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The Cantor-Schreuder-Berstien theorem |
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| Nov12-12, 11:11 AM | #1 |
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The Cantor-Schreuder-Berstien theorem
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? |
| Nov12-12, 12:12 PM | #2 |
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| Nov12-12, 02:47 PM | #3 |
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Mentor
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What do you mean with "homomorphism" in the first place?
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| Nov13-12, 02:27 AM | #4 |
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The Cantor-Schreuder-Berstien theorem
What I mean by homomorphism is a function f:(G,.)->(H,*) where f(g.g')=f(g)*f(g')
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| Nov13-12, 02:42 AM | #5 |
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Mentor
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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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