Alternating group implies every element is automorphic to its inverse

Statement

Let $n$ be any natural number. Let $A_{n}$ denote the Alternating group (?) of degree $n$, i.e., the group of even permutations on a set of size $n$. Then, $A_{n}$ is a Group in which every element is automorphic to its inverse (?). In other words, for every element $g\in A_{n}$, there is an automorphism of $A_{n}$ that conjugates $g$ to $g^{-1}$.

Related facts

Related facts about alternating groups

• Alternating group implies any two elements generating the same cyclic subgroup are automorphic
• Classification of ambivalent alternating groups: $A_{n}$ is ambivalent for precisely the values $n=1,2,5,6,10,14$.
• Classification of alternating groups having a class-inverting automorphism: $A_{n}$ has a class-inverting automorphism for precisely the values $n=1,2,3,4,5,6,7,8,10,12,14$.

Related facts about every element being automorphic to its inverse

• General linear group implies every element is automorphic to its inverse
• Projective general linear group implies every element is automorphic to its inverse
• Special linear group implies every element is automorphic to its inverse
• Projective special linear group implies every element is automorphic to its inverse
• Every element is automorphic to its inverse is characteristic subgroup-closed
• Normal subgroup of ambivalent group implies every element is automorphic to its inverse

Facts used

1. Symmetric groups are rational, or cycle type determines conjugacy class

Proof

Consider the group $S_{n}$, in which $A_{n}$ is a normal subgroup of index two. Since $S_{n}$ is a rational group, the elements $g$ and $g^{-1}$ are conjugate. Thus, there exists $x\in S_{n}$ such that $xgx^{-1}=g^{-1}$.

Since $A_{n}$ is a normal subgroup of $S_{n}$, conjugation by $x$ restricts to an automorphism of $A_{n}$.