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#1
Sep2106, 10:39 AM

P: 867

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 [tex]S = \{G_1,G_2,\dots\}[/tex] as opposed to a set of generator elements as for free groups. Then we define the free product [tex]\star_{i\in I} G_i = G_1\ast G_2\ast \dots[/tex] as the most general group containing all the [itex]G_i[/itex]'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 [itex](G_1 \ast_A G_2) \ast_A G_3[/itex] is isomorphic to [itex]G_1 \ast_A (G_2 \ast_A G_3)[/itex] where A is a subgroup of the three groups [itex]G_i[/itex] 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
Sep2106, 01:38 PM

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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*_{K}H we take the free group generated by G and H and add in the relataions that K imposes. 


#3
Sep2106, 09:09 PM

P: 867

So basically, if I want to show that my two groups [itex](G_1 \ast_A G_2) \ast_A G_3[/itex] and [itex]G_1 \ast_A (G_2 \ast_A G_3)[/itex] are isomorphic I have to show that there exists a map between the two that sets up a onetoone 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: [tex]j_i\,:\,G_i \rightarrow(\star_{i\in I} G_i)[/tex] from each group into the free product."? 


#4
Sep2106, 10:59 PM

P: 867

Free Product



#5
Sep2106, 11:19 PM

P: 867

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 [tex]f_1\,:\,A \rightarrow G_1[/tex] [tex]f_2\,:\,A \rightarrow G_2[/tex] [tex]f_3\,:\,A \rightarrow G_3[/tex] Step 3: Then there exist monomorphisms from each group G_i into the free product w.r.t. A: [tex]j_1\,:\,G_1\rightarrow G_1\star_A G_2[/tex] [tex]j_2\,:\,G_2\rightarrow G_1\star_A G_2[/tex] [tex]j_3\,:\,G_2\rightarrow G_2\star_A G_3[/tex] [tex]j_4\,:\,G_3\rightarrow G_2\star_A G_3[/tex] Note: I believe these homomorphisms make the composition of homo's i and j commutative: [tex]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[/tex] Step 4: There exist homo's: [tex]i_1\,:\,G_1 \rightarrow (G_1 \star_A G_2) \star_A G_3[/tex] [tex]i_2\,:\,G_1\star_A G_2 \rightarrow (G_1 \star_A G_2) \star_A G_3[/tex] [tex]i_3\,:\,G_3 \rightarrow (G_1 \star_A G_2) \star_A G_3[/tex] [tex]i_4\,:\,G_2\star_A G_3 \rightarrow G_1 \star_A (G_2 \star_A G_3)[/tex] I drew a picture of all this and it looked something like ......______________G_1 * (G_2 * G_3) ...../....................... ...G_1G_1 * G_2 ../....../.................... AG_2 ..................... ..\......\.................... ...G_3G_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? 


#6
Sep2206, 12:49 AM

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Regard G_{1} and G_{2} as disjoint. Let A_{1} list the generators of A as a subgroup of G_{1}, and let A_{2} list the generators of A as a subgropu of G_{2} such that A_{1} and A_{2} have a 11 correspondence f : A_{1} > A_{2}. Then G_{1}*_{A}G_{2} = < Gen(G_{1}) U Gen(G_{2})  Rel(G_{1}) U Rel(G_{2}) U {af(a) : a in A_{1}} >, no? You can then take the free product w.r.t A of this with G_{3}, and associativity of the free products follows essentially from associativity of set union. I guess this should work.



#7
Sep2206, 01:24 AM

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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 nonfree 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. 


#8
Sep2206, 02:08 AM

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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 [tex]\Pi_1(X)\star_{\Pi_1(X\cap Y)} \Pi_1(Y)[/tex] 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^1g1 and e: Pi_1(T)= < f,g  fg=gf> which is a presentation of ZxZ, as required. 


#9
Sep2206, 05:45 AM

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



#10
Sep2206, 10:47 AM

P: 867

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.
[tex]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[/tex] 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 [itex]G_1\star_A G_2[/itex] 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. If it is the way to go then I have to prove that I can get that presentation you wrote for the free product. 


#11
Sep2206, 12:21 PM

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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.



#12
Sep2206, 01:22 PM

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A correction to my post: in the relations, the third set in the union should be the set of a^{1}f(a) for a in A_{1}. 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
Sep2206, 08:46 PM

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#14
Sep2206, 09:36 PM

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Like I said, regard them as disjoint. Adding the a^{1}f(a) relations serves to simply identify the copy of A in G_{1} with the copy of A in G_{2}. 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
Sep2206, 09:42 PM

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#16
Sep2206, 09:45 PM

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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
Sep2206, 10:29 PM

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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
Sep2306, 01:57 AM

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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*? 


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