Group of Order 315

syme.gabriel · #1 $Old$ October 8th, 2008, 06:19 PM

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

ThePerfectHacker · #2 $Old$ October 10th, 2008, 02:46 PM

Quote:

Originally Posted by syme.gabriel $View Post$

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

I cannot finish this

. Maybe someone can finish this.

Let $C = G'$ , the commutator subgroup. To show $G$ is abelian it is sufficient to show $|C| = 1$ . Let $H$ be the unique Sylow $3$ -subgroup, then it is normal. Construct the group $G/H$ and note $|G/H| = 5\cdot 7$ . Since $7\not \equiv 1 (\bmod 5)$ it means $G/H$ is an abelian group. This implies that $C$ is contained in $H$ . But since $|H| = 3^2$ it means by Lagrange's theorem $|C| = 1,3,9$ .

Lemma: If $G$ is a group with $|G| = 3^2\cdot 7=63$ then $G$ is abelian.

Proof: There is exactly one Sylow $3$ -subgroup and exactly one Sylow $7$ -subgroup. Call them $M_1,M_2$ . Then $|M_1\cap M_2| = 1$ therefore $|M_1M_2| = 63$ i.e. $M_1M_2 = G$ . Furthermore, both $M_1,M_2$ are normal subgroups. It follows by isomorphism theorems that $G \simeq M_1 \times M_2 \simeq \mathbb{Z}_3 \times \mathbb{Z}_7$ .

Consider the Sylow $7$ -subgroups. By Sylow theorems there are either $1$ or $15$ of them. If there is just one, call it, $N$ then form $G/N$ . It follows that $|G/N| = 63$ . This is abelian by lemma. Thus, $N$ contains $C$ . This means by Lagrange's theorem $|C| = 1,7$ . But by the first paragraph it would force $|C|=1$ . Therefore, $G$ abelian. Thus, it is safe to assume that there are $15$ Sylow $7$ -subgroups.

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