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Chapter 9: Q 18 E (page 303)

Question: If G is a simple group of order 168, how many Sylow 7-subgroups does G have?

Short Answer

Expert verified

G have 8 Sylow 7-subgroups.

Step by step solution

01

Referring to Corollary 9.16 and Third Sylow Theorem

Corollary 9.16

Let G be a finite group and K is a Sylow p-subgroup for some prime p. Then, K is normal in G if and only if K is the only Sylow p-subgroup.

Third Sylow Theorem

The number of Sylow p-subgroups of finite group G divides $|G|$ and is of the form 1+pk for some non-negative integer k.

Given that G is a simple group of order 168.

02

Finding the Sylow 7-groups of group G

As given, G has an order $168 = 2^{3} . 3.7$ .

So, byThird Sylow Theorem, its Sylow 7-subgroups must be in the form of 1+7k and it must divide 168.

Therefore, Sylow 7-subgrous can be either 1 or 8.

Let us assume that G has only one Sylow 7-subgroup.

Therefore, byCorollary 9.16, it must be a normal subgroup in G.

However, G is a simple group which contradicts the result we got by assuming that G has only one Sylow 7-subgroup.

Therefore, G must have 8 Sylow 7-subgroups.

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Most popular questions from this chapter

Let $n \geq 3$ be a positive integer and let G be the set of all matrices of the forms

$(\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}) o r (\begin{matrix} - 1 & a \\ 0 & 1 \end{matrix})$ with $a \in ℤ_{n}$ .

Prove that G is a group of order 2n under matrix multiplication.

List all abelian groups (up to isomorphism) of the given order:30

Let $N, H$ be subgroups of a group $G$ . $G$ is called the semidirect product of $N$ and $H$ if $N$ is normal in $G$ , $G = N H$ , and $N \cap H = ⟨ e ⟩$ . Show that each of the following groups is the semidirect product of two of its subgroups: $S_{4}$

Question: Prove Theorem 9.23.

Let $N$ be a normal subgroup of $G$ , $a \in G$ , and $C$ the conjugacy class of $a$ in $G$ .

Prove that $a \in N$ if and only if $C ⫅$ $N$ .

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