1

The symmetric group of order $720=6!$, $S_6$, is the only symmetric group with an outer automorphism. This suggests the following:

Question: Is $A_6$ characteristic in $S_6$, where $A_6$ is the alternating group of order $360$?

Clearly, $A_6$ is normal in $S_6$, so if it is not characteristic, then some outer automorphism of $S_6$ must map some even permutation to an odd permutation.

If $n \ne 6$, then since all automorphisms of $S_n$ are inner, $A_n$ is characteristic in $S_n$.

Geoffrey Trang
  • 9,959
  • 4
    $A_6$ is the only subgroup of index $2$ in $S_6$, so perforce is characteristic. Also, the puter automorphisms of $S_6$ send transpositions to products of three transpositions, so the respect parity. – Arturo Magidin Feb 18 '22 at 15:11
  • 3
    Here's a different argument to what Arturo suggests, based on the simplicity of $A_6$. Let $\phi$ be an automorphism of $S_6$. Then $\phi(A_6)$ is a normal subgroup of $S_6$, and hence $\phi(A_6) \cap A_6$ is normal in $A_6$. Since $A_6$ is simple, this intersection is either $1$ or $A_6$, and it cannot be $1$ since $A_6$ and $\phi(A_6)$ are both index $2$ in $S_6$. Evidently this argument is more general: it works for any index $2$ simple subgroup of a group, except $\mathbf{Z}/2$. – Stephen Feb 18 '22 at 15:15

0 Answers0