This is exercise 3.3 of Serre's book on Representation theory of finite groups: Let $G$ be a finite abelian group and $G'$ the set of irreducible characters of $G$. It is easy to prove that $G'$ is also a finite abelian group of the same order as $G$ via the usual multiplication of characters. Also, for $x \in G$, it can be establish that the mapping $\Psi_x :\chi \to \chi(x)$ is an irreducible character of $G'$, so an element of $G''$, the dual of $G'$. This naturally defines a map $f$ from $G$ to $G''$, sending $x$ to $\Psi_x$ which I want to prove is an isomorphism. My question is:
Why is $f$ injective?
It is also easy to see that $f$ is a group homomorphism, so it is enough to show that its kernel is trivial. If $x$ is such that $\Psi_x = 1_{G''}$, then we know that $\chi(x)=1$ for every irreducible character $\chi$ of $G$, right? Why does this imply that $x=1$? This should be easy to prove, but I guess I am getting confused with the double dual and not seeing how to do it.
If $\chi(x)=1$ for every irreducible character of $G$, then $\rho(x)=1$ for every representation $\rho$ of $G$, since every representation is a direct sum of irreducible representations. Taking $\rho$ to be the regular representation, this implies $x=1$.
Alternatively, you can use the classification of finite abelian groups to explicitly construct $\chi$ such that $\chi(x)\neq 1$ if $x\neq 1$. Identify $G$ with a direct sum of cyclic groups, and choose a coordinate on which $x$ is nontrivial. The projection onto this coordinate is a homomorphism $\varphi:G\to\mathbb{Z}/n$ such that $\varphi(x)\neq 0$, and then composing with a faithful 1-dimensional representation of $\mathbb{Z}/n$ gives a 1-dimensional representation $\chi$ of $G$ such that $\chi(x)\neq 1$.