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I have been given a character table and I need to find from the table the centre of each character. I dont know how to do this. if someone could please explain how i can find the centre by looking at the character table.

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Are you just trying to find the centre of each character, or the centre of the whole group? –  user16299 Jan 12 '12 at 18:49
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Do you know what the centre of a character is? Can you tell us what your definition is? The definition I know of makes it obvious how to find it if you are given the character table! –  Mariano Suárez-Alvarez Jan 12 '12 at 18:50
    
thank you guys. sorry for the lack of clarity in the question. i think i have it figured out now. thanks again :) –  sarah jamal Jan 13 '12 at 3:17
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1 Answer

up vote 1 down vote accepted

Let $G$ be some finite group and $\text{irr}(G)$ the set of irreducible characters of $G$. For $\chi\in\text{irr}(G)$ define $\mathbf{Z}(\chi)=\left\{g\in G:|\chi(g)|=\chi(1)\right\}$ (called the center of the character). Then, it's a common fact that

$$\mathbf{Z}(G)=\bigcap_{\chi\in\text{irr}(G)}\mathbf{Z}(\chi)$$

So, if you are given a character table, then for each row you can look at the entries for which the modulus of that entry matches with the first entry of that row (assuming you are writing character tables with $\{1\}$ corresponding to the first column), and take the union over the conjugacy classes those entries sit below, call this union $Z_k$ if $k$ is the row we are in. Then, the above theorem says that $\mathbf{Z}(G)=Z_1\cap\cdots\cap Z_n$ if you have $n$ rows.

For proofs of the above statements you can see my blog post here, or for a more comprehensive source you can see Isaac's Character Theory of Finite Groups.

EDIT: Thankfully Yemon Choi has pointed out that you were just looking how to obtain the center of the CHARACTER from the character table, and not the center of the group. This is implicitly stated in the second paragraph of the above. Namely, to find $\mathbf{Z}(\chi)$, locate the row corresponding to $\chi$, and then take the union of the conjugacy classes lying above the row entries whose modulus equals $\chi(1)$.

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The OP's question seems to ask the different (and easier) question of determining each $Z(\chi)$, though it is not clear what definition she is using –  user16299 Jan 12 '12 at 18:48
    
@YemonChoi Thank you for pointing that out. I didn't even catch that--it's sort of an odd thing to ask. I have edited accordingly. –  Alex Youcis Jan 12 '12 at 18:50
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