# 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 $S_{3}$, $(1~{}2~{}3)$ and $(1~{}3~{}2)$ are conjugate, since

 $(2~{}3)(1~{}2~{}3)(2~{}3)=(1~{}3~{}2)$

but these two are not conjugate in $A_{3}$ (note that $(2~{}3)\notin A_{3}$).

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 $(1~{}2~{}3~{}4~{}5)$ are all conjugate in $A_{7}$, while the elements of the conjugacy class of $(1~{}2~{}3)(4~{}5~{}6)$ 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 $(1~{}2~{}3~{}4)(5~{}6)$, 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 $\mathcal{C}_{S}(\sigma)$ 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 $\mathcal{C}_{S}(\sigma)$ splits into $\lvert S_{n}:A_{n}C_{S_{n}}(\sigma)\rvert$ 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 $\lvert S_{n}:A_{n}\rvert$ 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 $\mathcal{C}_{S}(\sigma)$ are all conjugate in $A_{n}$ if and only if $A_{n}C_{S_{n}}(\sigma)=S_{n}$, which happens if and only if $C_{S_{n}}(\sigma)\nsubseteq A_{n}$, 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 $(a_{1}~{}a_{2}~{}\ldots~{}a_{2k+1})$ and $(b_{1}~{}b_{2}~{}\ldots~{}b_{2k+1})$, then

 $\displaystyle((a_{1}~{}b_{1})\ldots(a_{2k+1}~{}b_{2k+1}))(a_{1}~{}a_{2}~{}% \ldots~{}a_{2k+1})(b_{1}~{}b_{2}~{}\ldots~{}b_{2k+1})((a_{1}~{}b_{1})\ldots(a_% {2k+1}~{}b_{2k+1}))^{-1}=\\ \displaystyle(a_{1}~{}a_{2}~{}\ldots~{}a_{2k+1})(b_{1}~{}b_{2}~{}\ldots~{}b_{2% k+1})$

so the product of $(a_{1}~{}a_{2}~{}\ldots~{}a_{2k+1})$ and $(b_{1}~{}b_{2}~{}\ldots~{}b_{2k+1})$, 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 $A_{n}$ ConjugacyInAn 2013-03-22 17:18:04 2013-03-22 17:18:04 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 20M30