# Find $\tau \in S_5$ that satisfices $\sigma \tau = \tau \sigma$ but $\tau \notin A_5$ for $\sigma = (1\;2\;3\;4\;5)$ [duplicate]

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 $$\tau \notin A_5$$ ?

• Keep it (very) simple and think of single transpositions? Commented May 26, 2023 at 14:25
• @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 Commented May 26, 2023 at 14:28
• 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? Commented May 26, 2023 at 14:35
• you need to specify, with respect to which $\sigma$ do you want to find the centralizer! Commented May 26, 2023 at 14:36
• @MathFail of course that is my fault of a typo. Thank you for pulling my attention! Commented May 26, 2023 at 14:39

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 $$A_5$$.

• Yes, see also this post. It shows that the centralizer is exactly the subgroup $\langle \sigma\rangle=\{1,\sigma,\sigma^2,\sigma^3,\sigma^4\}$. Commented May 26, 2023 at 14:56
• yes, it is also mentioned in this post. Commented May 26, 2023 at 14:59
• It is even proved there (by the orbit-stabilizer theorem, as you did). Commented May 26, 2023 at 14:59
• @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 Commented May 26, 2023 at 15:11
• 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 Commented May 26, 2023 at 15:16