The question related to title has been appeared on mathstack, and different answers are also given.

What I want to consider here is the following.

Fact: A finite abelian group $G$ is isomorphic to the group $\hat{G}$ of its complex characters.

Q. Using the above fact, can we show easily that if $H\leq G$, $G$ finite abelian, then there exists a subgroup $K$ of $G$ with $G/K\cong H$?

  • In similar question here, the answers are using either basic structure theorem of abelian group, or other answer is avoiding characters, mentioned in note. I want to deduce from characters, but was not able to do it. Any hint could be sufficient.

  • In link here, consider exact sequence $1\rightarrow H\rightarrow G \rightarrow G/H\rightarrow 1$. Apply ${\rm Hom}(-, C^*)$, $$1\rightarrow {\rm Hom}(G/H,C^*)\rightarrow {\rm Hom}(G,C^*)\rightarrow {\rm Hom}(H,C^*).$$ Using Fact one can see that we have sequence $1\rightarrow \widehat{(G/H)}\rightarrow \widehat{G}\rightarrow \widehat{H}$ but the last map was not stated to be surjective; I couldn't deduce here that $\widehat{H}\cong H$ is quotient of $\widehat{G}\cong G$.


The map is actually surjective, as can be seen for instance considering cardinality.


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.