Representation of an abelian groups

You want to prove that the map $$ x \mapsto (\chi \mapsto \chi(x)) $$ is injective. I assume that you have already proved that this is a homomorphism. So all you need to show is that the kernel of this map is trivial. That is, you want to show that $$ \ker (x \mapsto (\chi \mapsto \chi(x))) = \{e\} $$ where $e$ is the identity in $G$. Now the kernel is exactly all $x$ such that $$ \chi \mapsto \chi(x) $$ is the trivial map from $\hat{G}$ to $\mathbb{C}^\times$. If $x$ is in the kernel, then $\chi(x) = 1$ for all $\chi \in \hat{G}$. That is, $x$ is in the intersection of the kernels of all the characters $\chi: G\to \mathbb{C}^\times$. Hence $x$ is the identity.