Let $G$ be a finite group.
We can define an action of $G\times G$ on the group algebra $\mathbb{C}[G]$ in the following way:
If $x \in \mathbb{C}[G]$ then $(g,h)\cdot x=gxh^{-1}$.
Now, what about the character of $\mathbb{C}[G]$ as $G\times G$-module? Can I express it in function of the character of $\mathbb{C}[G]$ as $G$-module?
Thank You for Help!
As john mangual says, this is essentially Frobenius reciprocity. You're considering three representations: the left and right-regular representations of $G$ on $\Bbb{C}[G]$, and the bi-regular representation of $G\times G$ on $\Bbb{C}[G]$. Write $\lambda$ for the left-regular representation of $G$, $\rho$ for the right-regular representation, and $\pi$ for the bi-regular representation of $G\times G$. To make the notation clearer, let's write $G\times G$ as $G\times G'$, where $G$ is the factor acting via $\lambda$ and $G'$ the factor acting via $\rho$.
So clearly $\pi\downharpoonright_G\simeq\lambda$ and $\pi\downharpoonright_{G'}\simeq\rho$. Then Frobenius reciprocity says that, for any representation $\sigma$ of $G'$, there is a natural isomorphism
$$\mathrm{Hom}_{G\times G'}(\pi,\mathrm{Ind}_{G}^{G\times G'}\sigma)\simeq\mathrm{Hom}_G(\lambda,\sigma)$$
(and that this is true if we reverse the order of the arguments in each side, as well as being true for $G'$ and $\rho$). In particular, if we know the decompositions of $\lambda$ and $\rho$ into irreducibles, then we can, more or less, read off the decomposition of $\pi$ into irreducibles from these.
So suppose we have a decomposition of $\lambda$ into irreducibles as $\lambda\simeq\lambda_1\oplus\cdots\oplus\lambda_n$, so that $\rho$ is given by the corresponding dual representations: $\rho\simeq\lambda_1^*\oplus\cdots\oplus\lambda_n^*$. Then, (by Frobenius again) noting that $\mathrm{Ind}_G^{G\times G'}\ \lambda_i$ is going to look like a sum of $\lambda_i\otimes\lambda_j^*$ and doing a bit of work with Schur's lemma to find multiplicities, you'll see that
$$\pi\simeq\bigoplus_{i=1}^n\lambda_i\otimes\lambda_i^*.$$
Then we can just read off the character of $\pi$ as
$$\chi_\pi(g,g')=\sum_{i=1}^n\chi_{\lambda_i}(g)\chi_{\lambda_i^*}(g').$$