Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question
    
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
2  
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

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.