow A \cong C \times B$$

Thank you very much in advance!


2 Answers 2


You need to define the splitting map $g\colon \mathbb Z \to G$ (so that $fg$ is the identity). It should be clear how to do this, pick $x \in G$ such that $f(x) = 1$ and then define $g(n) = nx$.

Then you show that $g$ is injective and $G = \ker f \times \mathrm{im} \ g$. It's a faily elementary argument, just show that the two subgroups have trivial intersection and generate $G$.

As for your more general question the answer is no. For example $\mathbb Z/4$ has subgroup $\{0, 2\}$ isomorphic to $\mathbb Z/2$. The quotient is also isomorphic to $\mathbb Z/2$ but $\mathbb Z/4$ is not isomorphic to $\mathbb Z/2 \times \mathbb Z/2$.

On the other hand if $B \simeq \mathbb Z^n$ for some $n$ then it is true that $A \simeq A/B \times B$. To use more advanced language, the reason is that $\mathbb Z^n$ is a free $\mathbb Z$-module, hence projective, hence surjections to it are split. Basically, with $\mathbb Z^n$ you will always be able to define the splitting map $g$ and carry out the remainder of the argument above.

  • $\begingroup$ I am so very sorry for not understanding this immediately. Here is my point of confusion. I cannot see how $G = \ker(f) \times im(g) \implies G \cong \ker(f) \times \Bbb Z$, since $im(g) \subset G$ so you would get $G \cong \ker(f) \times im(g) \subset \ker(f) \times G$ $\endgroup$ Commented Apr 16, 2013 at 6:32
  • $\begingroup$ $g\colon \mathbb Z \to G$ is injective so $\mathrm{im} \ g$ is isomorphic to $\mathbb Z$. $\endgroup$
    – Jim
    Commented Apr 16, 2013 at 6:49
  • $\begingroup$ I have tried to show that $\ker(g)$ is trivial, but I am failing to do so. My problem is that the $x$ you chose above might have finite order so you would have $nx=0$ with $n \neq 0$ thus the kernel will not be trivial. I cannot find a method to by-pass this problem. $\endgroup$ Commented Apr 19, 2013 at 4:54
  • $\begingroup$ If $f(x) = 1$ then what is $f(nx)$? Can $nx$ be zero? $\endgroup$
    – Jim
    Commented Apr 19, 2013 at 6:03
  • $\begingroup$ I just realised my stupid mistake! Just one more thing. How can I show that any element of $G$ is expressible as a product of two elements, the first of which is from $\ker(f)$ and the second from $im(g)$? $\endgroup$ Commented Apr 19, 2013 at 6:23

Jim's answer gives you everything you need, but let me try to draw the bigger picture in other words.

What you are looking at is a short exact sequence, i.e. an exact sequence of the form

$$0\to A\xrightarrow{f} B\xrightarrow{g} C\to 0$$

This means you have three abelian groups and all arrows are group homomorphisms. Exactness means that at any given point in the sequence the kernel of the exiting map coincides with the image of the entering map. In particular this means that the kernel of $f$ is zero, i.e $f$ is injective. Similarly the image of $g$ is the whole of $C$, so $g$ is surjective. Moreover the kernel of $g$ is the image of $f$. But since $f$ is injective you can identify $A$ with its image, so $Ker(g)\cong A$.

In your intial example you have $B=G$, $C=\mathbb Z$ and $A=ker(G\to \mathbb Z)$.

Now a short exact sequence is called split if one of the following equivalent properties hold:

a) $$B\cong A\oplus C$$ b) There exists a group homomorphism $i:C\to B$ such that $g\circ i=id$

c) There exists a group homomorphism $p:B\to A$ such that $p\circ f=id$.

So the direct way of showing that $G\cong \mathbb Z\oplus ker(G\to \mathbb Z)$ would be constructing a map $\mathbb Z\to \mathbb G$ which satisfies property b).

A more general argument uses the definition of projectiveness:

An $R$-module $P$ (all abelian groups are $\mathbb Z$-modules) is called projective if for any surjective $R$-module homomorphism (homomorphism of abelian groups) $f:N\to M$ and any $R$-module homomorphism $g:P\to M$ there exists a lift $h:P\to N$ such that $f\circ h=g$.

Now you can show that $P$ is projective if and only if any short exact sequence of the form $$0\to A\xrightarrow{f} B\xrightarrow{g} P\to 0$$ splits.

You can also show that all free modules (the abelian group $\mathbb Z$ is clearly free as a module over itself) are projective.


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