I have given the following problem to solve;

(i) Prove any two cycles in Sn of the same length are conjugate in $S_n$ for any $n\geq 3$.

(ii) Is the same true in $A_n$ for $n\geq 3$?

(iii) Prove any two cycles of length three are conjugate in $A_n$ if $n\ge 5$.

I have answered the first part of the question accordingly;

If two elements are conjugate, they satisfy the following definition; $\sigma_2 = \tau^{-1}\sigma_1\tau$ for some $\tau \in S_n$. And as $\sigma_1, \sigma_2$ are of equal length, we can express them thusly;

$(b_1 b_2 b_3...b_k) = \tau^{-1} (a_1 a_2 a_3...a_k) \tau$

$\implies (b_1 b_2 b_3...b_k) = (\tau^{-1} a_1 \tau)(\tau^{-1} a_2 \tau)...(\tau^{-1} a_k \tau)$


$b_1=\tau^{-1} a_1 \tau, b_2=\tau^{-1} a_2 \tau...b_k=\tau^{-1} a_k \tau$

Showing any two cycles in $S_n$ are conjugate for some $\tau$.

Now for the second part, I thought I could use the argument that $A_n \subset S_n$ to say that the same would be true for $A_n$. But I spoke to a few people answering the same question and seemed to believe the statement was false. Could some one shed some light on the problem please?

And also, if possible would someone be able to give some assistance to the final part of this question as I do not really know who to approach this, due to the similarities to the previous part of the question.

Thank you for your time.

  • 1
    $\begingroup$ A simple counting argument will settle part ii). In $S_5$ there are $24$ 5-cycles - all of them also elements of $A_5$. But the size of a conjugacy class is always a factor of the order of the group, so they cannot form a single conjugacy class in $A_5$ as $24\nmid 60$. See Andreas Caranti's answer for an extension of this reasoning. Actually Tim Raczkowski's example can be explained in this way as well. In $S_4$ there are $8$ 3-cycles. As $8\nmid 12=|A_4|$ they cannot form a single conjugcay class of $A_4$. $\endgroup$ – Jyrki Lahtonen Apr 29 '15 at 14:09

The problem is that you need to find $\tau\in A_n$ when your construction only guarantees $\tau\in S_n$.

Start with $n\ge 5$. It suffices to show that any $3$-cycle $(a\,b\,c)$ is conjugate to $(1\,2\,3)$. We yould like to let $\tau(1)=a$, $\tau(2)=b$, $\tau(3)=c$ and otherwise $\tau$ is a bijecting between $\{4,\ldots, n\}$ and $\{1,\ldots,n\}\setminus\{a,b,v\}$. If we can pick $\tau$ like this, then indeed $\tau(1\,2\,3)\tau^{-1}=(a\,b\,c)$. However, it can happen that our first pick produces a $\tau\notin A_n$. Fortunately, if we replace $\tau$ with $\tau\circ (4\,5)$ then the latter is in $A_n$.

We needed $4$ and $5$ to play with, but we do not have them available for $A_4$. And indeed in $A_4$, the $3$-cycles $(1\,2\,3)$ and its inverse $(1\,3\,2)$ are not conjugate: We must have $\tau(4)=4$ because $\tau$ can map fixpoints only to fixpoints. So we may view $\tau$ (as well as $(1\,2\,3)$ and $(1\,3\,2)$) as element of $A_3$. As $A_3$ is abelian, two different elements are not conjugate.


$\newcommand{\Size}[1]{\lvert #1 \rvert}$First, a general fact. Suppose $H$ is a subgroup of index $2$ of the group $G$. Let $G$ act on a set $X$. Let $x \in X$. Then for the orbit $x H$ of $x$ under $H$, and writing $G_{x}, H_{x}$ for the stabilizers, we have $$ \Size{x H} = \Size{H : H_{x}} = \Size{H : G_{x} \cap H} = \Size{H G_{x} : G_{x}} = \frac{\Size{G : G_{x}}}{\Size{G : H G_{x}}} = \begin{cases} \Size{x G} & \text{if $G_{x} \not\le H$,}\\ \dfrac{1}{2} \Size{x G} & \text{if $G_{x} \le H$.}\\ \end{cases}. $$

So (ii) is a matter of checking whether the centralizer of a $k$-cycle $\alpha$ ($k$ odd) contains odd permutations or not. This is definitely the case if $k \le n - 2$ (as there will be a $2$-cycle centralizing $\alpha$), and not the case when $k = n -1, n$ (as the centralizer of $\alpha$ is $\langle \alpha \rangle$).


For the second question, consider $A_4$ and the cycles $[123]$ and $[124]$. It is true that $[124]=[34][123][34]^{-1}$, but there is no even permutation $\tau$ for which $[124]=\tau[123]\tau^{-1}$.


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