# Am I computing the centralizer of $(1, 2, 3)$ in $A_4$ correctly?

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?

• NB: $$(123)=(231)=(312)$$ and $$(132)=(213)=(321).$$ Nov 24, 2021 at 17:14
• @Shaun So if I exclude the permutations (132), (213), and (321) from my set, then I will get the correct answer, right? Nov 24, 2021 at 17:17
• 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? Nov 24, 2021 at 17:18
• @FernandoTorres Your answer is correct - It's just that at times you've written the same element multiple times. Nov 24, 2021 at 17:19
• @Shagchi Got it, so the answer $\{(), (123), (132)\}$ would be the correct, simplified answer, right? Nov 24, 2021 at 17:22

$$(123)=(231)=(312)$$
$$(132)=(213)=(321).$$