Kernel and centralizer of irreducible characters

[syj]

Homework Statement

Find the kernel of all irreducible characters of G, when given the character table.
Find the centralizer of each irreducible character of G, when given the character table.
Find Z(G) (the centralizer of G) for the same character table.

Homework Equations

I know that ker($\chi$)={g $\in$G| $\chi$(g)=$\chi$(1)} where $\chi \in$ Irr(G).

I also know that Z($\chi$)={g $\in$ G | |$\chi$(g)|=$\chi$(1)}

and I think Z(G) is the intersection of all Z($\chi$)I hope you can the gist of the question.

The Attempt at a Solution

The question I have to answer pertains to a given character table of S4.
I do not want an answer to this particular table.
If someone could please give me an example using another character table so I can follow.

 

[micromass]

So, what did you try already?? If you tell us where you're stuck then we'll know where to help!

 

[syj]

Ok
the character table I am given for S₄ is as follows.
I am really green at this latex stuff and so I will try to list it as a table as best as I can:

[g] 1A 2A 2B 3A 4A

$\varphi$₁ 1 1 1 1 1
$\varphi$₂ 1 -1 1 1 -1
$\varphi$₃ 2 0 2 -1 0
$\varphi$₄ 3 1 -1 0 -1
$\varphi$₅ 3 -1 -1 0 1

so i have:
ker($\varphi$₁)={1A,2A,2B,3A,4A}
ker($\varphi$₂)={1A,2B,3A}
ker($\varphi$₃)={1A,2B}
ker($\varphi$₄)={1A}=ker$\varphi$₅

and
Z($\varphi$₁)={1A,2A,2B,3A,4A}
Z($\varphi$₂)={1A,2A,2B,3A,4A}
Z($\varphi$₃)={1A,2B}
Z($\varphi$₄)={1A}=Z($\varphi$₅)

so Z(G)={1A}

i apologise for the horrible table. i could not get the numbers to align. *so embarrassed*

 

[micromass]

Ok
the character table I am given for S₄ is as follows.
I am really green at this latex stuff and so I will try to list it as a table as best as I can:

[g] 1A 2A 2B 3A 4A

$\varphi$₁ 1 1 1 1 1
$\varphi$₂ 1 -1 1 1 -1
$\varphi$₃ 2 0 2 -1 0
$\varphi$₄ 3 1 -1 0 -1
$\varphi$₅ 3 -1 -1 0 1

so i have:
ker($\varphi$₁)={1A,2A,2B,3A,4A}
ker($\varphi$₂)={1A,2B,3A}
ker($\varphi$₃)={1A,2B}
ker($\varphi$₄)={1A}=ker$\varphi$₅

and
Z($\varphi$₁)={1A,2A,2B,3A,4A}
Z($\varphi$₂)={1A,2A,2B,3A,4A}
Z($\varphi$₃)={1A,2B}
Z($\varphi$₄)={1A}=Z($\varphi$₅)

so Z(G)={1A}

i apologise for the horrible table. i could not get the numbers to align. *so embarrassed*

Seems correct! However, there is a small notational thingy. The 1A, 2A, 2B, 3A, 4A are conjugacy classes, right? But the kernel, centralizer and centres do not consist out of conjugacy classes, but ouy of elements of G. So instead of saying

ker($\varphi$₁)={1A,2A,2B,3A,4A}

you have to say that the kernel is the elements from these classes. In this case $ker(\varphi)=S_4$.

 

[syj]

ok,
so if 1A = 1
2A = (1 2)
2B = (1 2)(3 4)
3A = (1 2 3)
4A = (1 2 3 4)

then z($\varphi$₃)={1, 2, 3,4} ?

 

[micromass]

ok,
so if 1A = 1
2A = (1 2)
2B = (1 2)(3 4)
3A = (1 2 3)
4A = (1 2 3 4)

No, that is not true. 2A contains much more elements than that! 2A is an entire conjugacy classs, remember!
The same for the rest.

then z($\varphi$₃)={1, 2, 3,4} ?

I cannot see how you came to that conlusion. z($\varphi$₃) must contain elements of S₄...

 

[syj]

ok
i think i understand ...
would Z($\varphi$₃)={1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ?

 

[micromass]

ok
i think i understand ...
would Z($\varphi$₃)={1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ?

Yes, that is correct!

 

[syj]

woooo hooooo !
thanks a mil
im gona tackle the representation of a group question now.
still not sure how to get that proof sorted out.