Hi: Let G be a group, A, B subgroups, |G| = 120, |A| = |B| = 60, A and B normal in G. Is there any reason why the following should be true?: |A : A \cap B| < 3. Thanks.

EDIT: f: G/B --> G/A, f(xB) = xA. Then f is a homomorphism and is onto, and its kernel is AB/B. By the homomorphism theorem (G/B)/(AB/B) isomorphic to G/A. Therefore G/A isomorphic to (G/B)/(A/A \cap B). Hence |(G/B)/(A/A \cap B)| = |G/A| and |B| = |A \cap B| from which |B : A \cap B| = 1. But I possitively know there is something wrong with this, for in some cases |B : A \cap B| = 2.