Automorphism Group

1. Oct 22, 2006

Oxymoron

If you have two groups, $G_1$ and $G_2$ and A is a common subgroup, then you can form the free product of $G_1$ and $G_2$ amalgamated over A. Denote this free product by $G_1 \star_A G_2$.

Q1: Now I have read that you can associated a tree, T, to $G_1 \star_A G_2$. Is this true?

Q2: What is $\mbox{Aut}(\Gamma)$? Is it the collection of all isomorphic homomorphisms $\varphi$ from the tree to itself?

Q3: Does it make sense to think that there should be a homomorphism from the free product $G_1 \star_A G_2$ to $\mbox{Aut}(\Gamma)$?

2. Oct 22, 2006

matt grime

A1. Probably. You can associate a tree to anything.

A2. You've not defined what Gamma is. Aut(X) is always the isomorphisms of X whatever X is.

A3. See A2.

3. Oct 23, 2006

Oxymoron

$\Gamma$ is the tree associated to $G:=G_1\star_A G_2$.

So are you saying that $\mbox{Aut}(\Gamma)$ is the collection of all isomorphisms

$$\varphi\,:\,\Gamma \rightarrow \Gamma$$

from the tree into itself?

But if I have a tree, $\Gamma$, then an automorphism $\varphi$ on $\Gamma$ is a map which preserves edges and vertices, is it not?

The question is can I mix the idea of tree/graph automorphisms with group automorphisms?

I understand a group automorphism is a group isomorphism from a group to itself. So it looks like this

$$\phi\,:\,G \rightarrow G$$

The collection of all such maps is the group $\mbox{Aut}(G)$. Now is it true that for every group $G$ there exists a natural group homomorphism $G \rightarrow \mbox{Aut}(G)$? Im sure this is correct.

But then I read a theorem somewhere that if I have the natural map

1. $$G \rightarrow \mbox{Aut}(G)$$

there there exists a map

2. $$G \rightarrow \mbox{Aut}(\Gamma)$$

I would just like to know how 1. relates to 2.

Last edited: Oct 23, 2006
4. Oct 23, 2006

HallsofIvy

Staff Emeritus
Yes, to every a in G, associate the automorphism that maps x to ax.

5. Oct 23, 2006

Hurkyl

Staff Emeritus
The map x --> ax is not an automorphism. (because it's not a homomorphism)

The map x --> x^(-1) a x is, though.

6. Oct 23, 2006

Oxymoron

Right, this is the conjugate map is it not?

$$\varphi_C(a) \,:\, a \mapsto gag^{-1}$$

and I checked that this is indeed a homomorphism since

$$\varphi_C(ab) = gabg^{-1} = gag^{-1}gbg^{-1} = \varphi_C(a)\varphi_C(b)$$

$$\varphi_C^{-1}(a) = g^{-1}(gag^{-1})g = a$$

I guess the idea is that the conjugation map of group elements, a, is an automorphism of the group. But the conjugation map is of the form

$$\varphi_C\,:\,G \rightarrow G$$

not

$$\varphi_C\,:\,G \rightarrow\mbox{Aut}(G)$$

The conjugation is an automorphism so

$$\mbox{Conjugation} \in \mbox{Aut}(G)$$

But then what is stopping me from constructing a map which takes an element a of some group G and mapping it to its conjugation:

$$\varphi_C(a) = gag^{-1}$$

Then surely, isn't this map of the form

$$\varphi\,:\,G \rightarrow \mbox{Aut}(G)$$??

That is, $\varphi$ takes an element a from a group G and maps it to the conjugation of a (which, in turn, is an automorphism). So in this case $\varphi$ is a map from a group element to an automorphism. Is this right?

Last edited: Oct 23, 2006
7. Oct 23, 2006

Oxymoron

And since conjugation is a group action (is it not?) then this G-action induces a homomorphism

$$\varphi\,:\,G \rightarrow \mbox{Aut}(G)$$

is this where the meaning behind the part in italics comes from?

8. Oct 23, 2006

Oxymoron

If my last two posts are correct then do all G-actions induce a homomorphism $G \rightarrow \mbox{Aut}(G)$? I guess they must, since once you have a G-action on a group G you automatically have a natural way to construct a homomorphism into the group of automorphisms of G.

So the question remains: If I have a tree $\Gamma$ (not a group!) which is in someway associated with the amalgam-decomposition $G:=G_1\star_A G_2$ with its natural G-action, then can I put meaning behind a map of the form

$$G \rightarrow \mbox{Aut}(\Gamma)$$?

I mean, from what we have discussed, $\mbox{Aut}(\Gamma)$ is the group of automorphisms of the tree, that is, the group of bijective homomorphisms from the tree onto itself:

$$\phi\,:\,\Gamma\rightarrow\Gamma$$

An example of such a bijective homomorphism is the trivial one: The map which takes each vertex and edge to themselves. Another could be the map which takes each vertex to the next and the last to the first. A third could be the inverse of this.

If I am correct, these types of maps are automorphisms of the tree $\Gamma$.

The current question reads like this:

Let $A$ be a common subgroup of $G_1$ and $G_2$ and let $G:=G_1\star_A G_2$. Then suppose we have the tree, $\Gamma$, associated to this amalgam-decomposition of $G$ with its natural G-action (which induces the homomorphism $G \rightarrow \mbox{Aut}(\Gamma)$). Then explain the part in parenthesis???

9. Oct 23, 2006

matt grime

You called that T in the first post

yes

yes. and?

No. One is an automorphism of a tree the other a graph. It is of course perfectly possible that in the association of Gamma to your amalgam that you fail to define that any automorphism of the underlying group induces an automorphism of the tree and vice versa.

Of course G is isomorphic to the set of inner automorphisms given by conjugation.

Since you have failed to define Gamma for us we cannot say. If you tell us what the tree associated to the group is we might be able to help.

10. Oct 23, 2006

matt grime

No. A G action on itself is not necessarily an automorphism, it is just a map on the underlying set of G.

Let g be in G. Define an automorphism f_g of G by f_g(h) = ghg^{-1} for all h. The map

g--> f_g is the thing you're struggling to define as the map G --> Aut(G)

The rest of the post remains unanswerable, as it has since post 1, since you've not told us what this mysterious association is between trees and amalgams. I can guess, and can think of a reasonalbe one, but it is up to you to tell us what it is.

11. Oct 23, 2006

AKG

A G-action on a set X is, by definition (at least in my book) a homomorphism G -> S(X), where S(X) is the set of bijections X -> X. If X is a group, Aut(X) is a subset of S(X). So a G-action on G itself is a homomorphism f: G -> S(G), and if there is a natural homomorphism g: S(G) -> Aut(G) then I suppose you can say that the action induced a natural homomorphism G -> Aut(G), namely gof.

I don't know how you've defined Aut(T), for a tree T, but I would assume that a G-action on T would, by my book's definition, be a homomorphism from G to Aut(T). All you have to do is:

a) clearly define the natural G-action on a tree T
b) clearly define Aut(T)

12. Oct 23, 2006

Oxymoron

I swear, all I know about my tree $\Gamma$ is that it is that tree associated to the amalgam-decomposition of $G:=G_1\star_A G_2$ with its natural G-action (which induces a homomorphism $G\rightarrow\mbox{Aut}(\Gamma)$. This is all I am told. And I basically just want to work out what $G\rightarrow\mbox{Aut}(\Gamma)$ is and what kind of map it is, how it came to be, and possibly an example.

EDIT: I have a theorem which states: Let $G:=G_1\star_A G_2$ be an amalgam of two groups. Then there exists a tree $\Gamma$ (and exactly one, up to isomorphism) on which G acts, with fundamental domain a segment whose vertices are P and Q and whose edges are $y$ and $\bar{y}$ such that $G_P = G_1$, $G_Q = G_2$, and $G_y = A$ and their respective stabilizers.

Perhaps this tells me exactly what the tree is.

Last edited: Oct 23, 2006
13. Oct 23, 2006

matt grime

Why say that? The map from G to AUt(G) is just the identification of G with the inner automorphisms.

14. Oct 23, 2006

matt grime

Then you're a bit stuffed, aren't you?

Let me suggest how to make Gamma. Either its edges or vertices will be labelled by elements of the amalgam. Since G acts on G by left multiplication, amongst other things, G will act on Gamma. It will be a bijection on either the edges or the vertices since G's action on itself will be a bijection on the underlying set.

For example, there is the obvious way to get a tree from F_2 the free group on 2 elements, and it is the topologists television aerial. Do you need me to describe it? It is the covering space of the bouquet of two circles.

Start with a node for the identity. There are 4 edges out of this node, one each to nodes labelled a,b,A and B where we use A for a^-1 (neat trick someone told me recently).

For a, say, there are edges back to e=aA, and edges to modes labelled aa, ab, aB, then repeat.

Clearly the map x-->gx on the labelling of the nodes is an automrophism of the tree.

15. Oct 23, 2006

matt grime

In general, fix a set of generators of G, the vertices of G will probably be labelled by elements of G, and there will be an edge from x to y x=gy for some generator g of G. That is one way to do it. There could be others. We don't know. Ask the person how set the question, perhaps?

16. Oct 23, 2006

Oxymoron

Maybe. Although the actual question asks

Show, by example, that $G \rightarrow \mbox{Aut}(\Gamma)$ need not be surjective or injective

although I didn't want to ask this just yet (or at all!), hoping that perhaps by discussing the ingredients (the automorphisms, etc...), the solution to the question stated here would come naturally. Unfortunately, this didn't happen, and now you are saying that I need the tree to be defined explicitly (for the other questions I have been asking) but maybe not for the actual question. So I thought I'd post the actual question before getting stuck on something that may not be worth getting stuck at. However, if the tree needing to be defined explicitly is still required to answer the question of whether that map need be surjective/injective then we will continue with that problem and I will ask my lecturer what's up with that.

17. Oct 23, 2006

matt grime

There are many trees on which G acts. Infinitely many, in fact. Strictly speaking it is a proper class, since we can easily define non-isomoprhic ones for each ordinal number.

All we know is that T/Gamma is a tree. It is defined in some way so that it has an action of G on it in a certain way. We don't know how. (Any time G acts as bijections on some X you get a map from G to Aut(X))

I can probably prove that the map from G to Aut(T/Gamma) is not surjective quite trivially. Aut(T/Gamma) will just be far too big, and there won't even be a set isomorphism from G to it, never mind a group isomorphism.

I can also probably show why it won't be injective. If there are two different g's with the same g action then the map is not injective. For instance, if G acts on a group G by conjugation, then the identification g-->f_g is injective if and only if the centre of G is trivial. So if G is abelian, the action of G on itself by conjugation is always trivial, and the map fromG to Aut(G) sends every element to the identity mapping.

18. Oct 23, 2006

AKG

It was in response to the question "do all G-actions on G induce a homomorphism G to Aut(G)?" not "does the particular G-action on G of conjugation induce a homomorphism G to Aut(G)?"

19. Oct 23, 2006

AKG

Could you define fundamental domain, GP, GQ, and Gy?

My book gives a construction of a graph associated with a group. In general, this won't be a tree:

Start with a group G and a set of generators X for G. For simplicity, assume X has no self-inverse elements. Then the associated graph is the ordered pair (V,E) where:

V = G [or rather, the underlying set of G]
E = {(g,z) | g in G and either z in X or z-1 in X}

Note that the edges of the graph are labelled by the ordered pairs (g,z), but this does not mean that they are the edges with initial point g and final point z. (g,z) is just a labelling. The initial point of (g,z) is defined to be g, and the final point is defined to be gz. The reverse of (g,z) is defiend to be (gz,z-1). So vertices g and h of the graph have an edge between them iff there is a generator x in X such that either gx = h or g = hx. g in G acts on (V,E) by sending vertex v to gv, and sending edge (h,z) to (gh,z).

This graph is connected. If G is free over X, then the graph described above is also a tree (but I'm not sure if the converse is true).

20. Oct 24, 2006

Oxymoron

Right. Should I remember that the map $G\rightarrow\mbox{Aut}(\Gamma)$ is induced by the G-action on the tree. So to show that the induced homomorphism $G\rightarrow\mbox{Aut}(\Gamma)$ need not be surjective I need to come up with an example which is not surjective.

So, what properties must the map $G \rightarrow \mbox{Aut}(\Gamma)$ hold? Well, just the properties of a homomorphism, no? Which means all it has to do is preserve the structure of the group G. But $\mbox{Aut}(\Gamma)$ is the collection of all automorphisms of the tree $\Gamma$. If I could somehow show that the set of all automorphisms of the tree was bigger than G then the trivial homomorphism $G \rightarrow \mbox{Aut}(\Gamma)$ would suffice to show that it need not be surjective. Is this right?

Is this an example which shows that the map $G \rightarrow \mbox{Aut}(\Gamma)$ is a homomorphism which is not injective? - or the map $G \rightarrow \mbox{Aut}(G)$ is not injective?

Definition

Let $G$ be a group acting without inversion on a graph $\Gamma$. A fundamental domain of $\Gamma$ mod $G$ is a subgraph $T$ of $\Gamma$ such that $T \rightarrow G\backslash \Gamma$ is an isomorphism.

Last edited: Oct 24, 2006
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