# Subgroup of an abelian group isomorphic to a given quotient group [duplicate]

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STATEMENT: Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroups that is isomorphic to $G/H$.

QUESTION: Could someone offer a proof using dual groups. I have found one use the fundamental theorem of finitely generated abelian groups, but I cannot seem to find on using dual groups.

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## 1 Answer

Let $S^1$ be the circle group. Consider the exact sequence:

$$0\to H\to G \to G/H \to 0$$

Applying the functor $\operatorname{Hom}(-,S^1)$, we get another exact sequence:

$$0\to \operatorname{Hom}(G/H,S^1) \to \operatorname{Hom}(G,S^1) \to \operatorname{Hom}(H,S^1)$$

Now, it so happens that $\operatorname{Hom}(G,S^1) \cong G$ for a finite abelian group $G$—though I don't know if there's a good way to prove that without using the fundamental theorem. But if you accept it, then this sequence gives us a (non-canonical) injection $G/H\to G$.

• I'm not too familiar with the circle group or functors. Do you mind showing the isomorphism Hom$(G,S1)≅G$ – Enigma Oct 3 '14 at 14:53
• @S.Sheng The point is that $S^1$ has a unique cyclic subgroup of every finite order. So if $G$ is a finite cyclic group, say $G\cong \mathbb{Z}/n$, then $nG = 0$ so $\operatorname{Hom}(G,S^1) \cong \operatorname{Hom}(\mathbb{Z}/n,\mathbb{Z}/n) \cong \mathbb{Z}/n$. We then extend this isomorphism to all direct sums of finite cyclic groups. But the fact that every finite abelian group is a direct sum of cyclic groups is basically the fundamental theorem. There cannot be a canonical isomorphism (as there is no canonical section of $G\to G/H$). – Slade Oct 3 '14 at 14:58
• @S.Sheng Oh, and the only thing that's really important about all the functor business it that if $A\to B$ is a surjection, then $\operatorname{Hom}(B,C)\to \operatorname{Hom}(A,C)$ is an injection (for any $C$). This can be proved directly. – Slade Oct 3 '14 at 15:10
• Got it. Thanks! – Enigma Oct 3 '14 at 16:50