$(123),(213)$ not conjugate in $A_3$

Question

Let $n\geq 3$ be an odd number. Then I want to show that $(123\ldots n)$ and $(2134\ldots n)$ are not conjugate in $A_n$.

I know that they are conjugate in $S_n$ via $(12)$. If $M(\sigma,\rho):=\{\tau\in G\mid \tau\sigma\tau^{-1}=\rho\}$ (where $\sigma,\rho$ in an arbitrary group) I know that $|M(\sigma,\rho)|=|C(\sigma)|=|C(\rho)|$ where $C(\ldots)$ denotes the centralizer. For $\sigma=(123\ldots n)$ I further know that $C(\sigma)=<\sigma>$ in $G=S_n$.

Do you have any hints for me? Thanks in advance!

Answer 1

Suppose $\sigma\in S_n$ is such that $$\sigma^{-1}(1234\ldots n)\sigma = (2134\ldots n)$$ Then given the value of $\sigma(1)$ we can reconstruct the entire $\sigma$ by knowing what $\sigma$ must take the cycle into. So there are at most $n$ such $\sigma$s.

On the other hand, you know that $(12)$ is a possible $\sigma$, so $$ \sigma = (1234\ldots n)^k(12) $$ works for every $k=0,1,2,\ldots,n-1$. They are all different, so these are all the $\sigma$s that work. Since they are all odd, none of them are in $A_n$.

Answer 2

Here is a sketch of a proof. I'll let you fill out the details.

1.) Let $C_G(\pi)$ denote the centralizer of a permutation with respect to the group $G$, and $K_G(\pi)$ it's conjugacy class, again with respect to the group $G$. We know $K_{S_n}((1,2,\dots,n))$ is the set of $n$ cycles in $S_n$, of which there are $(n-1)!$. Hence by orbit stabilizer, $|C_{S_n}(1,2,\dots,n)|=n$. From this, conclude that $C_{S_n}(1,2,\dots,n)=\langle(1,2,\dots,n)\rangle$.

2.) From the above, conclude that $C_{A_n}(1,2,\dots,n)=\langle(1,2,\dots,n)\rangle$, and hence $K_{A_n}(1,2,\dots,n)$ contains exactly half of the $n$ cycles in $A_n$, again by orbit stabilizer.

3.) Show that any $n$ cycle is conjugate, in $A_n$, to at least one of $(1,2,\dots,n)$ or $(2,1,3,4,\dots,n)$, using the fact that these specific cycles are conjugate by $(1,2)$ and the fact that all $n$ cycles are conjugate in $S_n$ to any other $n$ cycle.

4.) Conclude $K_{A_n}(1,2,\dots,n)\cup K_{A_n}(2,1,3,\dots,n)$ consists of all the $n$ cycles.

5.) Conclude $(1,2,\dots,n)$ is not conjugate to $(2,1,3,\dots,n)$ in $A_n$.

Answer 3

If $\tau\sigma\tau^{-1}=\rho=(12)^{-1}\sigma(12)$ in $S_n$, then, multiplying by $\tau$ on the right and by $(12)$ on the left, you get that $(12)\tau$ commutes with $\sigma$. But you know that the centralizer of $\sigma$ is just the subgroup generated by $\sigma$ and therefore is a subgroup of $A_n$. So you get that $(12)\tau$ is an element $\pi$ of $A_n$. So if $\tau$ were an element of $A_n$, you'd get that $(12)=\pi\tau^{-1}$ is a product of two elements of $A_n$ and therefore is itself in $A_n$, which is clearly false. So $\tau$ cannot be in $A_n$.

What's really going on here is that $M(\sigma,\rho)$ is not only equinumerous with $C(\sigma)$ but is a coset of $C(\sigma)$ in $S_n$. Since $C(\sigma)=\langle\sigma\rangle$ in included in $A_n$, each of its cosets is either included in $A_n$ or disjoint from $A_n$. You know one member $(12)$ of $M(\sigma,\rho)$ that is outside $A_n$, so it follows that all members of $M(\sigma,\rho)$ are outside $A_n$.

Finally, let me mention that the special case in the title of your question is too special. $A_3$ is commutative, so no two distinct elements are conjugate in it.