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The dimension of $Z (k [G])$ is the number of conjugacy classes of $G$

Statement

Let $G$ be a finite group and let $k$ be a field. Then

\dim_{k} Z (k [G])

is exactly the number of conjugacy classes of $G$ .

Facts & Assumptions

Given: A finite group $G$ and a field $k$ .

[L1]

For a finite group, the class sums indexed by the conjugacy classes of $G$ form a basis of $Z (k [G])$ (For a finite group, the class sums form a basis of $Z (k [G])$ ).

Proof

technique · direct

1.1

By [L1], there is one basis vector of $Z (k [G])$ for each conjugacy class of $G$ .

L1given

2.1

The dimension of a finite-dimensional vector space is the number of vectors in any basis, so step 1.1 identifies $\dim_{k} Z (k [G])$ with the number of conjugacy classes.

step 1.1givenalgebra∎

Depends on

For a finite group, the class sums form a basis of $Z(k[G])$

Used by

If k is algebraically closed and char k ∤ |G|, the number of irreducible representations of G equals the number of conjugacy classes Theorem

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Webb, A Course in Finite Group Representation Theory, Lemma 3.4.2 (standard reference, not scraped)

The dimension of Z(k[G])Z(k[G]) is the number of conjugacy classes of GG