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From my understanding, the centralizer of a permutation $p$ can be computed by including the identity permutation $()$ and then finding all the equivalent ways to represent $p$ (which can be done by rearranging the order of the disjoint cycles and rearranging the elements within the disjoint cycles).

So what I get that the centralizer of $(1, 2, 3)$ is the set:

$S = \{(), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 2, 1), (3, 1, 2)\}$

Is that correct?

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    $\begingroup$ NB: $$(123)=(231)=(312)$$ and $$(132)=(213)=(321).$$ $\endgroup$
    – Shaun
    Nov 24, 2021 at 17:14
  • $\begingroup$ @Shaun So if I exclude the permutations (132), (213), and (321) from my set, then I will get the correct answer, right? $\endgroup$
    – Fernando Torres
    Nov 24, 2021 at 17:17
  • $\begingroup$ The computed centralizer is correct but I'm not sure if I understand how you arrive at it - as in whether you're using the right result to deduce $S$. To be precise, what constraints do you place on $p$ and the underlying group? $\endgroup$
    – RichoddAsscraft
    Nov 24, 2021 at 17:18
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    $\begingroup$ @FernandoTorres Your answer is correct - It's just that at times you've written the same element multiple times. $\endgroup$
    – RichoddAsscraft
    Nov 24, 2021 at 17:19
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    $\begingroup$ @Shagchi Got it, so the answer $\{(), (123), (132)\}$ would be the correct, simplified answer, right? $\endgroup$
    – Fernando Torres
    Nov 24, 2021 at 17:22

1 Answer 1

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Sort of.

First of all, as I said in the comment above,

$$(123)=(231)=(312)$$

and

$$(132)=(213)=(321).$$

To find the centraliser of a permutation, there is a nice result: conjugation by permutations preserves cyclic structure.

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