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.
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 $M=\mathbb{Z}/n\mathbb{Z}$?
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
the homomorphism is injective when $A=\mathbb{Z}$ and $K=\mathbb{Q}/\mathbb{Z}$;
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.
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 $\phi(g)=1$.
