conjugacy in $A_n$

Recall that conjugacy classes in the symmetric group $S_n$ are determined solely by cycle type. In the alternating group $A_n$, however, this is not always true. A single conjugacy class in $S_n$ that is contained in $A_n$ may split into two distinct classes when considered as a subset of $A_n$. For example, in $S_3$, $(123)$ and $(132)$ are conjugate, since

$$(23)(123)(23)=(132)$$

but these two are not conjugate in $A_3$ (note that $(23)\notin A_3$).

Note in particular that the fact that conjugacy in $S_n$ is determined by cycle type means that if $\sigma \in A_n$ then all of its conjugates in $S_n$ also lie in $A_n$.

The following theorem fully characterizes the behavior of conjugacy classes in $A_n$:

Theorem 1.

A conjugacy class in $S_n$ splits into two distinct conjugacy classes under the action of $A_n$ if and only if its cycle type consists of distinct odd integers. Otherwise, it remains a single conjugacy class in $A_n$.

Thus, for example, in $S_7$, the elements of the conjugacy class of $(12345)$ are all conjugate in $A_7$, while the elements of the conjugacy class of $(123)(456)$ split into two distinct conjugacy classes in $A_7$ since there are two cycles of length $3$. Similarly, any conjugacy class containing an even-length cycle, such as $(1234)(56)$, splits in $A_7$.

We will prove the above theorem by proving the following statements:

• A conjugacy class in $S_n$ consisting solely of even permutations (i.e. that is contained in $A_n$) either is a single conjugacy class or is the disjoint union of two equal-sized conjugacy classes when considered under the action of $A_n$.

• If $\sigma \in A_n$, then the elements of the conjugacy class of $\sigma$ in $S_n$ (which is just all elements of the same cycle type as $\sigma$) are conjugate in $A_n$ if and only if $\sigma$ commutes with some odd permutation.

• $\sigma \in S_n$ does not commute with an odd permutation if and only if the cycle type of $\sigma$ consists of distinct odd integers.

Throughout, we will denote by ${𝒞}_S(\sigma )$ the conjugacy class of $\sigma$ under the action of $S_n$.

To prove the first statement, note that conjugacy is a transitive action. By the theorem that orbits of a normal subgroup are equal in size when the full group acts transitively, we see that if $\sigma \in A_n$, then ${𝒞}_S(\sigma )$ splits into $|S_n:A_n{C}_{S_n}(\sigma )|$ classes under the action of $A_n$ (recall that $C_G(x)$, the centralizer of $x$, is simply the stabilizer of $x$ under the conjugation action of $G$ on itself). But since $|S_n:A_n|$ is either $1$ or $2$, we see that the conjugacy class of $\sigma$ either remains a single class in $A_n$ or splits into two classes.

Note also that the elements of ${𝒞}_S(\sigma )$ are all conjugate in $A_n$ if and only if $A_n{C}_{S_n}(\sigma )=S_n$, which happens if and only if $C_{S_n}(\sigma )⊈A_n$, which in turn is the case if and only if some odd permutation is in the centralizer of $\sigma$, which means precisely that $\sigma$ commutes with some odd permutation. This proves the second statement.

To prove the third statement, suppose first that $\sigma$ does not commute with an odd permutation. Clearly $\sigma$ commutes with any cycle in its own cycle decomposition, so if $\sigma$ contains a cycle of even length, that is an odd permutation with which $\sigma$ commutes. So $\sigma$ must consist solely of [disjoint] cycles of odd length. If two of these cycles have the same length, say $(a_1{a}_2\mathrm{\dots }{a}_{2k+1})$ and $(b_1{b}_2\mathrm{\dots }{b}_{2k+1})$, then

$$\begin{array}{c}((a_1{b}_1)\mathrm{\dots }(a_{2k+1}{b}_{2k+1}))(a_1{a}_2\mathrm{\dots }{a}_{2k+1})(b_1{b}_2\mathrm{\dots }{b}_{2k+1}){((a_1{b}_1)\mathrm{\dots }(a_{2k+1}{b}_{2k+1}))}^{-1}=\hfill \\ \hfill (a_1{a}_2\mathrm{\dots }{a}_{2k+1})(b_1{b}_2\mathrm{\dots }{b}_{2k+1})\end{array}$$

so the product of $(a_1{a}_2\mathrm{\dots }{a}_{2k+1})$ and $(b_1{b}_2\mathrm{\dots }{b}_{2k+1})$, and thus $\sigma$, commutes with the product of $2k+1$ transpositions, which is an odd permutation. Thus all the cycles in the cycle decomposition of $\sigma$ must have different [odd] lengths.

To prove the converse, we show that if the cycles in the cycle decomposition all have distinct lengths, then $\sigma$ commutes precisely with the group generated by its cycles. It follows then that if all the distinct lengths are odd, then $\sigma$ commutes only with these permutations, which are all even. Choose $\sigma$ with distinct cycle lengths in its cycle decomposition, and suppose that $\sigma$ commutes with some element $\tau \in S_n$. Conjugation preserves cycle length, so since $\tau$ commutes with $\sigma$ and $\sigma$ has all its cycles of distinct lengths, each cycle in $\tau$ must commute with each cycle in $\sigma$ individually.

Now, choose a nontrivial cycle ${\tau }_1$ of $\tau$, and choose $j\in \tau$ such that $\sigma$ moves $j$ (we can do this, since $\sigma$ can have at most one cycle of length $1$ and the cycle length of $\tau$ is greater than $1$). Let ${\sigma }_1$ be the cycle of $\sigma$ containing $j$. Then ${\tau }_1$ commutes with ${\sigma }_1$ since $\tau$ commutes with $\sigma$, so ${\tau }_1$ is in the centralizer of ${\sigma }_1$, and it is not disjoint from ${\sigma }_1$. But the centralizer of a $k$-cycle $\rho$ consists of products of powers of $\rho$ and cycles disjoint from $\rho$. Thus ${\tau }_1$ is a power of ${\sigma }_1$. So each cycle in $\tau$ is a power of a cycle in $\sigma$, and we are done.

Title	conjugacy in $A_n$
Canonical name	ConjugacyInAn
Date of creation	2013-03-22 17:18:04
Last modified on	2013-03-22 17:18:04
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 20M30