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 $A_n$.