outer automorphisms of $S_6$

$$\begin{array}{|l|c|c|} \hline \text{cycle structure} & \text{number of permutations} & \text{order} \\ \hline 6 & 120 & 6 \\ 5+1 & 144 & 5 \\ 4+2 & 90 & 4 \\ 4+1+1 & 90 & 4 \\ 3+3 & 40 & 3 \\ 3 + 2 + 1 & 120 & 6 \\ 3 + 1 + 1 + 1 & 40 & 3 \\ 2 + 2 + 2 & 15 & 2 \\ 2 + 2 + 1 + 1 & 45 & 2 \\ 2 + 1 + 1 + 1 + 1 & 15 & 2 \\ 1 + 1 + 1 + 1 + 1 + 1 & 1 & 1 \\ \hline \end{array}$$

In trying to understand outer automorphisms of $S_6$ at the most concrete level of arithmetic, it appears to me that the $15$ generators whose cycle structure is $2+1+1+1+1$ get mapped to the $15$ permutations whose structure is $2+2+2$ (which also generate the group). And the $40$ permutations whose structure is $3+1+1+1$ similarly get interchanged with the $40$ whose structure is $3+3$; likewise the $120$ whose structure is $3+2+1$ with the $120$ whose structure is $6$.

But apparently the set of $90$ whose structure is $4+2$ is invariant, as is the set of $90$ whose structure is $4+1+1$.

• Does some web page somewhere walk through this? I would think the routine arithmetic would have been figured out.
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I take it you've seen en.wikipedia.org/wiki/… already. Maybe polyomino.f2s.com/david/haskell/outers6.html or math.ucr.edu/home/baez/six.html ? –  mt_ Mar 22 '12 at 20:26
Perhaps groupprops.subwiki.org/wiki/S6#Up_to_automorphism this is of some help? –  MJD Mar 22 '12 at 20:42
I don't know that I've seen the one from John Baez before. He quotes Chris Henrich: "a permutation of 123456 induces a permutation of ABCDEF. A two-cycle such as (12) induces a product of three disjoint two-cycles such as (AB)(CD)(EF)". So the set of six things getting permuted is different in the two cases. It had irritated me that there seems to be no natural choice of correspondence between the $(2+1+1+1+1)$-structured permutations and the $(2+2+2)$-structured ones. This takes that into account rather neatly. –  Michael Hardy Mar 22 '12 at 22:34
BTW, it is clear that $4+2$ and $4+1+1$ are preserved, because the first is in $A_6$ and the second is not. –  Derek Holt Mar 23 '12 at 13:58
@Derek: I think it's worth indicating why $A_6$ is invariant under outer automorphisms (namely it's the commutator subgroup). –  Qiaochu Yuan Mar 23 '12 at 18:21

I think the best source for understanding the outer automorphisms of $S_6$ is the following paper:
Combinatorial Structure of the automorphism group of $\mathbf{S_6}$ by T.Y. Lam and David B. Leep, Expositiones Mathematicae 11 (1993), no. 4, pp. 289-903.
It describes exactly how to think about the exchange map between $2$-cycles and "pseudo-transpositions" $(ab)(cd)(ef)$, and it gives nice ways to interpret them combinatorially. I highly recomment it.
I think the best way to see where this outer automorphism comes from is to find a copy of $S_5$ in $S_6$ which is not a point stabilizer. The coset action then induces a map from $S_6$ to itself, which is the outer automorphism. It is also not too hard to show for other values of $n$, every $S_{n-1}\subset S_n$ is a point stabilizer. –  user641 Mar 23 '12 at 2:48