# Subgroup of a finite abelian group is isomorphic to a quotient

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 $\widehat{G}\cong G$.

## 1 Answer

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