TITLE: Character of $\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$
QUESTION [5 upvotes]: I've been given the following two questions, and for both I'm really unsure what to do (I'm a beginner with the theory of characters). 
Let $G$ be a finite group, $\mathbb{C}G$ its group algebra over $\mathbb{C}$. $\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$ forms a $\mathbb{C}G$-module with the action defined, for $a,b \in \mathbb{C}G$, $f \in \text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$, by $(af)(b) = f(ba)$. 
With this module structure defined on $\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$, firstly I want to find $\chi_{\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})}(g)$ for all $g \in G$. Secondly, in the case that $G = S_3$, I want to express $\chi_{\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})}(g)$ in terms of simple characters. Could someone please clearly describe how to do this.

REPLY [2 votes]: The character is just the character 
$$ \rho(g) = \begin{cases} 0 & g\neq1\\ |G| & g=1\end{cases}$$
In other words, the character of the regular representation.  Two ways to see this:

Pick the basis for $Hom_\mathbb{C}(\mathbb{C}G,\mathbb{C})$ which consists of the functions $f_g$, where for $h\in G$
$$ f_g(h) = \begin{cases} 1 & g=h\\ 0 & g\neq h\end{cases}$$
Now to compute the trace $\chi(k)$ with $k\in G$, you need to know when $f_g(hk)=f_g(h)$, and that's easy.
$\mathbb{C}G$ has character $\rho$, the regular representation.  $\mathbb{C}$ has character $\psi$, the trivial representation. We also have the isomorphism $Hom_\mathbb{C}(\mathbb{C}G,\mathbb{C})\cong \mathbb{C}G^\ast\otimes\mathbb{C}$, with associate character $\bar{\rho}\cdot\psi=\rho$.