# Defining a ring homomorphism in proving $|C| \cdot \frac{\chi(C)}{\dim V}$

Lemma irreducible representation $\rho: G \rightarrow GL(V)$, $C$ conjugacy class, then $$|C| \cdot \frac{\chi(C)}{\dim V}$$ is an algebraic integer.

In the start of this proof we have:

Define a ring homomorphism

$$\varphi:\mathbb{C}[G] \longrightarrow Hom (V,V)$$ $$g \mapsto \rho(g)$$ $$\sum \alpha_g g \longrightarrow \sum \alpha_g \rho(g)$$

How do we know it is a ring homomorphism. I can see some sort of multiplication will be one operation but what is the other?

Then the proof proceeds by

(this $\varphi$) induces $$\mathbb{C}[G]^G \longrightarrow Hom(V,V)^G$$ which is equivalent to $$\mathbb{Z}(\mathbb{C}[G]) \longrightarrow Hom_G(V,V)$$

I cannot see how $\mathbb{C}[G]^G =\mathbb{Z}(\mathbb{C}[G])$ holds. Ive tried $$h \cdot \sum_{g \in G}\alpha_g g=\sum_g \alpha(h\cdot g)$$ but cannot see how this would help.

I cannot see how $Z(\Bbb C[G])=\Bbb C[G]^G$ holds.
To be central in $\Bbb C[G]$ it's enough to commute with every $g\in G$. Commuting with $g$ is the same as being a fixed point under conjugation by $g$.
• @sandstone If you didn't know what the action of $G$ on $\Bbb C[G]$ is then that is one of the very first things you should have been asking about! Go back into whatever you're reading to verify this. – whacka Apr 10 '15 at 14:44