## The Center of a Character (Pt. II)

**Point of post: **This is a continuation of this post.

We now show some connections between the center of a character and its kernel. Namely:

**Theorem:** *Let be a finite group and a character of . Then: a) is cyclic, b) and c) if is an irreducible character then .*

**Proof: **

**a) **We first claim that . To see this we merely note that if and only if but and so if and only if from where the conclusion follows. It follows then from the first isomorphism theorem that . But, is a finite multiplicative subgroup of and thus clearly cyclic.

**b) **Evidently one has that (this is immediate since each for is a scalar matrix). Thus, one has that for any (since ) and one has that so that and so and thus so that . Since was arbitary the conclusion follows.

**c) **If is irreducible and it clearly follows that and so by Schur’s Lemma it follows that and so .

The real interesting thing now comes from the next theorem:

**Theorem: ***Let be a finite group. Then,*

* *

* *

**Proof: **Clearly if then we have clearly from Schur’s lemma that with and so for every .

Conversely, suppose that for every . It follows then from the comments of the previous theorem that for every . Thus, for any one has that or that . It follows then that . But, recalling that every normal subgroup is the realization of some intersection of the ‘s, and so in particular is the intersection of some of the ‘s we may conclude that and so and so and commute. Since was arbitrary it follows that and so the conclusion follows.

**References:**

1. Isaacs, I. Martin. *Character Theory of Finite Groups*. New York: Academic, 1976. Print.

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