For $g_1,g_2$ in finite abelian group with $g_1\neq g_2$, there is $\phi\in\mathrm{Hom}(G,\mathbb C^\times)$ such that $\phi(g_1)\ne\phi(g_2)$

Question

I am having trouble with the following problem, taken from Herstein's Topics in Algebra 2nd Edition:

If $G$ is an abelian group, let $\hat G$ be the set of all homomorphisms of $G$ into the group of nonzero complex numbers under multiplication. If, $\phi_1,\phi_2\in \hat G$, define $\phi_1 \cdot \phi_2$ by $\phi_1 \cdot \phi_2(g) = \phi_1(g) \phi_2 (g)$ for all $g \in G.$

Show that, given $G$ is also finite and $g_1\neq g_2$ are in $G$, there is a $\phi \in \hat G$ with $\phi (g_1) \neq \phi (g_2)$.

I have already proved that $\hat G$ is abelian and that for $\phi \in \hat G$ we have $\phi(g)$ a root of unity for every $g \in G$. I also know that $G$ is isomorphic to $\hat G$.

What I am basically proving is that each $\phi \in \hat G$ is one-to-one. Is that correct? How can I leverage the fact that $G$ and $\hat G$ are isomorphic to show that each element of $\hat G$ is in fact an isomorphism?

Answer 1

The steps of a proof can be the following:

1. It suffices to show that for $g\in G\setminus\{1\}$ there exists $\phi$ such that $\phi(g)\neq 1$.

2. It suffices to prove the statement for cyclic groups due to the structure theorem for finitely generated abelian groups.

3. Handle the cyclic case.