Classification of ambivalent alternating groups

Statement

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

Note that in the proof, we show that in each of these cases, for every element in the alternating group, there is an element of order two in the alternating group conjugating it to its inverse. Thus, we show that the set of $n$ for which $A_{n}$ is strongly ambivalent is precisely the same: $n=1,2,5,6,10,14$.

Related facts

Related facts about alternating groups

• Classification of alternating groups having a class-inverting automorphism: This turns out to be $n=1,2,3,4,5,6,7,8,10,12,14$.
• Alternating group implies every element is automorphic to its inverse
• Finitary alternating group on infinite set is ambivalent

Related facts about symmetric groups

• Symmetric groups are rational
• Symmetric groups are rational-representation
• Symmetric groups are ambivalent

General information pages

• Linear representation theory of symmetric groups
• Linear representation theory of alternating groups

Facts used

1. Criterion for element of alternating group to be real

Proof

By fact (1), 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. Note also that if it is conjugate to its inverse, we can choose as our conjugating element an element of order two: the product of transpositions described above.

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 $n$.