Center of Group algebra finitely generated

Question

I am going through a proof in Jean-Pierre Serre's german version of "Linear representations" and have the following theorem here.

Let $\rho$ be an irreducible representation of the finite group G of degree d, $\chi$ the character of $\rho$ and K a conjugacy class of G. Then $\frac{1}{d} \sum\limits_{s \in K} s$ is an algebraic integer.

The proof goes as follows:

Let $Z(\mathbb{C}[G])$ and $Z(\mathbb{Z}[G])$ be the centers of $\mathbb{C}[G]$ and $\mathbb{Z}[G]$.

It is clear that $e_K = \sum\limits_{s \in K} s \in Z(\mathbb{C}[G])$.

Furthermore $e_K \in Z(\mathbb{Z}[G])$, which is a finitely generated $\mathbb{Z}$-module implying that $e_K$ is an algebraic integer.

This is the first point that I don't get. Why can you say that $Z(\mathbb{Z}[G])$ is a finitely generated $\mathbb{Z}$-module? If $\mathbb{Z}[G]$ is finitely generated, does that imply $Z(\mathbb{Z}[G])$ being finitely generated ?

Then the proof continues :

The representation $\rho$ defines a homomorphism $\rho: \mathbb{C}[G] \longrightarrow End(W)$. If you reduce $\rho$ to the Center $Z(\mathbb{C}[G])$, you can apply Schur's Lemma, so that you get multiplications of W as the image.

That means $\rho(e_K) = \lambda \cdot 1$, where $\lambda$ is an algebraic integer.

Now, the point I don't understand here is why you have to reduce $\rho$ to the Center of $\mathbb{C}[G]$ and why is $\lambda$ an algebraic integer ?

My thoughts on this were:

If $e_K$ is an algebraic integer, there is a polynomial $f$ with $f(e_K) = 0$. Applying $\rho$, gives $0 = \rho(f(e_K))$, but I don't know how I can conclude from there that $\rho(e_K)$ is the root of any polynomial (probably $f$ again?)

Answer 1

First, I assume your definition of "algebraic integer" is "element $a$ of a $\mathbb{Z}$-algebra $A$ for which $\mathbb{Z}[a]$ is a finitely-generated $\mathbb{Z}$-module." The usual definition also requires $a$ to be a complex number, which clearly isn't the case for your elements.

Why can you say $Z(\mathbb{Z}[G])$ is a finitely-generated $\mathbb{Z}$-module.

Indeed, we can find an explicit basis using the following reasoning: First,

$$ x\in Z(\mathbb{Z}[G]) \iff (\forall g\in G)\, gx=xg \iff (\forall g\in G)\, gxg^{-1}=x. $$

Writing $x=\sum c_gg$, we see this is equivalent to $c_g$ being constant on conjugacy classes, so we have a $\mathbb{Z}$-basis $\{e_{\small K}\mid K\subset G~ \textrm{conjugacy class}\}$ for $Z(\mathbb{Z}[G])$.

If $\mathbb{Z}[G]$ is finitely generated, does that imply $Z(\mathbb{Z}[G])$ being finitely generated ?

Yes, that is also another way to see it.

Now, the point I don't understand here is why you have to reduce $\rho$ to the Center of $\mathbb{C}[G]$

In general if $x\in\mathbb{C}[G]$, the linear operator $\rho(x)$ of $W$ will not simply be multiplication by a scalar. If we know $x$ is central, then $T=\rho(x)$ is a map such that $\rho(g)T=T\rho(g)$ for all $g\in G$, in other words it is an automorphism of $W$, an element of $\hom_G(W,W)$, and Schur's lemma states the only automorphisms when $W$ is irreducible (over $\mathbb{C}$) are multiplication by scalars.

why is $\lambda$ an algebraic integer ? [...] If $e_{\small K}$ is an algebraic integer, there is a polynomial $f$ with $f(e_{\small K})=0$. Applying $\rho$, gives $0=\rho(f(e_{\small K}))$, but I don't know how I can conclude from there that $\rho(e_{\small K})$ is the root of any polynomial (probably $f$ again?)

Since $\rho:\mathbb{C}[G]\to\mathrm{End}(W)$ is an algebra homomorphism, we have $\rho(f(e_{\small K}))=f(\rho(e_{\small K}))$. Then we know $f(\rho(e_{\small K}))=f(\lambda\cdot\mathrm{Id})=f(\lambda)\cdot\mathrm{Id}=0$ so we get $f(\lambda)=0$.

Answer 2

I'm doing the same thing, but I just can't see how it must be that easy to see that \begin{align} \{e_K |~ K \subset G~ conjugacy~ class \} \end{align} is a basis for $Z(\mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even after Kenny's respond I don't see that...

My approach to this is:

Be $x \in Z(\mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g \in G$. Writing $x = \sum\limits_{h \in G} \lambda_h h$, we get $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b \in K$ (same conjugacy class) you get $\sum\limits_{s \in K} \mu_s s = \mu\sum\limits_{s \in K} s$, since \begin{align} a = hbh^{-1} \Rightarrow \mu_a = \mu_b ~\forall a,b\in K \end{align} So all elements in the same conjugacy class have the same scalar.

But from here I don't really know how I can generate the $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$ with the $e_K$ ...

I would be really thankful, if someone could give me quite a basic proof that even I understand...