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Math Help - Group of Order 315

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    Let G be a group of order 315 and G contains a normal Sylow subgroup. Show that G must be abelian.
    Last edited by syme.gabriel; October 15th 2008 at 03:43 AM.
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    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.
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