# Free Products II

1. Oct 31, 2006

### Oxymoron

Question:

Let $A$ be a common subgroup of $G$ and $H$. Show that if $G \trianglelefteq (G \star_A H)$ then $G = A$.

But before we go into this problem I'd like to ask a few, hopefully simple, questions.

1] What does the $\trianglelefteq$ symbol mean? Does it just mean that G is a normal subgroup of $G \star_A H$? What is the difference between this and $\triangleleft$?

2] So is this question basically asking prove that G = A if G is a normal subgroup of the free product amalgamated over A?

Last edited: Nov 1, 2006
2. Nov 1, 2006

### AKG

1) Wherever you found this question, the symbols should be defined. Everytime I've seen that symbol, it has indeed meant that something is a normal subgroup of another group. In theory, $X \trianglelefteq Y$ should mean that X is a normal subgroup of Y, and $X \triangleleft Y$ should mean that X is a proper normal subgroup of Y. But just as you often see people write $X \subset Y$ to mean that X is a subset of Y, and not necessarily a proper one, people will also tend to write $X \triangleleft Y$ to mean that X is a normal subgroup of Y, and not necessarily a proper one.

2) It's asking you to prove that G = A if G is a normal subgroup of the free product of G with H amalgamated over A.

3. Nov 1, 2006

### Oxymoron

How exactly am I going to be able to show this anyway? How am I meant to be able to prove that G = A if G is a normal subgroup of the amalgam G*H? Am I meant to show that they are isomorphic and therefore equal?

If $G \trianglelefteq G \star_A H$ then G is invariant under conjugation, by definition. In other words, for each group element $n \in G$ and each $g \in G\star_A H$ then the element $gng^{-1}$ is still in G.

But by another definition, if $G \trianglelefteq G \star_A H$ then $Gg = gG$ for all $g \in G\star_A H$. That is, the left and right cosets of G coincide.

This is what it means for G to be a normal subgroup of the free product of G and H amalgamated over a common subgroup A. Now I have to show that G = A. Which means we have to work out how A comes into this picture.

Last edited: Nov 1, 2006
4. Nov 1, 2006

### Oxymoron

I have just noticed something that could be important. Let me know what you think of its relevance: One may form the amalgam of G and H along A by means of group homomorphisms f and g and this is written

$$G \star_A H$$

Then, if $A = \{e\}$ the corresponding amalgam is called the free product of G and H amalgamated over A.

So does this mean that wherever I have said "...free product...", I have been indicating my choice of A? I mean, is the only time one can talk of the "free product" when $A = \{e\}$? I don't think I have ever needed this, which leads me to believe that whenever I have said "free product" I have actually meant "the amalgam".

Back on topic: Since we are constructing the free product (with amalgamation) of two groups G and H the we have implicitly two group homomorphisms:

$$\varphi\,:\,A \rightarrow G$$

and

$$\psi\,:\,A \rightarrow H$$

because these are necessary for the construction of the amalgam (getting the terminolgy right now!) $G \star_A H$.

Last edited: Nov 1, 2006
5. Nov 1, 2006

### matt grime

*The* free product is indeed just when H={e}. It is a common abuse to refer to an amalgam as a free product, strictly speaking it is a quotient of the free product by adding in the relations from H. But we have told you all this before.

6. Nov 1, 2006

### Oxymoron

I just had a thought. If I can show that G acts (on some object) in precisely the same way that A acts (on that same object) then will G = A?

That'd be right! Forgive me, but Im really struggling with this stuff.

7. Nov 1, 2006

### matt grime

No. let G act trivially on anything, and A act trivially on anything. Perhaps you want 'faithful' action. But 'the same' is imprecise.

Just use the definitions, the universality arguments.

8. Nov 1, 2006

### Oxymoron

First we need to know what an inversion is. Let $\Gamma$ be a graph on which a group G acts. An inversion is a pair consisting of an element g in G and an edge y of $\Gamma$ such that $gy = \bar{y}$ (where a bar over the edge indicates the back-tracking). If no such pair exists then we say that the group G acts without inversion. Basically, this says that there is an orientation of the graph preserved by the group.

Note: Once a group acts without inversion on a graph $\Gamma$ then we can define the quotient graph $G \backslash \Gamma$ of the action by G on the graph. The vertex and edge sets of $G \backslash \Gamma$ are the quotient sets of the vertex and edges sets of $\Gamma$ under the action of G respectively.

Next I will define a fundamental domain. Let G be a group acting without inversion (so that it preserves orientation of the graph) on a graph $\Gamma$. A fundamental domain of the quotient graph $\Gamma\mbox{mod}G$ is a subgraph $\Gamma'$ of $\Gamma$ such that $T \rightarrow G\backslash\Gamma$ is an isomorphism.

Basically, a fundamental domain arises when you have a group G acting on a tree. The fundamental domain for G is a subtree of $\Gamma$ such that the vertex set of the fundamental domain contains exactly one vertex from each G-orbit in the vertex set of $\Gamma$. The fundamental domain is a set of representatives of the orbits.

Now suppose that G is a group acting without inversion on a graph $\Gamma$ and let T be a segment of the graph with vertices P and Q and edges $y$ and $\bar{y}$. Suppose that this segment, T, is actually a fundamental domain of $\Gamma\mbox{mod}G$, and let $G_P$, $G_Q$, and $G_y = G_{\bar{y}}$ be the stabilizers of the vertices and edges of the segment (fundamental domain). There is a theorem which says that the graph must actually be a tree!

Now, if we construct the amalgam $G\star_A H$ of two groups (with A a common subgroup of them both). Then there exists precisely one tree $\Gamma$ (up to isomorphism) one which G acts, with fundamental domain a segment, T, whose vertices are P and Q and whose edges are $y$, and $\bar{y}$, such that $G_P = G$, $G_Q = H$, and $G_y = A$ are their respective stabilizers.

So this unique (up to isomorphism) tree $\Gamma$, is associated with this amalgam via its natural G-action - this is what I mean by the natural G-action.

To answer the question then, if I suppose that $\Gamma$ is the tree associated with the amalgam $G \star_A H$ with its natural G-action, then surely if I hypothesize that, in fact, G is a normal subgroup of the amalgam, then G actually acts trivially on $\Gamma$!! And from this I can claim that G actually equals A! Can I not?

9. Nov 1, 2006

### Oxymoron

That was my original plan. But I got stumped because I realized I didn't know how to prove a group G equals another group A. I figured that G = A if there exists an isomorphism $\varphi\,:\,G \rightarrow A$. Since, from the definition of an amalgam I already have half of this isomorphism, namely, the homomorphism associated with G and A (and also with H and A).

10. Nov 1, 2006

### matt grime

Sure you do. G is contained in A is contained in G is contained in A. Either the first 3 of those or the last three of those will do. Note, it is no good showing G is isomorphic to A which is what you're thinking of: isomorphism and equality are very different. Even in finite groups. For instance, in S_n acting on 1,..,n the stabilizer of the element 1 is not equal to the stabilizer of the element 2, though they are obviously isomorphic.

11. Nov 1, 2006

### Oxymoron

So I have to show

$$G \subset A$$

and

$$A \subset G$$

Then this becomes a problem about comparing elements, does it not?

If A and G are groups how does one go about showing that A is contained in G? I didn't think you could show that a group was contained in another group, I thought this tactic only worked with sets. Perhaps I could use the fact that since I have an amalgam, then there is a homomorphism $f\,:\,A \rightarrow G\star_A H$ and $f'\,:\,G\rightarrow G\star_A H$. Perhaps I will need to use some presentation stuff, no?

Last edited: Nov 1, 2006
12. Nov 1, 2006

### matt grime

You take something in A and show it is in G, etc... A group is a set, by the way.

Just think about it for a second - if G is normal in the amalgam, then hgh^-1 is in G for any h in H, right? I.e. hgh^-1 = k for some k in G. The right hand side is purely a word in elements of G, and the LHS isn't. So it is a relation, and must come from something to do with A...

13. Nov 1, 2006

### mathwonk

my knee jerk reflex is to use the m,apping property of a free product with amalgamation. (which i learned from mauslander, as matt will appreciate.)

14. Nov 3, 2006

### Oxymoron

...right, because normal subgroups are invariant under conjugation.

Since G is normal in the amalgam, then G is invariant under conjugation, that is, for every g in G, and h in the amalgam, hgh^-1 is still an element in G, no matter what h from the amalgam we choose. So performing conjugation on an element g in G, we essentially get another element from G, you call it k. Since k is in G we know that it is some word represented by elements of G, ie.

$$k = g_1\cdot g_2 \cdot \dots \cdot g_n$$

But now you say that $hgh^{-1}$ is not a word made up of elements of G. Is that because there are h's in there?

I dont exactly see why if k is a word made up of only elements in G, and $hgh^{-1}$ is not, then it is a relation. Does this mean that h~g if $hgh^{-1}$ equals a word made of only elements of G?

So we have an amalgam (a group), $G \star_A H$ and a subgroup (that is normal) G. We now have an equivalence relation on G by $h~g$ if and only if $hgh^{-1} \in G$. The equivalence classes are the double cosets (which we talked about before) of G in the amalgam. One of them is G itself. They all have the same number of elements. And since G is a normal subgroup, then the set of all double cosets is itself a group.

I still dont see where the subgroup A is going to magically appear. The only way I know that a subgroup A, such as this one (being a subgroup of G), could come into the picture is if I construct the free group F(A). I would do this by using a representative from A, namely a, and forming all finite words consisting of a and its inverse, $a^{-1}$. Two words are equivalent if one is obtained from the other via replacing instances of $aa^{-1}$ or $a^{-1}a$ by the empty syllable. This is an equivalence relation. By the universal mapping property there exists a function

$$f \,:\, A \rightarrow G$$

and then there exists a unique group homomorphism

$$j\,:\,F(A) \rightarrow G$$

such that $j(a) = f(a)$.

Other than this, I dont see anyway A is going to affect what I do with G.

Last edited: Nov 3, 2006
15. Nov 4, 2006

### matt grime

Let's forget the question. Think about the free group on two generators g and h. Are the two expressions hg and hg equal? No, of course not. Why? Because if they were then hgh^-1 = g. If they were that would be a relation, and there are no relations in F_2. Look up the definition *again* of what an amalgam is. The free product of G and H modulo an equivalence relation (the relations coming from A as a common subgroup). By definition the only way that I can get a relation amongst the elements of G *_A H is if it comes from one in G, one in H or one in A. Since the relation involves both elements of G and of H it must be a relation coming from A.

16. Nov 8, 2006

### Oxymoron

Sounds good to me! (of course it is good! - you say).

The aim is to show that G is contained in A is contained in G. And we know that since G is normal in the amalgam then hgh^-1 is in G for any h in H, in other words, hgh^-1 = k for some k in G.

If I can show that somehow all the k's are in A then I have shown one inclusion. Is this where you are heading? Because hgh^-1 is a relation that comes from A? Im still struggling to find how A is going to relate to G. The last thing you said was that the relations are going to come from A, but what does this mean?

17. Nov 9, 2006

### matt grime

Again, go and think about how we construct the amalgam. The group is words in the generators with the relations to simplify. The relations are those from the relations that define G, those that define H, and the only times you can simplify words that contain elements of both H and G is if the letters are in H. Have you even tried to look at any small examples to see what this looks like? Just suppose that A is some abelian subgroup of G and H. In general the words gh and hg will not be the same. If g and h are in A, which is abelian, then gh=hg. See? A relation from A.

You know A is a subgroup of G, you just need to show that there is nothing in G-A.

Last edited: Nov 9, 2006