G, a finite abelian and cyclic group, isomorphic to set of homomorphisms of G into C\{0}

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 $$i=1,..n$$.

• Isn't the problem assuming that $G$ has only one cyclic factor? Nov 16 '16 at 0:38
• @WillDana oh right, of course, and that immediately shows that $\hat G$ is cyclic, since it is abelian and each of its elements is a root of unity. Nov 16 '16 at 0:45
• Every map is defined by where it sends a generator of $G$. How many complex numbers can be the image of such a generator? Nov 16 '16 at 2:26
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?