Since H is cyclic, so is Aut(H). Since H is normal in G, we have N_G(H) = G. Thus,

G/C_G(H) = N_G(H)/C_G(H) < Aut(H).

Hence G/C_G(H) is abelian.

Now, G/G' is the largest abelian quotient of G in the sense that if G/K is abelian then G' < K.

So G' < C_G(H).