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Centralizers and Center for S_3

  • #1
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Homework Statement


For ##S_3## compute the centralizers of each element and find the center.

Homework Equations




The Attempt at a Solution


I found that ##Z (S_3) = \{ 1 \}##, ##C_{S_3} (1) = S_3##, ##C_{S_3} ((13)) = \{1, (13) \}##, ##C_{S_3} ((12)) = \{1, (12) \}##, ##C_{S_3} ((23)) = \{1, (23) \}##, ##C_{S_3} ((123)) = \{1, (123) \}##, ##C_{S_3} ((132)) = \{ 1, (132)\}##.

However, I found these by direct computation of each pair of elements to see if they commute. Would there be a faster way to do this? Would Lagrange's theorem help reduce the number of calculations I needed to do?
 

Answers and Replies

  • #2
12,690
9,235
The first line is more or less obvious, so there are only the 3-cycles which might have a different result. As they are inverse to each other, only one centralizer will actually have to be calculated: ##C_{S_3}(123)=\{\,1,(123),(123)^{-1}=(132)\,\}##, which suffices to show that ##(12) \notin C_{S_3}(123)##. So all in all, only $$
(12)(123)(12) \neq (123)
$$
has to be shown.

I understand, that you could say: "What about ##S_n\,?## Well, ##S_n## is a semidirect product. I haven't checked or in mind how centralizers behave with the operation of the subgroup on the normal subgroup, but this is where I would start. Centralizers doesn't seem to be normal, so that could by a problem. The center is easier, as it is a normal subgroup and the ##A_n## are simple for ##n > 4##.
 

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