The Cantor-Schreuder-Berstien theorem

1. Nov 12, 2012

gottfried

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. Nov 12, 2012

lavinia

it implies nothing except that there is a bijection between them. No structure needs to be preserved by the map.

3. Nov 12, 2012

micromass

Staff Emeritus
What do you mean with "homomorphism" in the first place?

4. Nov 13, 2012

gottfried

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

5. Nov 13, 2012

micromass

Staff Emeritus
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 $F_3$ (free group on three generators) as a subgroup of $F_2$ by considering the subset $\{a^2,ab,b^2\}$ as generators.