What is the Automorphism Group really?

[Poopsilon]

So if a group can be interpreted as the symmetry of some object, abstract or tangible, than should I think of its automorphism group as the symmetry of the group itself? And so then the automorphism group of the automorphism group of the group is the symmetry of the symmetry of the group, ad nauseam?

I've been looking ahead at Galois Theory and it seems like these automorphism groups play a pretty central role so I'd like to know how to think about them. What is the significance of the fact that Aut(D_4) is isomorphic to D_4 itself? Why would that be the case? Is there some property of the square that would hint at why this is so?

 

[galoisjr]

The automorphism group is the just the collection of mappings from any group to itself. It is in fact a group.

 

[micromass]

To be honest, the group of automorphisms of a group isn't as interesting as the automorphisms of a geometric solid. The only use I've seen for the automorphism group is to calculate semidirect products...

 

[fourier jr]

I've been looking ahead at Galois Theory and it seems like these automorphism groups play a pretty central role so I'd like to know how to think about them.

In Galois Theory the automorphism group contains automorphisms of a field extension, not a group. There's a big (aka extension) field & a small one, and the automorphisms that you'd be interested in are the ones that move elements of the big field & leave the small field alone (or "fixed"). The simplest example is probably where the ground field is $\mathbb{R}$ & the extension is $\mathbb{C}$. In that case there are two automorphisms that leave $\mathbb{R}$ fixed, the identity $a+ib \mapsto a+ib$ and complex conjugation $a+ib \mapsto a-ib$. So the automorphism group has two elements, & is isomorphic to $\mathbb{Z}_2$. There's other stuff about polynomials & their roots, some linear algebra & normal subgroups but that's where automorphisms come into it. If you've read anything about Galois Theory you've probably encountered Galois groups, which are certain kinds of automorphism groups. In most cases I think the ground field is $\mathbb{Q}$ and the extension is an nth root (called a radical extention), a root of unity (cyclotomic extension) or some combination of those things.

 

[Poopsilon]

@fourier jr, ok that makes since thanks, now I am curious why is it of interest to us to look at automorphisms of a field extension which leave the ground field fixed?

@micromass, isn't the automorphism group of a geometric solid just what we normally call a group, such as a dihedral group?

 

[fourier jr]

@fourier jr, ok that makes since thanks, now I am curious why is it of interest to us to look at automorphisms of a field extension which leave the ground field fixed?

that's the linear algebra part; notice in the $\mathbb{R}$ & $\mathbb{C}$ example you can think of {1, i} as a basis for $\mathbb{C}$ over $\mathbb{R}$, so in other words it's a 2-dimensional vector space, or in the language of field theory, a degree-2 extension. in linear algebra it would be written $dim_{\mathbb{R}}\mathbb{C} = 2$ but since it's field theory, it's written $[\mathbb{C}:\mathbb{R}] = 2$. exactly the same idea, just a different situation. "scalars" come from the ground field & "vectors" come from the extension. I guess historically it was used to figure out whether or not cerain polynomials over the rationals had formulas similar to the quadratic formula, so in that case the ground field would have been the rationals.

 

[lavinia]

So if a group can be interpreted as the symmetry of some object, abstract or tangible, than should I think of its automorphism group as the symmetry of the group itself? And so then the automorphism group of the automorphism group of the group is the symmetry of the symmetry of the group, ad nauseam?

I've been looking ahead at Galois Theory and it seems like these automorphism groups play a pretty central role so I'd like to know how to think about them. What is the significance of the fact that Aut(D_4) is isomorphic to D_4 itself? Why would that be the case? Is there some property of the square that would hint at why this is so?

The group of automorphisms of a group may- but I am not sure - be interpretable as homeomorphisms of the Eilenberg Maclane space of the group. In the case of a flat Riemannian manifold they may correspond to isometries of the manifold.

If this is true the one does get an ad nauseum sequence of groups and symmetries of their Eilenberg Maclane spaces.

The interesting case may be the outer automorphism group.

 

[Deveno]

an interesting fact about automorphisms is this:

the group Aut(G) has a subgroup Inn(G), the group of inner automorphisms.

an inner automorphism is of the form: x-->gxg^-1, and furthermore Inn(G) = G/Z(G), so how many inner automorphisms a group has (sort of) measures how "non-abelian" G is.

in the case of D4, D4/Z(D4) is of order 4, so there are 4 inner automorphisms of D4 (for example, () and (1 3)(2 4) lead to the same inner automorphism, the identity map).

since D4 is "half-abelian", one might expect that D4 has outer automorphisms, and this turns out to be true.

letting D4 = <r,s>, where r = (1 2 3 4), s = (1 3), we see that any element of Aut(D4) must either map r-->r, or r-->r^3, and that s must map to either s, rs, r^2s, or r^3s (we can't have s-->r^2, because if φ in Aut(D4) mapped r-->r, then φ(rs) = φ(r)φ(s) implies that φ(rs) = r^3, but e = φ(e) = φ((rs)^2) ≠ φ(rs)φ(rs) = r^6 = r^2. similarly if φ(r) = r^3, then φ(rs) must be r, but e = φ(e) = φ((rs)^2) ≠ φ(rs)φ(rs) = r^2).

it is natural to ask whether φ:r-->r, s-->rs, and ψ:r-->r^3,s-->s are automorphisms. it is instructive to verify this yourself (they are).

φ^2 is the same as conjugation by r, and ψ is the same as conjugation by s. in this case Inn(D4) = <φ^2,ψ>. but there is no inner automorphism corresponding to φ (the inner automorphisms of D4 correspond to the cosets Z(D4), rZ(D4), sZ(D4), (rs)Z(D4), and each of these cosets have order 2 in D4/Z(D4), so any inner automorphism of D4 is of order 2), which is of order 4.

an isomorphism between Aut(D4) and D4 is given explicitly by:

φ --> r
ψ --> s

in Galois theory, the group of automorphisms of an extension field E which leave fixed a base field F, is a special group, and is closely conected to a group of permutations of roots of a polynomial which factors in E. in the example given above, the group {1C,*} where:

1C: a+bi--->a+bi (the identity automorphism)
*:a+bi--->(a+bi)* = a-bi (complex conjugation automorphism)

corresponds to S2, where the polynomial in question is x^2+1, and the set of roots is {i,-i} (the identity automorphism corresponds to the identity permutation () on the set of roots, and the conjugation automorphism corresponds to the only transposition, (1 2), which sends i-->-i, and -i-->i).