Classification of ambivalent alternating groups

Statement

The alternating group $A_{n}$ is an ambivalent group for precisely the following choices of $n$: $n=1,2,5,6,10,14$.

Facts used

• Splitting criterion for conjugacy classes in the alternating group

Proof

Overall plan

If $g\in A_{n}$ has the property that its conjugacy class in $S_{n}$ does not split in $A_{n}$, then $g$ is conjugate to $g^{-1}$ in $A_{n}$ (because they're conjugate in $S_{n}$). Thus, it suffices to check whether every element whose conjugacy class does split inside $A_{n}$, is conjugate to its inverse.

By the splitting criterion for conjugacy classes, it suffices to look at those even permutations that arise as products of cycles of distinct odd lengths. Further, in order to determine whether such a product of cycles is conjugate to its inverse in $A_{n}$, it suffices to find one permutation that conjugates this cycle to its inverse. If that one permutation is even, then the element is conjugate to its inverse. If that one permutation is odd, then the element is not conjugate to its inverse in $A_{n}$.

Criterion for determining whether an element is conjugate to its inverse

For a cycle of odd length $(a_{1},a_{2},\dots ,a_{r})$, the product of $(r-1)/2$ transpositions $(a_{j},a_{r+2-j})$ conjugates this cycle to its inverse. Thus, if a permutation is a product of cycles of odd lengths $r_{1},r_{2},\dots ,r_{k}$, then there is a product of $\sum (r_{i}-1)/2$ transpositions that conjugates this to its inverse.

The upshot: a product of cycles of distinct odd lengths $r_{1},r_{2},\dots ,r_{k}$ is conjugate to its inverse if and only if $\sum (r_{i}-1)/2$ is even. Equivalently, it is conjugate to its inverse if and only if the number of $r_{i}$s that are congruent to $3$ modulo $4$ is even.

What this boils down to for $n$

Thus, the problem reduces to the following: for what $n$ can we write $n=\sum _{i=1}^{k}r_{i}$ in such a way that all $r_{i}$ are distinct, and the number of $r_{i}$ that are congruent to $3$ modulo $4$ is odd? These are precisely the $n$ for which $A_{n}$ is not ambivalent.

We quickly see the following:

• $n=4d+3$ can be written in this form, because we can take $k=1,r_{1}=4d+3$.
• $n=4d+4$ can be written in this form, because we can take $k=2,r_{1}=4d+3,r_{2}=1$.
• $n=4d+9$ can be written in this form, because we can take $k=3,r_{1}=1,r_{2}=5,r_{3}=4d+3$.
• $n=4d+18$ can be written in this form, because we can take $k=4,r_{1}=1,r_{2}=5,r_{3}=9,r_{4}=4d+3$.

The only cases left are $n=1,2,5,6,10,14$, and it is readily seen that a decomposition into $r_{i}$ of the above form is not possible for these $n$.