If a group is abelian, then it is equal to its center by definition (the center of a group is the set of elements that commute with everything - if G is abelian, then every element commutes with everything).
Here is a proof that G/Z(G) cyclic implies G is abelian
Assume that G/Z(G) = <X> for some X in G/Z(G). So, X = xZ(G) for some x in G. If g is in G, then gZ(G) = (xZ(G))k = xkZ(G) for some integer k. So, g = xkz for some z in Z(G). If h is another element of G, then similarly we can write h = xry for some y in Z(G). Thus, gh = xkzxry = xkxrzy = xrxkyz = xryxkz = hg. Thus G is abelian.