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 $\hat G$ is an abelian group under the operation defined
I tried to take advantage of the fact that since $G$ is abelian, we must have, $\forall g_1,g_2 \in G$
$\phi_1 \cdot \phi_2(g_1g_2) = \phi_1(g_1g_2) \phi_2 (g_1g_2)=\phi_1(g_2g_1) \phi_2 (g_2g_1)=\phi_1 \cdot \phi_2(g_2g_1)$,
but this has not gotten me any closer to showing that,
$\phi_1 \cdot \phi_2(g_1g_2)=\phi_2 \cdot \phi_1(g_1g_2)$.
Is this the wrong direction to proceed from? This comes from the section that introduces the Fundamental Theorem of Finite Abelian Groups, but I don't see how I can leverage that for this problem.