Free Product

1. Oxymoron

870
It seems that I am having difficulty understanding some of the definitions given to me concerning Free Groups.

I just finished a thread on free groups where I learnt that a free group is one that is constructed from an underlying basis set that is enlarged so that one may form all finite words - then declare that two words are equivalent if and only if inverses and identity cancellation would make them so.

Then I discovered that all free groups have a universal mapping property - that is, any function from the basis set S into a group G has a unique extension to a homomorphism of the free group F(S) into G.

I've now come across the concept of a free product and as far as I can tell we adopt a similar approach to their construction as we do for free groups. Let me explain...

First, let S be a collection of groups

$$S = \{G_1,G_2,\dots\}$$

as opposed to a set of generator elements as for free groups. Then we define the free product

$$\star_{i\in I} G_i = G_1\ast G_2\ast \dots$$

as the most general group containing all the $G_i$'s as subgroups. And likewise, the free product satisfies a universal mapping theorem: Given any collection of homomorphisms from each factor group into some common group W, there is a unique extension to a homomorphism of the free product into W.

Now I am not going to pretend that I am an expert at any of this, in fact, quite the opposite. I want to be able to "prove that $(G_1 \ast_A G_2) \ast_A G_3$ is isomorphic to $G_1 \ast_A (G_2 \ast_A G_3)$ where A is a subgroup of the three groups $G_i$ and where amalgams are formed with respect to the respective inclusions of the group A are isomorphic."

In order to do so I will need a lot of help to understand:

What does it mean to be isomorphic in this sense?

What on Earth is an amalgam?

Is the free product of two groups A and B necessarily free? Is the free product free if and only if A and B are free?

I know this is a lot of stuff to do, but if anyone can help flatten out the learning curve with this, I will rest easy.

Thanks.

2. matt grime

9,396
The same as any other sense. Indeed, there is only one sense in which two groups are isomorphic. However, in this case you can appeal to the universal property to make life easier.

Ah, now this is interesting. The free product of G and H is the set of all words in the elements of G and H and with the only relations those in G and those in H. Thus the free product of the integers with the integers is the free group on two generators. Now, sometimes it is nice to use the fact that G and H may have a subgroup in common. Thus to create the amalgam of G and H along K, i.e. G*KH we take the free group generated by G and H and add in the relataions that K imposes.

, of course not. Free groups have no torsion, free products will have at least as much torsion as A and B.

Yes. You should prove this yourself. I know for a fact you know all that is required to show that F free and F=G*H implies G and H are free, and the converse is trivial.

3. Oxymoron

870
Ah, I had a feeling that property might come in handy, but I wasnt sure.

So basically, if I want to show that my two groups $(G_1 \ast_A G_2) \ast_A G_3$ and $G_1 \ast_A (G_2 \ast_A G_3)$ are isomorphic I have to show that there exists a map between the two that sets up a one-to-one correspondence between the elements of each group in such a way that it respects the group operation.

Ok, so let me get this straight. The free product of groups has a universal mapping property - does this mean to say: "The free product of groups comes equipped with a family of homomorphisms:

$$j_i\,:\,G_i \rightarrow(\star_{i\in I} G_i)$$

from each group into the free product."?

4. Oxymoron

870
I really, really like that explanation! Thank you.

Could you possibly remind me what torsion means? Is it a set of elements? Or is it another group? Or both?

Last edited: Sep 22, 2006
5. Oxymoron

870
Step 1: Three groups G_1, G_2, and G_3 and a subgroup A of all three.
Step 2: There exist homo's from A into each group G_i

$$f_1\,:\,A \rightarrow G_1$$
$$f_2\,:\,A \rightarrow G_2$$
$$f_3\,:\,A \rightarrow G_3$$

Step 3: Then there exist monomorphisms from each group G_i into the free product w.r.t. A:

$$j_1\,:\,G_1\rightarrow G_1\star_A G_2$$
$$j_2\,:\,G_2\rightarrow G_1\star_A G_2$$

$$j_3\,:\,G_2\rightarrow G_2\star_A G_3$$
$$j_4\,:\,G_3\rightarrow G_2\star_A G_3$$

Note: I believe these homomorphisms make the composition of homo's i and j commutative:

$$A \rightarrow_{i_1} G_1 \rightarrow_{j_1} G_1\star_A G_2 = G_1\star_A G_2 \rightarrow_{j_1^{-1}} G_1 \rightarrow_{i_1^{-1}} A$$

Step 4: There exist homo's:

$$i_1\,:\,G_1 \rightarrow (G_1 \star_A G_2) \star_A G_3$$
$$i_2\,:\,G_1\star_A G_2 \rightarrow (G_1 \star_A G_2) \star_A G_3$$

$$i_3\,:\,G_3 \rightarrow (G_1 \star_A G_2) \star_A G_3$$
$$i_4\,:\,G_2\star_A G_3 \rightarrow G_1 \star_A (G_2 \star_A G_3)$$

I drew a picture of all this and it looked something like
......______________G_1 * (G_2 * G_3)
...../.......................|
...G_1--G_1 * G_2---|-|
../....../..................|..|
A-G_2 ...................|..|
..\......\.................|...|
...G_3--G_2 * G_3 -|....|
.....\________________(G_1 * G_2) * G_3

and I have to prove that there exists an isomorphism between the two free product on the right hand side of this diagram?

Last edited: Sep 21, 2006
6. AKG

2,585
Regard G1 and G2 as disjoint. Let A1 list the generators of A as a subgroup of G1, and let A2 list the generators of A as a subgropu of G2 such that A1 and A2 have a 1-1 correspondence f : A1 -> A2. Then G1*AG2 = < Gen(G1) U Gen(G2) | Rel(G1) U Rel(G2) U {af(a) : a in A1} >, no? You can then take the free product w.r.t A of this with G3, and associativity of the free products follows essentially from associativity of set union. I guess this should work.

7. matt grime

9,396
An element is (or has) torsion if its order is finite. Free groups have no torsion elements.

Just think in terms of presentations and relations. A free group has generators and no relations amongst the generators, a non-free group has some relations. In G*H, an relations on elements of G remain relations on those elements as considered elements in G*H.

I was just using 'torsion' as one example of a relation (x^n), I could have talked about commutation of elements: if x and y commute in G, they commute in G*H.

Last edited: Sep 22, 2006
8. matt grime

9,396
An example of alagam product: the fundamental group of the torus is ZxZ.

This is more generally called the seifert van kampen theorem.

Given a topological space, T, and two (open?) subsets X and Y with XnY path connected, then the fundamental group of T is

$$\Pi_1(X)\star_{\Pi_1(X\cap Y)} \Pi_1(Y)$$

Let T be the torus, X be T\{pt} and Y some small disc around the point. T\{pt} is homotopic to the bouquet of two circles (or two circles pinched together at a point like this oo) and its fundamental group is F_2 generated by two loops g and h. The fundamental group of Y is trivial, and the XnY is homotopic to a circle so its fundamental group is Z generated by z, which is a loop around {pt}. (Draw a picture; it helps to remember T is the unit square with opposite sides identified.)

Thus we need the amalgam of F_2 with {e} the trivial group along Z. We need to see what relations this imposes. Well, the homomorphism of Z to {e} sends z to e, and the homomoprhism from Z to F_2 sends z to fgf^-1g^-1, so in the amalgam the/a presentation is, since we must identity fgf^-1g-1 and e:

Pi_1(T)= < f,g | fg=gf>

which is a presentation of ZxZ, as required.

Last edited: Sep 22, 2006
9. mathwonk

9,671
implicit in matts hints is the fact that if two groups both have the same universal mapping property then they are isomorphic.

10. Oxymoron

870
So I guess my little diagram is correct. I was hoping that it was because that is what I am visualising and aiming to show that there is an isomorphism between the two graphs on the right hand side of that diagram.

This presentation you have

$$G_1 \star_A G_2 = \langle \mbox{Gen}(G_1) \,\cup\, \mbox{Gen}(G_2) \,|\, \mbox{Rel}(G_1)\,\cup\,\mbox{Rel}{G_2}\,\cup\,\{af(a)\,:\,a \in A_1\}\rangle$$

is a little strange to me. How did you manage to get the relations for the free product to be the union of the relations for G_1 and G_2? Same for the relators. I mean, this presentation tells me that the generators of the free product $G_1\star_A G_2$ is the set of all generators from G_1 and G_2 and the relators of the free product is the set of all relators from G_1 and G_2 and some more.

Are you using the fact that once you get a presentation for the free product $G_1\star_A G_2$ you can make the free product of it and the remaining group G_3 and get a new presentation for $(G_1 \star_A G_2)\star_A G_3$. Then all you have to work with are unions of generators and relators, switch them around a bit using union laws, get a new presentation for the free product and work backwards to show that this new presentation (now with the gens and rels moved around) is the same as $G_1 \star_A (G_2\star_A G_3)$. I dont even know if this will work, but if it does, what you are essentially doing is showing the associative property using the associative property of unions. Hmm...interesting.

If it is the way to go then I have to prove that I can get that presentation you wrote for the free product.

What about relations on elements in H? Do they remain relations on elements in G * H?

Last edited: Sep 22, 2006
11. matt grime

9,396
Well, obviously they do. I just picked G for the sake of it. The free product is totally symmetric. Just look at the defintion. G*H=H*G.

Last edited: Sep 22, 2006
12. AKG

2,585
A correction to my post: in the relations, the third set in the union should be the set of a-1f(a) for a in A1. Anyways, I'm not sure exactly what you're asking me because it seems that you understand what I'm saying. If F = < S1 | R1 > and G = < S2 | R2 > are disjoint, then isn't it clear that F*G = < S1 U S2 | R1 U R2 >?

13. Oxymoron

870
Yes, but I have no way to ensure that my groups G_1, G_2, and G_3 are disjoint. All I know is that they are groups and A is a subgroup.

14. AKG

2,585
Like I said, regard them as disjoint. Adding the a-1f(a) relations serves to simply identify the copy of A in G1 with the copy of A in G2. The free product is generally defined only on disjoint groups, but you can always regard any two groups as disjoint. Earlier, matt mentioned Z*Z. If you don't regard the elements of the first copy as distinct from those of the second, then you just get Z.

15. Oxymoron

870
Oh I see!! So if you have two groups that are not necessarily disjoint, you can always regard them as being disjoint if you have a subgroup of both groups present?

16. AKG

2,585
Well, you can always regard any two groups as disjoint. If there is no common subgroup of both present, however, then there must be no common element of both, so they already are disjoint. But the idea behind being able to regard two groups as disjoint has nothing to do with whether or not they share any subgroups.

17. mathwonk

9,671
If G1 and G2 are two groups, theirm free product would be some group such that group maps out of it correspond to pairs of group maps out of G1 and G2.

If G1 and G2 share a common subgroup H, then their free product amalgamated over H, would be a group such that group maps out of it correspond to apirs of groups maps out of G1 and G2 that agree on their common subgroup H.

18. matt grime

9,396

We appear, by our laziness, to have gotten you confused between the difference betwee equal and isomorphic. Whenever we are taking groups G and H and forming their amalgamated product they are just groups. And when we say there is a subgroup in common we mean there is a third group K and an injection from K into G and one into H. We do not mean that in any sense they have to overlap, i.e. that GnH is not empty.

As AKG pointed out, when I did Z*Z, you had no problems with the things that are bothering you in the quoted posted despite the two groups being isomorphic.

I don't see why you think you can regard groups as disjoint *if* they have elements in common. Perhaps you mean *even if*?

19. mathwonk

9,671
a subobject of X is as matt said technically an equivalence clas of injections A--->X where A--->X is equivalent to B--->X iff there is an isomorphism A--->B--->A such that the compositions A--->B--->X and B--->A--->X equal respectively the injections A--->X and B--->X.

oddly, with this definition, note that being a subobject is not transitive, but it is still useful.

20. mathwonk

9,671
for some purposes a paint can is useful. i.e. if G,K,H are groups, paint all elements of G red, all elements of K blue, and all elements of H green. now they are all disjoint since the elements are different colors.

if there are given injections f--->G and g--->K, remove the images of the maps and consider the set G-f(H) union H union K-g(H).

our goal is to enlarge this set to a certain big group so that the first two sets form a subgroup and also the last two sets.

we do this by working with the original disjoint groups, and finally we find a solution into which it is possible to map this union injectively, so that the restriction to each subgroup containing H is a homomorphism.

pretty bad, sorry. matt said it better.

but as mathematicians we are careleslike this all the time.

we say that R^2 = RxR contains two copies of R as axes, but really the axes are literally {0}xR and Rx{0}. we dont care, we think Rx{0} is so much like R that we identify them.