Talk on extensible automorphisms

Plan of the talk: An automorphism of a group is termed extensible if it can be extended to an automorphism for every bigger group containing it. All inner automorphisms are clearly extensible. Is the converse true?

The talk is divided in two parts. The first part focuses on the Problem: it lays out definitions of homomorphism, automorphism and inner automorphism, and then examines in excruciating detail the (easy) proof that every inner automorphism is extensible. The second part is focused on a partial solution: it introduces ideas of linear representation theory, and (with a bit of hand-waving) proves a partial converse.

Prerequisites for the talk: A clear understanding of the definition of group. Knowledge of homomorphism, automorphism etc. is helpful but these terms will anyway be defined clearly in the talk. For the second part of the talk, linear representation theory is helpful to know, though you should be able to follow the talk if you're willing to take my word for things.

Transcript of part one of the talk
Vipul: So I'll assume that everybody knows what a group is, and I'll define a homomorphism of groups.

Audience person: Could we skip that?

Vipul: No, it's very important.

Audience person (other): Good.

Vipul: So a homomorphism is a map that preserves the group structure. Group structure has three aspects: the multiplication, the inverse map and the identity element, so the homomorphism must preserve all three.

A homomorphism preserves words. So $$f(ab^2cd^{-3}) = f(a)f(b)^2f(c)f(d)^{-3}$$. So, a homomorphism also preserves equations. So if $$ab^2 = cd$$ then $$f(a)f(b)^2 = f(c)f(d)$$.

But a homomorphism doesn't preserve inequations. So if $$ab \ne cd$$ we might still have $$f(a)f(b) = f(c)f(d)$$. Basically, that's because the homomorphism may not be injective. So, you cannot pull equations back via homomorphisms.

Now, there's one special example of a homomorphism we've all seen, and that's the inclusion of a subgroup in a group.

So homomorphisms preserve structure, but this preservation of structure is one-way. They could lead to some kind of collapse. So now let's introduce a certain kind of homomorphism that preserves structure two-way:

So automorphisms preserve structure both ways. They preserve equations, and also inequations. Automorphisms can be thought of as symmetries of a group. I'll get to this view later.

Am I going too fast?

Audience member: No no.

Vipul: So what I'm going to do now is something miraculous, remarkable. You may have seen it before and not realized how wondrous it is.

This is called conjugating $$x$$ by $$g$$.

Now, I'm thinking of this, not as an operation that takes two variables and gives an answer, but rather, as an operation with one parameter, the conjugating element $$g$$, and one variable, the group element $$x$$. In other words, for any fixed value of $$g$$, I get a map $$c_g$$ from $$G$$ to $$G$$ that feeds on $$x$$ and spits out $$gxg^{-1}$$.

And now for the really remarkable theorem:

Notice the  any  here, that's again important. There are many different kinds of groups, and this results says that whatever group you pick, and whatever element you pick in it, you get an automorphism. So let's do this proof.

So we've already done the hard work. And if you think of it, it's remarkable: the fact that a simple formula is guaranteed to always yield an automorphism. So now let's reap the rewards of this with a cute theorem.

Now, this proof really relied on the fact that inner automorphisms arose from a formula, and that formula is guaranteed to work in $$H$$, a totally new group over which we have no control.

So this motivates a definition.

Now, remember (pointing at Chalkboard (1,1)) that the inclusion of a subgroup is just a special case of a homomorphism. So instead of extending, may be we can talk of pushing forward via a homomorphism:

So let's prove the analogous theorem for this for inner automorphisms. Let's see how we can modify the previous theorem. We cannot now choose $$\sigma'$$ as conjugation by $$g$$. Any guesses as to how we should choose $$\sigma'$$?

Audience member: $$f(g)$$?

Vipul: Any other guesses?

(No guesses)

Vipul: So at least we have unanimity here. So let's complete the proof.

So this might seem like some silly manipulation, but it goes back to what I said at the beginning. A homomorphism preserves words. The particular form of the word $$gag^{-1}$$ doesn't really matter, what matters for the proof is that there is a word.

There's a rule. A rule that is guaranteed to work. And because we have the rule, we can extend or push forward wherever we want.

So now how about pulling back? There's a little problem here, which is that we may not be able to pull back the element doing the conjugation. So we need to add the surjectivity condition.

So we prove this essentially by picking an arbitrary inverse image for $$g$$ under the homomorphism.