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#19
Sep2306, 12:11 PM

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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. 


#20
Sep2306, 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 Gf(H) union H union Kg(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. 


#21
Sep2606, 01:22 AM

P: 867

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: 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? 


#22
Sep2606, 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).



#23
Sep2606, 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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