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…


One thought on “Class sums

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s