dual isomorphism

Question

Let $C$ be the category of finitely generated abelian groups. If $M$ is a finitely generated abelian group, then define its dual as $M^* = \operatorname{Hom}(M,\mathbb{Q}/\mathbb{Z})$.

Now I want to show that $M$ and $M^{**}$ are canonically isomorphic.

I know that we need to use the structure theorem of abelian groups, but don't get how to use it to show isomorphisam.

Answer 1

The statement should be about finite abelian groups as Jeremy Rickard remarks.

The structure theorem tells you that $M$ is a direct sum of finite cyclic groups. Thus you are reduced to prove the statement for a finite cyclic group.

Can you prove it for $M=\mathbb{Z}/n\mathbb{Z}$?

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In the previous part it is taken for granted the existence of a canonical homomorphism $M\to M^{**}$, which exists in a much more general context. If $A$ is any commutative ring and $K$ is a fixed $A$-module, we can define $M^*=\operatorname{Hom}_A(M,K)$ and a canonical homomorphism $M\to M^{**}$ is easy to define like for vector spaces. The task here is to prove that

1. the homomorphism is injective when $A=\mathbb{Z}$ and $K=\mathbb{Q}/\mathbb{Z}$;

2. the homomorphism is surjective when $M$ is a finite abelian group.

Note that 1 holds for any abelian group (because $\mathbb{Q}/\mathbb{Z}$ is a cogenerator. Instead 2 doesn't hold generally, but it does for finitely generated modules using, as hinted, the structure theorem.

Answer 2

Saying that $M$ and $M^{**}$ are \emph{canonically} isomorphic means that the ismorphism $\phi :M\rightarrow M^{**}$ is constructed in a natural way, without any choice of basis or an identification of $M$ with a product of cyclic groups. Then the structure theorem can be used to prove that $\phi$ is an isomorphism, but the construction should not rely on the presenting $M$ as a product of cyclic groups.

The isomorphism $\phi :M\rightarrow M^{**}$ is as follows. $\phi(g) : M^* \rightarrow \mathbb{Q}/\mathbb{Z}$ sends $\eta \in M^*$ to \begin{equation} \phi(g)\eta=\eta(g) \end{equation} To prove that $\phi$ is an isomorphism you first use the structure theorem, thus you present $M$ as a product of cyclic groups, and you know that $\mathbb{Z}_N^*\cong \mathbb{Z}_N$ (you just send $x\in \mathbb{Z}_N$ to $\chi_x\in \mathbb{Z}_N^*$ given by $\chi_x(y)=\frac{xy}{N}$). Thus it is enough to prove that $\phi$ is injective. Suppose $\phi(g)=1$. This means that for any $\eta \in M^*$ $$ \eta(g)=1 $$ which implies $g=1$. Indeed $g\neq 1$, then let $n>1$ be the order of $g$, and we have $$ \left\{g^k \ | \ k=0,...,n-1\right\} \cong \mathbb{Z}_n $$ and with the same map used above to show that $\mathbb{Z}_N^*\cong \mathbb{Z}_N$ we see that there are elements $\eta \in M^*$ acting non-trivially on $g$, contradicting $\phi(g)=1$.