Abelian group C*-algebras Let $G$ be a locally compact Abelian group $C^*$-algebra, then $C^*(G)$ is an Abelian $C^*$-algebra, so $C^*(G)$ is isomorpohism with the $C_0(X)$ for some locally compact Hausdorff space $X$, here $X$ is the homomorphism(or multiplicative map) from $C^*(G)$ to $C$.
But I see that we can let $X=\hat G=\operatorname{Hom}(G,T)$ which is the dual group of $G$, here $T$ is the one-dimensional torus, how to explain this?
I think in $L^1(G)$, we can let $g^*=g^{-1}$, then treat every element of $G$ as a unitary element, so the image of the map from $C^*(G)$ to $C$ are also unitary. Am I right? Or there are any better explainations?
 A: I know this response is late, but I was not active at the time you asked the question. I am sure by this time you have a better understanding of this material, however, I am going to answer the question as you posed it. 
1) Construction of the Group C$^*$ algebra. 
Let G be a locally compact group e.g a countable discrete group, we represent G using the left-regular representation $\lambda : G \to \mathcal{U}(\ell^2(G))$. We have a canonical orthonormal basis on $\ell^2(G)$ the dirac masses, $\{\delta_g \}_{g \in G}$. Our unitaries act on the basis in the following way $\lambda_g (\delta_h)=\delta_{gh}$. It is clear that $\forall g \in G$, $\lambda_g$ is a unitary by the group axioms. Now, $C^*(G) : = \overline{span{\{\lambda_g\}_{g \in G}}}^{|| \cdot ||}$. Where the closure is in the operator norm. Moreoever, if G is abelian we have that $\lambda_g \lambda_h = \lambda_{gh}=\lambda_{hg}=\lambda_h \lambda_g$. It is clear that the abelianess on the unitaries passes to linear combinations and limits. (If this is not clear to you, then you should prove it. It is a good exercise.)
2) Now, we need a second fact. The Gelfand-Naimark Theorem, which states that given an abelian $C^*$ algebra, $\mathcal{A}$, then $\mathcal{A} \simeq ^{*} C_0(\triangle_{\mathcal{A}})$. Where $\triangle_{\mathcal{A}}$ is the maximal ideal space of your algebra, namely the set of $\phi: \mathcal{A} \to \mathbb{C}$ such that $\forall x,y \in \mathcal{A}$, $\phi(xy)=\phi(x)\phi(y)$. 
3) The last thing that you need to prove, and I would be willing to prove if you have not figured it out, is that $\triangle_{C^*(G)}$ is homeomorphic to  $\hat{G}$, which immediately gives the desired result.
