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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.

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I can see some sort of multiplication will be one operation but what is the other?

Rings have two operations: multiplication and addition.

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 $g$.

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  • $\begingroup$ How do we know that the action is conjugation? $\endgroup$ – Permian Apr 10 '15 at 8:39
  • $\begingroup$ @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. $\endgroup$ – whacka Apr 10 '15 at 14:44

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