1. ## Inner Automorphism Groups

How does one go about proving that a particular group is isomorphic to the set of its inner automorphisms? Specifically I'm looking at the symmetric group of order 3. I've done the tedious calculations conjugating the entire group by each element, and indeed there are 6 unique mappings from S3 to Inn(S3), but so what?

More specifically, here's what I've done:

Denote:
1 as the identity which fixes each element (1)(2)(3)
2 is the cycle (123)
3 is the cycle (132)
4 is (12)
5 is (13)
6 is (23)

The inner automorphism induced by the identity gives us the permutation (1)(2)(3)(4)(5)(6)
From this permutation, the other automorphisms are:
Induced by 2: (645)
3: (564)
4: (23)(465)
5: (23)(654)
6: (23)(546)

Aside from routine calculation showing that conjugation is one-to-one, onto, and operation preserving, I have no other work on this.

I feel like I can see the isomorphism laying here, I just don't know what bits and pieces I need to assemble in a proof. Or maybe I'm not even close?

Thanks

Anthony

2. ## Re: Inner Automorphism Groups

Hi,
Let G be a group. It looks to me as though you're forgetting the operation that makes Aut(G), the set of automorphisms of G, into a group -- namely composition of the automorphisms. So for x in G, define the automorphism I(x) of G to be conjugation by x. It is straightforward that the mapping I from G into Aut(G) is a homomorphism; just use the "product" in Aut(G), namely composition. So G/Ker(I) is isomorphic to the image, Inn(G). Here Ker(I) is easily seen to be the center Z(G) (the set of elements of G that commute with everybody). So always G/Z(G) is isomorphic to Inn(G) via the mapping I defined above. In particular, for S3, Z(S3) is the identity subgroup, so the isomorphism you're looking for is just I.