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Could you tell me how to show that $(123)$ and $(132)$ are not in the same conjugacy class in $A_4$?

I know that all 3-cycles can't be in the same class, because the order of each class must divide $|A_4| = 12$ and there are eight 3-cycles in $A_4$.

If $\sigma$ is a 3-cycle fixing $n$ and $τ \in A_4$ then $\tau \circ \sigma \circ \tau^{-1}$ is a 3-cycle fixing $τ(n)$. (And there are 3 such permutations). If we conjugate $(123)$ by $(132)$ we will get $(123)$ and vice versa. So it would appear that it only makes sense to conjugate $(123)$ by something with a $4$. If we conjugated $(123)$ by $(124)$, we would need $(124)(123)(124)^{-1}$ to send 4 to 4, but it doesn't. It turns out that such $\tau$ cannot have anything "common" with $(123)$ - what I mean by this is that there can't be $(12...), (23...), (31...)$ in $\tau$. So the only cycles that remain are $(134), (142), (243)$ - and they are indeed ion the same class as $(123)$.

As you see, this isn't the best way to solve this problem.

I also know that if $g \in A_n$ commutes with an odd permutation, then all permutations with the same cycle type as $g$ are in one conjugacy class, but I have no idea how to show that in $A_4$ no odd permutation commutes with $(123)$.

Could you make it more mature?

Thank you.

Bilbo
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    You can fairly easily see what a permutation must look like in order to conjugate $(123)$ to $(132)$, and then check that none of the possibilities are even permutations. – Tobias Kildetoft Oct 06 '13 at 10:11
  • Well. I know that such a permutation must stabilize one of $1,2,3$ and of course $4$, because $\sigma (123) \sigma^{-1} = (\sigma(1) \sigma (2) \sigma (3)) = (132)$. Do you think it's enough? – Bilbo Oct 06 '13 at 10:18
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    Since once you note that it must fix $2$ elements (and the identity does not work), you can conclude that it must be a transposition, yes, that is enough. – Tobias Kildetoft Oct 06 '13 at 10:20
  • Ok, thank you a lot :) – Bilbo Oct 06 '13 at 10:21

2 Answers2

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One way to see this is to work in $S_4$ instead. There $(1\,2\,3)$ and $(1\,3\,2)$ are conjugates, of course, since $(2\,3)^{-1}(1\,2\,3)(2\,3)=(1\,3\,2)$.

If we have a $\tau$ such that $\tau^{-1}(1\,2\,3)\tau=(1\,3\,2)$, it must factor as $\tau=\rho(2\,3)$ for some $\rho$, and we have $$(2\,3)^{-1}\rho^{-1}(1\,2\,3)\rho(2\,3)=(1\,3\,2)=(2\,3)^{-1}(1\,2\,3)(2\,3)$$ Canceling the $(2\,3)$s we find $\rho^{-1}(1\,2\,3)\rho=(1\,2\,3)$, and it's then easy to see that $\rho(1)$ determines $\rho$ and there are only three possibilities, which are all even permutations. So $\tau$ must be odd and so not in $A_4$.

This may or may not be easier to follow than Tobias' suggestion to brute-force the possible $\tau$s directly. Personally I think it's easier to think about possible stabilizers of a cycle than to keep things straight while trying to conjugate one cycle into another.

hmakholm left over Monica
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  • I've already accepted your answer, but now I see that I don't quite understand why $\tau = \rho (23)$. Could you explain that? Does it simply follow from from $(2,3)^{-1}(1,2,3)(2,3)=(1,3,2)$? – Bilbo Oct 06 '13 at 10:26
  • @Bilbo: In a group, everything factors through everything else. Whenever you have a $\tau$ you can just use $\tau(2,3)^{-1}$ as $\rho$. – hmakholm left over Monica Oct 06 '13 at 10:27
  • All right. Thank you. – Bilbo Oct 06 '13 at 10:29
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For an easy argument, you can use that the size of the conjugacy class $C_G(g)$ of $g\in G$ is $|G|/|Z_G(g)|$ where $Z_G(g)=\{\,h\in G\mid gh=hg\,\}$ is the centraliser of$~g$ (this is a special case of the orbit-stabiliser theorem). The centraliser contains at least the powers of $g=(123)$, which here gives $|Z_{A_4}(g)|\geq3$ and therefore $|C(g)|\leq12/3=4$. So as soon as we have $4$ conjugates of $g$ there cannot be more of them; one easily sees that $\{(123),(134),(142),(243)\}\subseteq C_{A_4}(g)$, and hence this must be all.

Marc van Leeuwen
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