Number of Automorphisms of $S_6$

Question

Trivial question about counting the number of automorphisms of $S_6$:

I know that for all $n \geq 3$, $Z(S_n)=1$, so Inn$(S_n) \cong S_n$.

I also know that $S_6$ has nontrivial outer homomorphisms, Out$S_n \cong \mathbb{Z}_2$.

Does this mean there are $S_n + \mathbb{Z}_2$ automorphisms in total, since Inn$(S_n)$ and Out$(S_n)$ are disjoint sets? So, other symmetric groups (excluding $S_2$) have $n!$ elements, while $S_6$ has $6! + 2$.

Answer 1

We have $\operatorname{Out}(S_6)=\operatorname{Aut}(S_6)/\operatorname{Inn}(S_6)\cong C_2$ and hence $$ |\operatorname{Aut}(S_6)|=|S_6|\cdot |C_2|=6!\cdot 2. $$ Here we have used that $|G/N|=\frac{|G|}{|N|}$.

Answer 2

$\mathrm{Out} (S_n) $ is not a group of automorphisms, but is the quotient of the automorphism group by the subgroup of inner automorphisms. Thus $$|\mathrm{Out}(S_n) |=|\mathrm{Aut} (S_n) |/|S_n|$$ so you can find the order of the automorphism group this way.

Essentially an outer automorphism is an equivalence class of automorphisms $f(x) $ where $f(x) \sim \sigma f(x) \sigma^{-1}$ for any $\sigma$.

Answer 3

The group $S_6$ has an outer automorphism that is unique up to composition with an inner automorphism. Hence, the group $S_6$ has 1440 automorphisms.