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Free Product

by Oxymoron
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Sep23-06, 12:11 PM
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mathwonk's Avatar
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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.
Sep23-06, 12:24 PM
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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:H--->G and g:H--->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.
Sep26-06, 01:22 AM
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Thankyou Matt and MW for convincing me. I was treating groups as "sets" and was rightly confused.

My last question is this: AKG posed a possible way to show the associativity of free groups with respect to a subgroup. But does this show that [itex](G_1 \star_a G_2) \star_A G_3[/itex] is isomorphic to [itex]G_1 \star_A (G_2 \star_A G_3)[/itex]? I was hoping use this:

Posted by Mathwonk:

Implicit in matts hints is the fact that if two groups both have the same universal mapping property then they are isomorphic.
Is AKG's proposed method of showing associativity using presentations part of the way to showing that they have the same universal property? Im not sure, in fact Im rather lost, from what I saw AKG's method showed that the two groups are in fact equal. But then...

Posted by Matt Grime:

We appear, by our laziness, to have gotten you confused between the difference betwee equal and isomorphic
And from this quote I am starting to doubt that AKG's method shows that [itex](G_1 \star_a G_2) \star_A G_3[/itex] and [itex]G_1 \star_A (G_2 \star_A G_3)[/itex] are isomorphic. Or does equal imply isomorphic in this case.

I believe that I must show they have the same universal mapping property to show that they are isomorphic and I believe this is the only way to do it. Is this right?
Sep26-06, 10:28 AM
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I think the two groups are equal. Unlike Cartesian products, where you use brackets that make a difference, i.e. the typical element of Gx(HxK) is (g,(h,k)) whereas the typical element of (GxH)xK is ((g,h),k), free products just contain strings of symbols so ghk is an element of both (G*H)*K and G*(H*K). The two groups are identical as sets, and their operations are identical, so they are identical as groups. Maybe the case is different with amalgams, I don't know. Keep in mind that I'm not entirely sure about my approach, I just thought it would be the simplest way (i.e. wouldn't involve constructing an isomorphism).
matt grime
Sep26-06, 10:32 AM
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I think you ought to take the same care initially as with cartesian products, this something in G*(H*K) are words in the symbols of G and H*K. Of course elements in H*K are just words in the symbols of H and K. thus clearly G*(H*K) and (G*H)*K have the same underlying objects. Now one merely needs to check the universality properties. The same thing applies when we replace * by an amalgam.

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