If G is a finite abelian group, prove that $o(G) = o(\hat{G})$ and G is isomorphic to $\hat{G}$.

Question

The below question comes from Herstein' "Topic in Algebra", 2nd edition. It has been asked before but I fail to understand what is available + my professor gives additional hints that I cannot see their connection with the question.

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

Question: If G is a finite abelian group, prove that $o(G) = o(\hat{G})$ and G is isomorphic to $\hat{G}$.

Here is the hint given:

Hints: Show the following two rules: i) If G and H are abelian groups with $G \cong H$, then $\hat{G} \cong \hat{H}$. ii) If $G_1$ and $G_2$ are abelian groups, then $\widehat{G_1 \times G_2} \cong \hat{G_1} \times \hat{G_2}$.

I have proved in an earlier question, that fix $\phi \in \hat{G}$, for all $g \in G$, $\phi(g)$ is a root of unity.
Unfortunately I am completely stumped on the question so I will just give some thoughts I have: (edits for my finished (?) thoughts)

• Since G is finite abelian, we may decompose G to cyclic groups, call them $A_i$. Proved in an earlier problem, that $\hat{A_i}$ is also cyclic, and $A_i \cong \hat{A_i}$. As a result, $G \cong \hat{G}$.

But somehow I still did not used the hints...

Any help appreciated.

Answer 1

You wrote "Since G is finite abelian, we may decompose G to cyclic groups, call them $A_i$, proved in an earlier problem, that $\hat{A}_i$ is also cyclic, and $A_i\cong \hat{A}_i$. As a result, $G\cong \hat{G}$."

Let us see what you used in your proof.

1. If a group $A$ is cyclic, then $A\cong \hat{A}$. (Hence it implies $A_1\times \cdots \times A_n \cong \hat{A_1}\times \cdots \times \hat{A_n}$ where $A_i$ are cyclic).
2. If G is finite abelian, it can be written as $G= A_1\times \cdots A_n$ where $A_i$ are cylic.

Then from 1) and 2) you conclude "$G\cong \hat{G}$". Here you are implicitly using the argument "$ \widehat{A_1\times \cdots \times A_n} \cong \hat{A_1}\times \cdots \times \hat{A_n}$". You have to prove this argument, which is hint ii. Also, note that the direct product of cyclic groups does not have to be cyclic.