# Finding an outer automorphism of $S_6$

From Wikipedia I found out about the fact that $S_6$ has outer automorphisms.

So the idea is that first you find a transitive copy of $S_5$ in $S_6$. This I was able to do, I found that for example $\langle (1 2 3 4 5),\ (1 5)(2 3)(4 6)\rangle \cong S_5$ .

This gives you a subgroup of index $6$ in $S_6$, let's call it $H$. And so we can use the left coset action or conjugation on $H$ to construct an automorphism of $S_6$ (for conjugation you have to show $[G : N_G(H)] = 6$ ). Then it turns out that this mapping does not send a transposition to its conjugate (another transposition) but to a product of three disjoint transpositions. This means that it is not an inner automorphism.

Okay, makes sense to me. But I have no idea how one should calculate the coset representatives for $H$ by hand, and how to determine how different elements of $S_6$ act on the cosets? I know it should be enough to determine how $(12)$ and $(123456)$ act on the cosets because $S_6$ is generated by these elements. I was able to brute-force a solution with GAP, but it is not a very satisfying way to do it..

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I think a large number of different-looking concrete realizations of the outer automorphism of $S_6$ exist. Notice that $S_6$ has a set of 15 generators that look like $(ab)$, and another set of 15 generators that look like $(ab)(cd)(ef)$. If I'm not mistaken, every outer automorphism must map that first set of 15 generators to the second. (But none of this answers the question.) – Michael Hardy Dec 11 '11 at 19:11
A very nice exposition and discussion of the outer automorphisms of $S_6$ can be found in Lam, T-Y and Leep, D.B. "Combinatorial structure on the automorphism group of S_6" Expositiones Mathematicae 11 (1993) no. 4, pp. 289-308 MR 1240362 (94i:2006). – Arturo Magidin Dec 11 '11 at 20:04
@Arturo: Thank you. I found the article and it seems very interesting (and elementary enough for me)! – Mikko Korhonen Dec 14 '11 at 22:45

If you look at the action of $(15)(23)(46)$, then since it is an element of $H$, it leaves the identity coset fixed. Therefore, the automorphism maps it to a permutation of the cosets that has a fixed point, so it cannot be a permutation of three 2-cycles.
Of course.. I kept trying to do things with the transposition because that was suggested and didn't think of this at all. So this shows that it is an outer automorphism. I would still be interested in knowing how to compute for the value of $(12)$ (for example) under this mapping? – Mikko Korhonen Dec 11 '11 at 18:55
@m.k. You have to calculate a list of coset representatives (e.g. make a list of $H$, take an element $g$ not in the list, apply it to every element), apply $(12)$ to the list and then check in which coset you landed. I can think of ways to reduce the computation load, but they are basically "brilliantly guess the representatives". – Phira Dec 11 '11 at 19:34