Why is the smallest subgroup of G containing A and B equal to G itself?

[blahblah8724]

In an example it says that, if |G| = 15 and G has subgroups A,B of G with |A| = 5 and |B| = 3, then A \cap B must equal \{e_G\} and the smallest subgroup of G containing both A and B is G itself. Could anyone explain why? Thanks!

 

[morphism]

Hint: Lagrange