# More automorphisms

1. Nov 20, 2004

### T-O7

Okay, so i'm having trouble understanding the following question:
Determine the group of automorphisms of $$S_3$$.

I understand that the automorphisms must match orders of the same element, and since there are three permutations of order 2 and two of order 3, there are 6 "possible" permutations. But I don't know where to go from here. I'm pretty sure there's a better way than to tediously go through all six possible automorphisms, and explicitly check whether each work or not. Am i missing something here?
(I mean, how do you know when you've determined the group?)

2. Nov 21, 2004

### matt grime

You only need to look at the elements of order two since you know where (123) goes once you know where (12) and (23) go - that is the point of automorphisms - and this doesn't involve much work.

Of course, you could just prove it has no outer automorphisms...

In this case, every permutation of the 3 elements of order two is an automorphism, obviously, and this determines every automorphism, equally clearly, and hence the group Aut(S_3) = S_3.

3. Nov 22, 2004

### matt grime

I was thinking that this may well need more explanation.

Every relabelling of the numbers 1,2,3 gives an automorphism of S_3 (and a relabelling of 1,..,n would give an automorphism of S_n). These correspond to the inner automorphisms - the ones where the group acts by conjugation. If you've done linear algebra it's a lot like a change of basis.

This tells us that there is a copy of S_3 inside Aut(S_3). Now all we need to do is show that Aut(S_3) has at most 6 elements and we are done.

Because S_3 has exactly 3 elements of order 2, and they generate S_3, then any permutation of them *might* be an automorphism, and any two automorphisms permute them in distinct ways, so there are at most 6 possible automorphisms, as we needed to show.

Hence S_3 <= Aut(S_3) <=S_3

so they are equal.

The reason I thought I needed to clarify this was that I suspected it wasn't clear why this didn't show that the Aut(S_n) was something that it wasn't.

It's important that n=3 here, so that all the elements of order 2 are of the same cycle type and generate the group.

In general there is always a map from G to Aut(G) given by $$x \to f_x$$ where $$f_x(y)=x^{-1}yx$$ is how the aut f_x acts on G.

Exercises:
1. Show this map from G to Aut(G) is a homomorphism
2. What is the kernel?
3. Hence conclude that when G=S_n it is an injection unless n=2

FACT it is an isomorphism on S_n except for n=2 and n=6.
Proof n=2 is easy since S_2 is abelian. n=6 is too hard for me to recall but essentially seems to be because 2*3=2+3+1. Have a look for a proof somewhere - there are some nice geometric ones available.