Talk on extensible automorphisms

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This is (approximately) the transcript of a talk given by Vipul Naik as part of the Graduate Student Talks for the Research Experiences for Undergraduates (REU) at the University of Chicago, on August 13, 2008

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.

Chalkboard (1,1):

Def: Homomorphism of groups: Given group $G, H$ a homomorphism from $G$ to $H$ is a map $f:G \to H$ such that:

• $f(ab) = f(a)f(b) \ \forall \, a,b \in G$
• $f(a^{-1}) = f(a)^{-1} \ \forall \ a \in G$
• $f(e) = e$

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.

Chalkboard (1,1):

Def: Homomorphism of groups: Given group $G, H$ a homomorphism from $G$ to $H$ is a map $f:G \to H$ such that:

• $f(ab) = f(a)f(b) \ \forall \, a,b \in G$
• $f(a^{-1}) = f(a)^{-1} \ \forall \ a \in G$
• $f(e) = e$
The inclusion of a subgroup in a group is an injective homomorphism: If $G \le H$, the inclusion of $G$ in $H$ is an injective homomorphism.

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:

Chalkboard (1,1):

Def: Homomorphism of groups: Given group $G, H$ a homomorphism from $G$ to $H$ is a map $f:G \to H$ such that:

• $f(ab) = f(a)f(b) \ \forall \, a,b \in G$
• $f(a^{-1}) = f(a)^{-1} \ \forall \ a \in G$
• $f(e) = e$

The inclusion of a subgroup in a group is an injective homomorphism: If $G \le H$, the inclusion of $G$ in $H$ is an injective homomorphism.

Definition: Automorphism of a group: An automorphism of a group $G$ is a bijective homomorphism from $G$ to $G$. The set of automorphisms of a group $G$ is denoted as $\operatorname{Aut}(G)$.

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.

Chalkboard (2,1):

Definition:Conjugation by an element: Define, for a group $G$, the operation of conjugation by an element:
$c_g(x) = gxg^{-1}$

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:

Chalkboard (2,1):

Definition:Conjugation by an element: Define, for a group $G$, the operation of conjugation by an element:
$c_g(x) = gxg^{-1}$
Theorem: For any group $G$ and any element $g \in G$, the function $c_g$ is an automorphism of $G$.

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.

Chalkboard (2,1):

Definition:Conjugation by an element: Define, for a group $G$, the operation of conjugation by an element:
$c_g(x) = gxg^{-1}$
Theorem: For any group $G$ and any element $g \in G$, the function $c_g$ is an automorphism of $G$.
Proof: We check the three conditions for a homomorphism:

• $c_g(a)c_g(b) = gag^{-1}gbg^{-1} = gaebg^{-1} = gabg^{-1} = c_g(ab)$
• $c_g(a^{-1}) = ga^{-1}g^{-1} = (g^{-1})^{-1}a^{-1}g^{-1} = (gag^{-1})^{-1} = c_g(a)^{-1}$
• $c_g(e) = geg^{-1} = e$

Next, we check that it is bijective. For this, it suffices to show that $c_(g^{-1})$ is a two-sided inverse for $c_g$:

• $c_{g^{-1}}(c_g(a)) = g^{-1}(gag^{-1})g = a$
• $c_g(c_{g^{-1}}(a)) = g(g^{-1}ag)g^{-1} = a$

Definition: Inner automorphism: An automorphism $\sigma$ of a group is termed an inner automorphism if there exists $g \in G$ such that $\sigma = c_g$. The set of inner automorphisms of $G$ is denoted $\operatorname{Inn}(G)$.

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.

Chalkboard (2,2):

Theorem: If $G \le H$, and $\sigma \in \operatorname{Inn}(G)$, there exists $\sigma' \in \operatorname{Inn}(H)$ such that the restriction of $\sigma'$ to $G$ is $\sigma$.
Proof: Since $\sigma$ is inner, there exists $g \in G$, such that $\sigma = c_g$.
$G \le H$, so $g \in H$.
Define $\sigma' = c_g$ (thought of as an element of $H$. Clearly, the restriction of $\sigma'$ to $G$ is $\sigma$.

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.

Chalkboard (1,2):

Definition: Extensible automorphism: $\sigma \in \operatorname{Aut}(G)$ is termed extensible if for any group $H$ containing $G$, there exists $\sigma' \in \operatorname{Aut}(H)$ such that the restriction of $\sigma'[itex] to [itex]G$ is $\sigma$.

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:

Chalkboard (1,2):

Definition: Extensible automorphism: $\sigma \in \operatorname{Aut}(G)$ is termed extensible if for any group $H$ containing $G$, there exists $\sigma' \in \operatorname{Aut}(H)$ such that the restriction of $\sigma'[itex] to [itex]G$ is $\sigma$.
Definition: Pushforwardable automorphism: $\sigma \in \operatorname{Aut}(G)$ is termed pushforwardable if for any homomorphism $f:G \to H$, there exists $\sigma' \in \operatorname{Aut}(H)$ such that $\sigma' \circ f = f \circ \sigma$.

Chalkboard (2,2) (modified dynamically)

Theorem: If $f:G \to H$ is a homomorphism, and $\sigma \in \operatorname{Inn}(G)$, there exists $\sigma' \in \operatorname{Inn}(H)$ such that the restriction of $\sigma'$ to $G$ is $\sigma$.

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.

Chalkboard (2,2) (modified dynamically)

Theorem: If $f:G \to H$ is a homomorphism, and $\sigma \in \operatorname{Inn}(G)$, there exists $\sigma' \in \operatorname{Inn}(H)$ such that $f \circ \sigma = \sigma' \circ f$. Proof: Since $\sigma$ is inner, there exists $g \in G$, such that $\sigma = c_g$.
Define $\sigma' = c_{f(g)}$.
Check: $(f \circ \sigma) (a) = f(\sigma(a)) = f(c_g(a)) = f(gag^{-1}) = f(g)f(a)f(g)^{-1} = c_{f(g)}(f(a)) = (\sigma' \circ f)(a)$

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.

Chalkboard (1,2):

Definition: Extensible automorphism: $\sigma \in \operatorname{Aut}(G)$ is termed extensible if for any group $H$ containing $G$, there exists $\sigma' \in \operatorname{Aut}(H)$ such that the restriction of $\sigma'[itex] to [itex]G$ is $\sigma$.
Definition: Pushforwardable automorphism: $\sigma \in \operatorname{Aut}(G)$ is termed pushforwardable if for any homomorphism $f:G \to H$, there exists $\sigma' \in \operatorname{Aut}(H)$ such that $\sigma' \circ f = f \circ \sigma$.
Definition: Quotient-pullbackable automorphism: $\sigma \in \operatorname{Aut}(G)$ is termed quotient-pullbackable if for any surjective homomorphism $f:K \to G$ there exists $\sigma' \in \operatorname{Aut}(K)$ such that $f \circ \sigma' = \sigma 'circ f$.

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

Chalkboard (2,2) (modified dynamically)

Theorem: If $f:K \to G$ is a homomorphism, and $\sigma \in \operatorname{Inn}(G)$, there exists $\sigma' \in \operatorname{Inn}(K)$ such that the restriction of $f \circ \sigma' = \sigma \circ f$.
Proof: Since $\sigma$ is inner, there exists $g \in G$, such that $\sigma = c_g$.
Define $\sigma' = c_k$ where $k \in f^{-1}(g)$.
Check: $(f \circ \sigma') (a) = f(\sigma'(a)) = f(c_k(a)) = f(kak^{-1}) = f(k)f(a)f(k)^{-1} = c_{f(k)}(f(a)) = c_g(f(a)) = (\sigma \circ f)(a)$