The Outer Automorphism of $S_6$

These are notes about the fact that $S_n$ is complete for all $n\ne 2,6$. Another title for these notes might be "How the Number 6 Misbehaves in a Natural Way".

Innies and Outies

Given a finite group $G$, it is natural to look at the group of its automorphisms $\text{Aut}(G)$, i.e. the group (under composition) of bijective homomorphisms from $G$ to itself. These automorphisms can be classified into two categories: inner and outer. An inner automorphism $\varphi :G\to G$ is an automorphism defined by conjugation of a fixed element $g\in G$. That is, $\varphi$ can be expressed as $$\begin{array}{rl}\varphi :G& \to G\\ \gamma & \mapsto g\gamma g^{-1}\end{array}$$ for some fixed $g\in G$. In this case, we write $\varphi$ as ${\varphi }_g$. It's easy to check that the collection of all inner automorphisms $\text{Inn}(G)$ of $G$ forms a normal subgroup of $\text{Aut}(G)$.

Define the group of outer automorphisms of $G$ by $\text{Out}(G):=\text{Aut}(G)/\text{Inn}(G)$. By somewhat abuse of notation, an element of $\text{Out}(G)$ is called an outer automorphism — really, these are classes of automorphisms. One can think of inner and outer automorphisms by the information they define: an inner automorphism is defined with information from within the group, whereas an outer automorphism is defined by some information coming from outside the group.

Already it is worth taking a quick moment for philosophy: to know the full symmetries of a group, one needs to have information from the inside and the outside. It would be nice, both for simplicity and computation, if no automorphisms of $G$ were outer. In this case, everything we might want to know about $G$, at least in terms of its symmetry, can be found within the group itself. This will motivate our definition of completeness.

The following lemma follows quickly from definitions:

Lemma 1. For a finite group $G$, $\text{Inn(G)}\cong G/Z(G)$.

Proof. Consider the map $$\begin{array}{rl}\mathrm{\Phi }:G& \to \text{Aut}(G)\\ g& \mapsto {\varphi }_g.\end{array}$$ By definition, the image of $\mathrm{\Phi }$ is $\text{Inn}(G)$. Note that $\mathrm{\Phi }$ is not, in general, injective. Suppose $g\in \ker \mathrm{\Phi }$. Then ${\varphi }_g=\text{id}$, i.e. $g\gamma g^{-1}=\gamma$ for all $\gamma \in G$. This is exactly to say that $g\in Z(G)$. In the other direction, any $g\in Z(G)$ yields ${\varphi }_g=\text{id}$. So $\ker \mathrm{\Phi }=Z(G)$.
By the first isomorphism theorem, $\text{Inn}(G)\cong G/Z(G)$.

Using the notation above, this can be summarized in the following exact sequence: $$\begin{array}{r}Z(G)\hookrightarrow G\stackrel{\mathrm{\Phi }}{\to }\text{Aut}(G)\twoheadrightarrow \text{Out}(G).\end{array}$$ Thus the kernel of $\mathrm{\Phi }$ is $Z(G)$ and the cokernel of $\mathrm{\Phi }$ is $\text{Out}(G)$. In this sense, the center of $G$ is dual to its group of outer automorphisms.

Complete Groups

Since it is generally hard to find a group's automorphism group, our best bet might be to just hope that it is something easy...if we're lucky, maybe it's trivial! This would be good for computation, but ends up being too simple: if $G$ is finite and $\text{Aut}(G)\cong 1$, then $G\cong 1$ or ${\mathbb{Z}}_2$.

Instead, we might want that $\text{Aut}(G)\cong G$. But observe that this still might get complicated — the isomorphism might not be obvious. Consider the dihedral group $D_8=⟨s,r|s^2=r^4=1,s{r}^{-1}=rs⟩$. Its center is $⟨r^2⟩$, and so the group of inner automorphisms has four elements. (...something here about how to see this automorhpism group...) But its outer automorphism group (think about the automorphisms of ${\mathbb{Z}}_4$) is isomorphic to ${\mathbb{Z}}_2$. It turns out that the whole automorphism group is isomorphic to $D_8$.

Then it's clear that the best case scenario is when $\text{Aut}(G)$ is naturally isomorphic to $G$, i.e. when $\mathrm{\Phi }$ is an isomorphism. In other words, $G$ has trivial center and trivial outer automorphism group. In this case, we say $G$ is complete. The terminology here makes sense — if $G$ is complete, then we know exactly what its automorphism group is, and, in some sense, the group is totally self-contained.

UNDER CONSTRUCTION