(a) Proof: Suppose G is an abelian group. Let xZ(G) be any coset in G/Z(G), where x is an element of G. We want to show that xZ(G) generates G/Z(G), i.e., for any coset yZ(G) in G/Z(G), there exists an integer k such that x^kZ(G) = yZ(G).
Since G is abelian, we have (xy)x^-1y^-1 = e for any x, y in G. This implies that xy = yx for any x, y in GZ(G), since x and y commute with elements of Z(G). Therefore, for any y in G, we can write y = x^kz for some integer k and some z in Z(G). Hence, yZ(G) = x^kZ(G), which shows that xZ(G) generates G/Z(G). Therefore, G/Z(G) is cyclic.
(b) Proof: Suppose G is not abelian and has order pq, where p and q are distinct primes. Then by Cauchy's theorem, G has elements of order p and q. Let x be an element of order p, and let y be an element of order q. Since p and q are distinct, we have x and y not commuting. Thus, xy != yx. This implies that G does not lie in Z(G). Since Z(G) is a subgroup of G, it follows that Z(G) is a proper subgroup of G. But G has prime order, so the only proper subgroup of G is the trivial subgroup. Therefore, Z(G) is trivial.
(c) Proof: Let G be a group such that [G:Z(G)] = 4. Then G/Z(G) has order 4, and hence is isomorphic to either C4 or C2 x C2. Suppose G/Z(G) is isomorphic to C4. Then there exists an element x in G such that xZ(G) has order 4 in G/Z(G). This implies that x^4 is in Z(G). Since [G:Z(G)] = 4, there are only four cosets in G/Z(G), namely Z(G), xZ(G), x^2Z(G), and x^3Z(G). By the pigeonhole principle, two of these cosets must be equal, say x^kZ(G) = x^lZ(G) for some integers k < l. Without loss of generality, we may assume that k = 0 and l = 2. Then x^2 is in Z(G), and hence x^2 commutes with every element of G. But this implies that x commutes with every element of G, since [G:Z(G)] = 4. Hence, x is in Z(G), which contradicts the fact that xZ(G) has order 4 in G/Z(G). Therefore, G/Z(G) must be isomorphic to C2 x C2.
(d) Proof: Suppose G is a group of order p^n for some prime p, and assume Z(G) is not the trivial subgroup. Then Z(G) has order p^k for some k < n, and hence G/Z(G) has order p^(n-k). Since Z(G) is a normal subgroup of G, G/Z(G) is a well-defined group. By Lagrange's theorem, the order of any element in G/Z(G) must divide p^(n-k). But G/Z(G) cannot have any elements of order p^(n-k), since if xZ(G) is an element of order p^(n-k), then x^(p^(n-k)) is in Z(G), which is a contradiction. Therefore, every element of G/Z(G) has order dividing p^(n-k-1).
Now, let xZ(G) and yZ(G) be two elements in G/Z(G). By the above argument, we know that (xZ(G))^p and (yZ(G))^p are both in Z(G). Since Z(G) is a subgroup of G, we have (xZ(G))^p(yZ(G))^p = (xy)^pZ(G) = (yx)^pZ(G) = (yZ(G))^p(xZ(G))^p. This shows that (xyZ(G))^p = (yxZ(G))^p, which implies that xyZ(G) = yxZ(G). Therefore, G/Z(G) is abelian. By part (a) of this problem, it follows that G is abelian as well.