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I'm having a really hard time understanding the way $A_n$ the alternating group behaves. I understand it means that $\forall \sigma \in A_n$ we have $sgn(\sigma) = 1$ but I'm not sure I understand the way the sgn function works.

Anyway the hint in the question is to look for some $\lambda \in S_5$ that satisfies $\lambda \sigma \lambda^{-1} = \tau \iff \lambda \notin A_5$

So first of all I know I need to find $\tau \in S_5$ such that $|\tau| = 5$ meaning it's cycle is the length of 5 because that is the only way to have $\tau$ conjugated to $\sigma$ but how can I make sure $\tau \notin A_5$ ?

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  • $\begingroup$ Keep it (very) simple and think of single transpositions? $\endgroup$ Commented May 26, 2023 at 14:25
  • $\begingroup$ @prets I must say I'm super unsure of how to work with this question especially because i don't seem to understand the definition of An $\endgroup$ Commented May 26, 2023 at 14:28
  • $\begingroup$ I don't understand the definition of $A_n$ given in your question either---how is $\mathrm{sgn}(\sigma)$ defined that it can take values other than $+1$ or $-1$, and how is the number of "intersections" of $\sigma$ when written in two-line form supposed to inform whether the permutation is even or not? $\endgroup$ Commented May 26, 2023 at 14:35
  • $\begingroup$ you need to specify, with respect to which $\sigma$ do you want to find the centralizer! $\endgroup$
    – MathFail
    Commented May 26, 2023 at 14:36
  • $\begingroup$ @MathFail of course that is my fault of a typo. Thank you for pulling my attention! $\endgroup$ Commented May 26, 2023 at 14:39

1 Answer 1

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First, we use Orbit-Stabilizer theorem in $S_5$,

$$|C_{S_5}(12345)|=\frac{|S_5|}{|cl_{S_5}(12345)|}=\frac{5!}{|cl_{S_5}(12345)|}$$

where $|cl_{S_5}(12345)|$ is the number of elements in the conjugacy class of $(12345)$ in $S_5$, we know:

$$|cl_{S_5}(12345)|=\frac{5!}{5^1\cdot 1!}=4!$$

Plug in and we get:

$$|C_{S_5}(12345)|=5$$

Next, we move to $A_5$. Since $cl_{S_5}(12345)$ splits in $A_5$, we get

$$|cl_{A_5}(12345)|=\frac{1}2|cl_{S_5}(12345)|=12$$

Again, we use Orbit-Stabilizer Theorem,

$$|C_{A_5}(12345)|=\frac{|A_5|}{|cl_{A_5}(12345)|}=\frac{5!/2}{12}=5$$

So all centralizers belong to $A_5$, and the answer is zero.

Remarks:

Actually, all the five centralizers form the cyclic group $\langle(12345)\rangle$, and they are all even permutations, hence all belong to $A_5$.

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  • $\begingroup$ Yes, see also this post. It shows that the centralizer is exactly the subgroup $\langle \sigma\rangle=\{1,\sigma,\sigma^2,\sigma^3,\sigma^4\}$. $\endgroup$ Commented May 26, 2023 at 14:56
  • $\begingroup$ yes, it is also mentioned in this post. $\endgroup$
    – MathFail
    Commented May 26, 2023 at 14:59
  • $\begingroup$ It is even proved there (by the orbit-stabilizer theorem, as you did). $\endgroup$ Commented May 26, 2023 at 14:59
  • $\begingroup$ @MathFail How come we compute $C_{S_5}(1\;2\;3\;4\;5)$ and then we plug in and get the answer for $C_{S_5}(1\;2)(3\;4)$. Maybe it is obvious but I think i am missing this transition $\endgroup$ Commented May 26, 2023 at 15:11
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    $\begingroup$ I also answered in this post for more details, and you can take a look how to find exact elements of centralizer. math.stackexchange.com/questions/366639/… @AsiMathStudent $\endgroup$
    – MathFail
    Commented May 26, 2023 at 15:16

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