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 cyclic, $\hat G$ is cyclic and $G$ and $\hat G$ are isomorphic.
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 $g \in G$.
My approach:
Since $G$ and $\hat G$ are both finite and abelian, we have $G=A_1 \times ...\times A_n$ and $\hat G=B_1 \times ... \times B_m$ where each $A_i$ and $B_i$ are cyclic.
Now I want to show that $n=m$ and that $A_i=B_i$ for each $i=1,..n$.
Is this a proper approach? If so, any suggestions on how I can go about this?
First note that for a finite cyclic group $G$ any homomophism to any other group $H$ is known if the image of a generator $g$ of $G$ is known since the images of $g^2,g^3, \ldots$ can be calculated. As you stated correctly the images in this case are all roots of unity. If these are taken as the image of a certain homomophism what have these elements in common (w.r.t. their order)? What can you say about an arbitrary homomorphism w.r.t. a fixed homomorphism that maps $g$ to a primitive root of unity?