I have the following short exact sequence: $$1\to \ker(f)\rightarrow G\rightarrow \mathbb{Z}\to 1$$ Where the first map is inclusion and the second map is $f$. And $G$ is an abelian group.

I'd like to known if I may write: $G=\ker(f)\oplus \mathbb{Z}$

I guess I need to prove that the sequence is split exact...Is that the case? Can anyone help? Thanks!

  • $\begingroup$ @Daniel: It is customary to make a thread on meta before starting a new tag; and it is also impolite to bump six old posts at once (four is my personal limit, and I usually stop at three). $\endgroup$
    – Asaf Karagila
    Commented Apr 5, 2013 at 0:02
  • $\begingroup$ @Daniel: Bring it up to meta first. I have no opinion whether or not this tag is useful, but others might have. $\endgroup$
    – Asaf Karagila
    Commented Apr 5, 2013 at 0:07

1 Answer 1


The splitting lemma tells you that $G$ is a direct sum of the other two groups iff the sequence splits (either on the left or on the right, since we are in the abelian case). In general there would be no reason for this to happen, however your case is special.

Pick an element $g\in f^{-1}(1)$ and define $\psi(1)=g$, extending to a homomorphism $\psi\colon\mathbb{Z}\to G$. It is easy to see that $f\circ \psi$ is the identity, so your sequence splits on the right, hence $G=\mathrm{ker}(f)\oplus \mathbb{Z}$.

The proof generalizes easily to the case where $\mathbb{Z}$ is replaced by any free abelian group (if we were dealing with modules over some ring, a projective module would've sufficed, but, as was pointed out in the comments, there is no distinction between free and projective $\mathbb{Z}$-modules).

  • $\begingroup$ Miha, the projective $\mathbb{Z}$-modules are exactly the free abelian groups (have a look at Lang's Algebra, Chap. III, Theorem 7.1). $\endgroup$
    – tj_
    Commented Oct 5, 2012 at 14:35
  • $\begingroup$ @tj_ Of course, submodules of free modules over a PID are free. I should've thought of that. Somehow I got distracted by thinking that $\mathbb{Z}$ should be a semisimple ring. $\endgroup$ Commented Oct 5, 2012 at 16:08

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