the fact that G/A is simple means that there are no normal subgroups of G containing A except G itself.
now if A is cyclic, show that CG
(A) is normal in G. now every element of A certainly commutes with every other (A is abelian). thus CG
(A) is a normal sbgroup of A, containing A, so CG
(A) = G.
now show that this implies A = Z(G) (containment of A in Z(G) is easy-see above). use the fact that G/A is non-abelian to show that if g is not in A, g does not commute with some member of G.