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The Cantor-Schreuder-Berstien theorem

  1. Nov 12, 2012 #1
    The Cantor-Schreuder-Berstien theorem states that if there exists a one-to-one 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. jcsd
  3. Nov 12, 2012 #2


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    it implies nothing except that there is a bijection between them. No structure needs to be preserved by the map.
  4. Nov 12, 2012 #3
    What do you mean with "homomorphism" in the first place?
  5. Nov 13, 2012 #4
    What I mean by homomorphism is a function f:(G,.)->(H,*) where f(g.g')=f(g)*f(g')
  6. Nov 13, 2012 #5
    OK, so you're talking about group homomorphisms. Well, in that case, a version of Cantor-Shroder-Bernstein 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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