Suppose G is a finite Abelian group that does not contain a

Suppose G is a finite Abelian group that does not contain a subgroup isomorphic to Zp Zp for any prime p. Prove that G is cyclic.

Solution

G is the direct product of its Sylow subgroups. If each is cyclic, then so is G, using repeatedly the fact Zn × Zm = Znm when gcd(n, m) = 1. Thus, we may assume G has a p-Sylow subgroup P which is not cyclic. By the fundamental theorem, P is a direct product of cyclic groups. Say P = ha1i × · · · × hari for some ai G. By Lagrange each has order a power of p, and clearly we can assume none have order 1. Furthermore, r 2 since P is not cyclic. If ai has order p ei , then the subgroup ha e11 1 , ae21 2 i is then a group generated by two elements of order p, hence is isomorphic to Z)p × Zp. It is a subgroup of G, and so is the desired subgroup. Conversely, if G contains a subgroup isomorphic to Zp × Zp for some prime, then G is not cyclic because, if it were, each subgroup would be cyclic, but this is false since Zp × Zp is not cyclic. Thus, G is not cyclic. But here G does not contain Zp × Zp which proves G is cyclic.