# A question on the decomposition of the group algebra $\mathbb{C}[G]$ of a finite group $G$

I am am very confused about a fundamental result in representation theory of finite groups. Please let me first introduce the setting.

Let $G$ be a finite group. The group algebra $\mathbb{C}[G]$ is a semisimple ring by Maschke's theorem. Hence, there is an isomorphism $$\mathbb{C}[G]\cong U_1\oplus\ldots\oplus U_n \hspace{20pt}\text{(1)}$$ for simple $\mathbb{C}[G]$ (always ignored in this question: left-)modules $U_i$ and those correspond to (wlog non-trivial) irreducible complex representations of $G$. In particular, we get a formula $|G|=|U_1|+\ldots+|U_n|$ by taking the dimension as complex vector spaces.

On the other hand the Artin-Wedderburn implies that the semisimple ring $\mathbb{C}[G]$ decomposes as $$\mathbb{C}[G]\cong M_1\oplus\ldots\oplus M_m\hspace{20pt}\text{(2)}$$ where $M_i$ are (wlog non-trivial) matrix rings over division rings $D_i$. It can be seen that $m$ equals the number of conjugacy classes of $G$ and the number of non-equivalent (and non-trivial) irreducible representations of $G$ and one has the formula $|G|=d_1^2+\ldots+d_m^2$ where the $d_i$ are the dimensions of the irreducible representations of $G$.

My question is basically:

How do the two decompositions (1) and (2) relate?

My guess is the following: In the decomposition (1), some of the $U_i$ may be isomorphic as $\mathbb{C}[G]$-modules. Group them together to obtain $$\mathbb{C}[G]\cong (U_{1}\oplus\ldots\oplus U_{j_1})\oplus\ldots\oplus(U_{j_{k-1}+1}\oplus\ldots\oplus U_{j_k})$$ (to keep the indexing simple, we assumed that in (1) the modules already ordered properly.). Then one can perhaps (?) show that $$U_{i_{k-1}+1}\oplus\ldots\oplus U_{i_k}\cong M_i$$ as $\mathbb{C}[G]$ modules. Is this the right way to relate (1) and (2)? Thank you.

## 2 Answers

If $M_n(D)$ is a matrix over a division ring $D$, then it is semisimple. If you consider the column space of the matrix then it is $D^{n\times 1}$ and hence is semisimple. Then you can write $M_n(D)=M_1\bigoplus M_2\bigoplus...\bigoplus M_n$, where $M_i$ is the matrix which has non-zero entries only in it's $i$-th column. Hence $M_n(D)$ is semi-simple. So by Maschke's theorem

$\mathbb{C}[G]=M_1^{n_1}\bigoplus ...\bigoplus M_n^{n_1}\bigoplus M_{1}^{n_2}\bigoplus ...\bigoplus M_n^{n_2}\bigoplus...\bigoplus M_1^{n_s}\bigoplus ...\bigoplus M_n^{n_s}$, where $M_i^{n_j}$ is the matrix corresponding to the matrix having only non-zero entries in the $i$-th column corresponding to the $n_j$-th matrix in the statement of the Artin-Wedderburn theorem and $s$ denote the number of conjugacy classs of $G$.

If you just compare the number of matrices then you will see that the expresion for $|G|$ is same from both the theorems.

• Thank you for your answer. Just that I understand you correctly: The connection between your answer and my "suggestion" (at the end of the question) is that your answer gives a reason why my "suggestion" is correct, does it? – user8463524 Dec 13 '15 at 12:51
• @jeffrey yes, that's correct. and in fact, now you have the concrete description of your $U_i$'s. – seeker Dec 13 '15 at 16:40
• please let me again ask about the logic of your proof: We start with the Artin-Wedderburn decomposition (this is (2) in my question). Then we split the summands $M_n(D)$ further into summands $D^{n\times 1}$. Do you mean $D^{n\times 1}$ is simple instead of semisimple? If you mean this, then we have a decomposition of $\mathbb{C}[G]$ into simple summands in this way. Now you can show that each simple module over $\mathbb{C}[G]$ is isomorphic to one of the summands above. Does this reproduce your line of reasoning? Thanks. – user8463524 Dec 16 '15 at 15:01
• @jeffrey Yes. I wanted to say that $D^{n\times 1}$ is simple and hence $M_n(D)$ is semi-simple. Rest of the reasoning is same as you spelled out in your comment. – seeker Dec 16 '15 at 17:26

The Artin-Wedderburn decomposition can thought of as a decomposition

$$\mathbb{C}[G] \cong \bigoplus_V V \otimes V^{\ast}$$

with respect to two $G$-actions, namely both the action on the left and the action on the right (although I may be off by either an inverse or a dual). Here $V$ runs over all complex irreps of $G$. When you only pay attention to the left action, each of the right action bits $V^{\ast}$ just become vector spaces, and you get

$$\mathbb{C}[G] \cong \bigoplus_V (\dim V) V.$$