Question

5-4.8. Show that an abelian group is the direct product of its p-Sylow subgroups for primes p dividing G

Answer 1

Solution. We proceed by induction on the number of prime factors in the order of the group. If G has order p n for some prime p and integer n, then G equals its Sylow p-subgroup, so is trivially the product of its Sylow subgroups. Assume that any abelian group with fewer than m distinct prime factors is isomorphic to the product of its Sylow subgroups, and let G be abelian with |G| = p i1 1 ... p im m with the pj’s distinct primes. Let H be the subgroup of G generated by the Sylow pj-subgroups, for 1 ≤ j ≤ m−1. Since G is abelian, every element of H has order prime to pm: H is generated by elements of order p k j for 1 ≤ j ≤ m−1, and the order of the product of commuting elements is the least common multiples of the orders of the factors. Thus the elements of H have order dividing p i1 1 ··· p im−1 m−1 . Because H has as subgroups the Sylow pj-subgroups of G, its order must be at least p i1 1 ... p im−1 m−1 . Since |H| divides |G| and since H has no elements of order p k m for any k, the order of H must be exactly p i1 1 ... p im−1 m−1 . By induction, then, this subgroup is isomorphic to the product of its Sylow subgroups: H ∼= P1 ×··· ×Pm−1, where Pj is the Sylow pj-subgroup. Now consider H together with Pm, the Sylow pm-subgroup. Since G is abelian, each of these subgroups is normal. By order considerations, the intersection of these subgroups is {1} – every element in Pm has order p i m for some i, while all the elements in H (except for 1) have order prime to pm. Finally, a counting argument shows that HPm = G: the elements {hk : h ∈ H, k ∈ Pm} are distinct, and this set has size |G|. Thus by the recognition theorem for direct products, G ∼= H ×Pm ∼= P1 ×··· ×Pm−1 ×Pm.