# Class sums

Let’s define a new concept that seems to be really important in algebraic number theory, that will help us peek inside some of the things we’ve been seeing.

Let $C_j$ be a conjugacy class in a finite group. Then we call $z_j=\sum_{g\in C_j}g$ a class sum (for pretty obvious reasons, it is the sum of all the elements in a conjugacy class).

Lemma: The number of conjugacy classes in a finite group G is the dimension of the center of the group ring. Or if we let r denote the number of conjugacy classes, then $r=dim_k(Z(kG))$.

We prove this by showing that the class sums form a basis. First, given a class sum, we show that $z_j\in Z(kG)$. Well, let $h\in G$, then $hz_j h^{-1}=z_j\Rightarrow hz_j=z_j h$, since conjugation just permutes elements of the conjugacy class, thus they live in the right place. They are also linearly independent since the elements of the sums $z_j$ and $z_k$ are disjoint (they are orbits which partition the group) if $j\neq k$.

Now all we need is that they span. Let $u=\sum a_gg\in Z(kG)$. Then for any $h\in G$, we have that $huh^{-1}=u$, so by comparing coefficients, $a_{hgh^{-1}}=a_g$ for all $g\in G$. This gives that all the coefficients on elements in the same conjugacy class are the same, i.e. we can factor out that coefficient and have the class sum left over. Thus $u$ is a linear combination of the class sums, and hence they span.

As a corollary we get that the number of simple components of $kG$ is the same as the number of conjugacy classes of $G$. This is because $Z(M_{n_i}(k))$ is the subspace of scalar matrices. So if there are m simple components, we get 1 dimension for each of these by our decomposition in Artin-Wedderburn and so $r=Z(kG)=m$.

Another consequence is that the number of irreducible k-representations of a finite group is equal to the number of its conjugacy classes.

The proof is just to note that the number of simple $kG$-modules is precisely the number of simple components of $kG$ which correspond bijectively with the irreducible k-representations, and now I refer to the paragraph above.

Now we can compute $\mathbb{C}S_3$ in a different way and confirm our answer from before. We know that it is 6 dimensional, since the dimension is the order of the group. We also know that there are three conjugacy classes, so there are three simple components, so the dimensions of these must be 1, 1, and 4. Thus $\mathbb{C}S_3\cong \mathbb{C}\times\mathbb{C}\times M_2(\mathbb{C})$.

If we want another quick one. Let $Q_8$ be the quaternion group of order 8. Then try to figure out why $\mathbb{C}Q_8\cong \mathbb{C}^4\times M_2(\mathbb{C})$.

So I think I’m sort of done with Artin-Wedderburn and its consequences for now. Maybe I’ll move on to some character theory as Akhil brought up in the last post…