Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following short exact sequence: $$\ker(f)\rightarrow G\rightarrow \mathbb{Z}$$ 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!

share|improve this question
@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). –  Asaf Karagila Apr 5 '13 at 0:02
@Daniel: Bring it up to meta first. I have no opinion whether or not this tag is useful, but others might have. –  Asaf Karagila Apr 5 '13 at 0:07
add comment

1 Answer

up vote 3 down vote accepted

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).

share|improve this answer
Miha, the projective $\mathbb{Z}$-modules are exactly the free abelian groups (have a look at Lang's Algebra, Chap. III, Theorem 7.1). –  tj_ Oct 5 '12 at 14:35
@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. –  Miha Habič Oct 5 '12 at 16:08
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.