If G/Z(G) cyclic, then (xZ(G))^n = G implies xy = yx for all x, y in G, so G abelian. Since G nonabelian, G/Z(G) not cyclic.
Here's a proof showing G is abelian if G/Z(G) is cyclic:
1. Assume G/Z(G) is cyclic. This means there exists an element gZ(G) in the cosets that generates the entire group under multiplication. So, for any coset aZ(G), (aZ(G))^n = gZ(G) for some positive integer n.
2. Translating back to G: Since multiplication inside a coset is the same as multiplication in G, we can rewrite the previous equation as (an)^n = g in G.
3. Applying commutativity: If G were nonabelian, that means at least one pair of elements (a, b) in G doesn't commute, i.e., ab ≠ ba. However, using equation 2 twice and the definition of a coset, we get:
b(an)^n = (ban)^n = (bgZ(G))^n = bgZ(G), and
(ab)^n = (ba)^n (since G/Z(G) is cyclic, any cyclic permutation within a coset gives the same result).
4. Contradiction: Combining these equations with the non-commutativity of a and b gives b(an)^n = (ab)^n. But because ab ≠ ba, the left and right sides are actually different elements in G. This contradicts the possibility of them being equal.
Therefore, as the contrapositive of the original statement, we've shown that if G is nonabelian, then G/Z(G) cannot be cyclic.