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Find two permutations that conjugate in $S_5$, but not in $A_5$.

I can't understand why is it possible - in order for two permutations to conjugate, they must have the same cycle structure.

If two permutations are conjugate in $S_5$, this means they have the same cycle structure, and therefore will have the same structure in $A_5$, and will be still conjugate in $A_5$...

What am I missing?

ChikChak
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  • See for example here. I think we have even better threads, but that came up first in my search. – Jyrki Lahtonen Aug 06 '18 at 05:21
  • What's happening is that the conjugacy class of $S_n$ may or may not split in two conjugacy classes in $A_n$. See here. – Jyrki Lahtonen Aug 06 '18 at 05:23
  • Here the corresponding exercise in Dummit & Foote is discussed. I won't pick a dupe target because my search is biased by my own history. But I suspect others explained this earlier and better and I simply can't find a link to those threads quickly. – Jyrki Lahtonen Aug 06 '18 at 05:26
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    Just for you to think concretely: Are the $3$-cycles $(1,2,3)$ and $(1,3,2)$ conjugate in $A_3$? – Ted Shifrin Aug 06 '18 at 05:36
  • @Ted Shifrin In general, I'm not sure how can I prove that something does not conjugate, rather than looking at the cycle structure or trying all possiblities, which obviously isn't a good idea. – ChikChak Aug 06 '18 at 05:48
  • You might think about orbits and stabilizers ... but for the $A_3$ question it should be immediate. – Ted Shifrin Aug 06 '18 at 05:49
  • @Jyrki Lahtonen In the first link you gave here, they show $(12345)$ and $(21345)$ as examples. But they don't show that for every $ \sigma \in A_5$ those two permutations don't conjugate, but only for specific permutations... – ChikChak Aug 06 '18 at 17:40
  • ChikChak, my favorite way of seeing that no permutation from $A_5$ will work is the following. There are $24$ 5-cycles in $S_5$. As $|S_5|=120$ we see that the centralizer of a 5-cycle has order $120/24=5$. But a 5-cycle is obviously centralized by its powers. So if $\alpha\in S_5$ is a 5-cycle, then $C_{S_5}(\alpha)=\langle \alpha\rangle$. Assume that $\beta$ is some other 5-cycle, and $\gamma\beta\gamma^{-1}=\alpha$, and $\delta\beta\delta^{-1}=\alpha$, – Jyrki Lahtonen Aug 06 '18 at 18:32
  • (cont'd) Then $\gamma\delta^{-1}$ commutes with $\beta$. Putting the pieces together it follows that $\gamma\delta^{-1}$ must be a power of $\beta$, say $\gamma=\beta^i\delta$. But as a 5-cycle $\beta^i$ is even. So if $\gamma$ is odd so is $\delta$. In other words, either all the permutations that take $\beta$ to $\alpha$ are even, or they are all odd. – Jyrki Lahtonen Aug 06 '18 at 18:37
  • @Jyrki Lahtonen So basically, every two $5$-cycle in $S_5$ cannot conjuagte in $A_5$? – ChikChak Aug 06 '18 at 19:08
  • No. I didn¨t say that! If you conjugate a 5-cycle by any even permutation you still get another 5-cycle. In fact $C_{A_5}(\alpha)=C_{S_5}(\alpha)\simeq C_5$. So the conjugacy class of $\alpha$ in $A_5$ has $60/5=12$ elements. In other words the class of 24 splits into two classes of 12. – Jyrki Lahtonen Aug 06 '18 at 19:38
  • I said that if some two 5-cycles are gotten from each other by conjugation, then the parity of that conjugating element is fixed. Because one of the conjugating permutations that takes $(12345)$ to $(21345)$ is odd, they all are. That's what the argument was trying to justify. – Jyrki Lahtonen Aug 06 '18 at 19:39
  • @Jyrki Lahtonen Why does the centralizer of $5$-cycle has order $120/24$? – ChikChak Aug 07 '18 at 05:21
  • Are you familiar with the orbit-stabilizer theorem? Or how the size of the centralizer determines the size of a conjugacy class? – Jyrki Lahtonen Aug 07 '18 at 05:23

2 Answers2

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As you said, being conjugate in $S_5$ is equivalent to having the same cycle structure. But this is not true in $A_5$. Two permutations that are conjugate in $A_5$ will have the same cycle structure, but the converse is not necessarily true. $x,y \in A_5$ and $y=gxg^{-1}$ where $g \in S_5 \backslash A_5$ then it implies $x$ and $y$ are conjugate in $S^5$ but they might not be conjugate in $A^5$.

Carl
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If $\alpha$ is a $5$-cycle in $S_5$ then the permutations in $S_5$ which commute with it are the powers of $\alpha$ which are all even. If $\beta$ is an odd permutation, and $\alpha'=\beta\alpha\beta^{-1}$ then $\alpha'$ is a conjugate of $\alpha$ in $S_5$ but not in $A_5$. For if $\alpha'=\beta'\alpha\beta'^{-1}$ for $\beta'\in A_5$ then $\beta^{-1}\beta'$ commutes with $\alpha$, so lies in $A_5$. That forces $\beta\in A_5$, a contradiction.

I like this geometric picture. $A_5$ is isomorphic to the rotation group of the regular icosahedron. That has a rotation of order $5$ with angle $2\pi/5$. Its square is a rotation of order $5$ with angle $4\pi/5$. These two rotations cannot be conjugate; these correspond to non-conjugate $5$-cycles in $A_5$.

Angina Seng
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