# Centralizers and Center for S_3

## Homework Statement

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

## 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?

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fresh_42
Mentor
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 $n > 4$.

• Mr Davis 97