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The CantorSchreuderBerstien theoremby gottfried
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#1
Nov1212, 11:11 AM

P: 119

The CantorSchreuderBerstien theorem states that if there exists a onetoone function from X to Y and the reverse then there exists a bijection between X and Y.
Does anybody know if this implies to Homorphisms. ie: If we can find an embedding between X and Y and the reverse does this imply that X and Y are isomorphic? 


#2
Nov1212, 12:12 PM

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P: 1,716




#4
Nov1312, 02:27 AM

P: 119

The CantorSchreuderBerstien theorem
What I mean by homomorphism is a function f:(G,.)>(H,*) where f(g.g')=f(g)*f(g')



#5
Nov1312, 02:42 AM

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P: 18,334

OK, so you're talking about group homomorphisms. Well, in that case, a version of CantorShroderBernstein does not hold. A counterexample is given by free groups. Indeed, we can see [itex]F_3[/itex] (free group on three generators) as a subgroup of [itex]F_2[/itex] by considering the subset [itex]\{a^2,ab,b^2\}[/itex] as generators.



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