Problem Statement: Prove that the one-dimensional characters of a group $G$ form a group under multiplication of functions, i.e. where the group operation is: $$(\chi\cdot\chi')(g)=\chi(g)\chi'(g)$$ This group is called the character group of $G$, and is often denoted by $\hat{G}$. Prove that if $G$ is abelian, then $|\hat{G}|=|G|$.
So, I completely understand how to prove that $\hat{G}$ is a group, but I am unsure about the second part: proving that if $G$ is abelian, then $|\hat{G}|=|G|$.
Clearly, $\hat{G}$ is abelian regardless of $G$ being abelian, since $\hat{G}\subset\mathbb{C}^\times$ and $\mathbb{C}^\times$ is commuative. So $G$ being abelian eliminates the possibility that $(\chi\cdot\chi')(gh)=(\chi\cdot\chi')(hg)$ with $gh\neq hg$.
So basically, to prove that $|\hat{G}|=|G|$, we must show that for any distinct $g,h\in G$, then $(\chi\cdot\chi')(g)\neq(\chi\cdot\chi')(h)$, but I am unsure how I should approach this type of proof.
I was thinking somehow prove that there is an isomorphism between $G$ and $\hat{G}$, but my professor said that constructing a homomorphism between the two groups would be too difficult without using material that we haven't learned yet.
Any suggestions for showing that $|\hat{G}|=|G|$ are appreciated!