1. The problem statement, all variables and given/known data Suppose the G is a group of order p^n, where p is prime, and G has exactly one subgroup for each divisor of p^n. Show that G is cyclic. 2. Relevant equations 3. The attempt at a solution First, the inductive basis holds for n=1. Now for each k>=1, we assume that the claim is true for a group of order p^k where k < n. We shall show that the claim is also true for a group of order p^n. Note that in a general case, the group of order p^n is not necessarily abelian. However, by the given inductive hypotheis, we show that the above group G of order p^n is abelian. We know that Z(G) is non-trivial for every nontrivial finite p-group and is a subgroup of G. From this I am suppose to use the G/Z(G) theroem to show that G/Z(G) is cyclic and thus that G is abelian. However, I don't see how I this is true, for example what if |G|=p^3 and |Z(G)|=p, then G/Z(G)= p^2, which does not mean that G/Z(G) is cyclic. (If I remember right G/Z(G) has to be trivial cyclic) From this I can use the Fundamental Theorem of Finite Abelian Groups to get the G is isomorphic to Zp+Z(p^n-1) or Zp and then conclude that it is isomorphic to Zp.