5
$\begingroup$

The question: If G is an abelian group and f is a surjective homomorphism from G to Z with kernel K, prove that G has a subgroup H such that H is isomorphic to Z.

By the first isomorphism theorem I know that G/K is isomorphic to Z (as Z is the image of f). So I figure I just have to find an H that is isomorphic to G/K, and I'm good (thanks to the transitivity of isomorphisms).

$\endgroup$
  • $\begingroup$ More interestingly, $G$ must have $\mathbb Z$ as a direct factor, at least when $G$ is finitely generated. $\endgroup$ – lhf Apr 29 '15 at 0:14
5
$\begingroup$

Hint: You only need to prove that $G$ contains an element of infinite order.

Solution:

If $\phi:G \to \mathbb Z$ is a surjective homomorphism, take $g \in G$ such that $\phi(g)=1$.

$\endgroup$
  • $\begingroup$ $K$ is a red herring. $\endgroup$ – lhf Apr 29 '15 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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