# 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.