Thread: Group of Order 315

Let be a group of order 315 and contains a normal Sylow subgroup. Show that must be abelian.

Originally Posted by syme.gabriel

Let be a group of order 315 and suppose that contains a normal 3-Sylow subgroup. Prove that is abelian.

I cannot finish this . Maybe someone can finish this.

Let , the commutator subgroup. To show is abelian it is sufficient to show . Let be the unique Sylow -subgroup, then it is normal. Construct the group and note . Since it means is an abelian group. This implies that is contained in . But since it means by Lagrange's theorem .

Lemma: If is a group with then is abelian.

Proof: There is exactly one Sylow -subgroup and exactly one Sylow -subgroup. Call them . Then therefore i.e. . Furthermore, both are normal subgroups. It follows by isomorphism theorems that .

Consider the Sylow -subgroups. By Sylow theorems there are either or of them. If there is just one, call it, then form . It follows that . This is abelian by lemma. Thus, contains . This means by Lagrange's theorem . But by the first paragraph it would force . Therefore, abelian. Thus, it is safe to assume that there are Sylow -subgroups.

But I do not know how to show that having 15 is impossible.