# Decomposition of group algebra $\mathbb{C}[G]$ [duplicate]

Let $G$ be a finite group. By Wedderburn theorem, we see that $\mathbb{C}[G]$ is product of matrix algebras.

"Linear Representations of Finite Groups" by Serre has an explanation for this result using representations...

As $G$ is a finite group, there are only finitely many distinct irreducible representations of $G$. Let $(\rho_i,W_i)$ for $1\leq i\leq h$ are those distinct irreducible representations. So, we have $\rho_i:G\rightarrow GL(W_i)$.

For $\sum_{g\in G}a_gg\in \mathbb{C}[G]$ we define

$$\rho_i\left(\sum_{g\in G}a_gg\right)=\sum_{g\in G}a_g\rho_i(g).$$

But this time, $\sum_{g\in G}a_g\rho_i(g)$ need not be in $GL(W_i)$ but it is definitely in $\rm{End}(W_i)$. so, we have $\rho_i: \mathbb{C}[G]\rightarrow \rm{End}(W_i)$ for $1\leq i\leq h$. As we are concerned with only finite dimensional representaions, we can assume $\dim(W_i)=n_i$ and then we have $\rm{End}(W_i)\cong M_{n_i}(\mathbb{C})$.

We now define another map

$$\rho:\mathbb{C}[G]\rightarrow \prod_{i=1}^h\rm{End}(W_i)\rightarrow \prod_{i=1}^hM_{n_i}(\mathbb{C})$$ as $\rho(\widetilde{g})=(\rho_1(\widetilde{g}),\cdots,\rho_h(\widetilde{g}))$. This is a well defined homomorphism and then he says that this is actually an isomorphism, giving another proof for structure theorem in a special case. As dimension of $\mathbb{C}[G]$ is $|G|$ and as dimension of $\prod_{i=1}^hM_{n_i}(\mathbb{C})$ is $\sum_{i=1}^hn_i^2=|G|$, it is enough if we can prove that $\rho$ is surjective or injective. He does this by proving $\rho$ is surjective.

Suppose not, then $\rho(\mathbb{C}[G])$ is a proper subspace of $\prod_{i=1}^hM_{n_i}(\mathbb{C})$ and so there exists a nonzero linear functional $\eta: \prod_{i=1}^hM_{n_i}(\mathbb{C})\rightarrow \mathbb{C}$ which vanish on $\rho(\mathbb{C}[G])$. Now, any linear map $\eta :\prod_{i=1}^hM_{n_i}(\mathbb{C})\rightarrow \mathbb{C}$ is of the form $\eta_1+\eta_2+\cdots+\eta_h$ where $\eta_i:M_{n_i}(\mathbb{C})\rightarrow \mathbb{C}$

We have $\eta(\rho(\widetilde{g}))=0$ for all $\widetilde{g}\in G$ i.e., $\eta(\rho_1(\widetilde{g}),\cdots,\rho_h(\widetilde{g}))=0$ i.e., $\eta_1(\rho_1(\widetilde{g}))+\cdots+\eta_h(\rho(\widetilde{g}))=0$. As $\widetilde{g}=\sum a_gg$ for some $a_g\in \mathbb{C}$ if necessary, after renaming $\eta_i$ we can say that

$$\eta_1(\rho_1(g))+\cdots+\eta_h(\rho(g))=0.$$

Now, i do not understand why is this not possible. I thing i have to consider traces and then end up with some relation as $$\sum_{i=1}^ha_i\chi_i=0$$ with at least one $a_i\neq 0$ (this depends on non zeroness of $\eta$ and this non zeroness of some $\eta_i$) and then say that as distinct characters are linearly independent we see that $a_i=0$ for all $i$.

But i am not sure how to write this.. What am i supposed to do for that $\eta_i$??

## 1 Answer

Serre suggests you argue as follows: each $\eta_i\circ \rho_i$ is a linear combination of the coordinate functions for $\rho_i$ (the "matrix form" of the representations mentioned in 2.2). But these coordinate functions are linearly independent, indeed they are shown in section 2.2 to be an orthogonal set with respect to the inner product introduced on functions $G \to \mathbb{C}$ there.

For another approach to surjectivity, you can try this. First show that the map $\rho_i : \operatorname{End}(W_i)$ is onto (this is a general fact about irreps of $\mathbb{C}$-algebras). Then show that as a $\mathbb{C}G$-module by left-multiplication, $\operatorname{End}(W_i) \cong W_i^{\oplus \dim W_i}$. Then prove, using the fact that the $W_i$ are pairwise non-isomorphic, that given any $m_1,\ldots,m_h$, any submodule of $$\bigoplus W_i^{\oplus m_i}$$ must have the form $U_1\oplus \cdots \oplus U_h$ for $U_i \leq W_i^{\oplus m_i}$. Thus if you have surjective maps $r_i : \mathbb{C}G \to W_i^{\oplus m_i}$, it follows $r=\bigoplus r_i : \mathbb{C}G \to \bigoplus W_i^{\oplus m_i}$ is surjective. Applying this to your situation gives the result you want.

• Even i could not see that there is a duplicate... Thanks... I would delete this question... Excuse me for making you to write this long answer... – user87543 Sep 23 '15 at 13:09