# The Cantor-Schreuder-Berstien theorem

by gottfried
Tags: theorem
 Mentor P: 18,334 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.