The set of all complex homomorphisms of abelian group $G$ has the same structure as $G$ 
If $G$ is an abelien group, let $\hat{G}$ be the set of all
  homomorphisms of $G$ into the group of nonzero complex numbers under
  multiplication. If $\phi_1, \phi_1\in \hat{G}$ define $\phi_1\cdot
 \phi_2$ by $(\phi_1\cdot \phi_2)(g)=\phi_1(g)\phi_2(g)$ for all $g\in
 G$.
If $G$ is a finite abelian group prove that $o(G)=o(\hat{G})$ and $G$
  is isomorphic to $\hat{G}$.

The case when $G$ is finite cyclic is easy because in this case $\hat{G} \cong G$. Let's consider the case when $G$ is abelian group of order $p^n$, $p$ - prime. 
By Fundamental theorem on finite abelian groups we have the following: $$G=\langle a_1 \rangle_{p^{n_1}}\times \langle a_2 \rangle_{p^{n_2}} \times \dots \times \langle a_k \rangle_{p^{n_k}},$$
where $n_1\geq n_2\geq \dots \geq n_k>0$ and $n_1+\dots+n_k=n$. 
Let $A_i=\langle a_i \rangle_{p^{n_i}}$ and $A_i$ are cyclic groups then $A_i\cong \hat{A_i}$. So we have the following $$G=A_1\times \dots \times A_k\cong \hat{A_1}\times \dots \times \hat{A_k}.$$
Then I was trying to think about $\hat{G}$ and I think that $\hat{G}=\hat{A_1}\times \dots \times \hat{A_k}$ which follows that $G\cong \hat{G}$. 
How to prove rigorously that $\hat{G}=\hat{A_1}\times \dots \times \hat{A_k}$? I am not able to prove it.
 A: First of all if $\lambda, \beta:G\to K$ are two group homomorphisms with $K$ abelian then we define $\lambda+\beta:G\to K$ by $(\lambda+\beta)(x)=\lambda(x)+\beta(x)$. This is the pointiwise addition of homomorphisms and note that you need $K$ to be abelian in order for it to be a group addition defined on the set of all homomorphisms $\text{Hom}(G,K)$.
We also assume that the group operation on the Cartesian product $G\times H$ is pointwise as well.

Let $G,H,K$ be three groups with $K$ abelian and consider two homomorphisms:
$$j_G:G\to G\times H$$
$$j_G(g)=(g,1)$$
$$j_H:H\to G\times H$$
$$j_H(h)=(1,h)$$
Define
$$\Phi:\text{Hom}(G\times H, K)\to\text{Hom}(G, K)\times\text{Hom}(H, K)$$
$$\Phi(\lambda)=(\lambda\circ j_G, \lambda\circ j_H)$$
And treat each $\text{Hom}$ as a group via pointwise addition.

Lemma. $\Phi$ is a group isomorphism.

Proof.
1) $\Phi$ is a homomorphism: Indeed,
$$\Phi(\lambda+\beta)=\big((\lambda+\beta)\circ j_G, (\lambda+\beta)\circ j_H\big)$$
Without the loss of generality lets consider the first coordinate only:
$$((\lambda+\beta)\circ j_G )(g)=(\lambda+\beta)(g, 1)=\lambda(g,1)+\beta(g,1)=\lambda\circ j_G(g)+\beta\circ j_G(g)$$
In other words
$$(\lambda+\beta)\circ j_G =\lambda\circ j_G+\beta\circ j_G$$
And analogously $(\lambda+\beta)\circ j_H=\lambda\circ j_H+\beta\circ j_H$ showing that 
$$\Phi(\lambda + \beta)=\Phi(\lambda)+\Phi(\beta)$$
2) $\Phi$ is invertible: Indeed, consider the function
$$\Theta:\text{Hom}(G, K)\times\text{Hom}(H, K)\to\text{Hom}(G\times H, K)$$
$$\Theta(\lambda, \beta):G\times H\to K$$
$$\Theta(\lambda, \beta)(g,h)=\lambda(g)+\beta(h)$$
I leave as an exercise that $\Theta$ is the inverse of $\Phi$. $\Box$
Now that you have this result you can easily generalize it to any (finite) number of components by simple induction.
