Theorem 24.2 p-Groups Have Nontrivial Centers Let G be a nontrivial finite group whose order is a power of a prime p. Then Z(G) has more than one element. PROOF First observe that cl(a) = {a} if and only if a E Z(G) (see Exercise 4). Thus, by culling out these elements, we may write the class equation in the form IGI = IZ(G)I + EIG:C(a)l, where the sum runs over representatives of all conjugacy classes with more than one element (this set may be empty). But |G:C(a)| = |G|/|C(a)l, so each term in EIG:C(a)) has the form pk with k > 1. Hence, IGI – 2IG:C(a)| = IZ(G)I, where each term on the left is divisible by p. It follows, then, that p also divides IZ(G)I, and hence [Z(G)| # 1.

Algebra and Trigonometry (6th Edition)
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ISBN:9780134463216
Author:Robert F. Blitzer
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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Determine which of the equations below could be the class equation
given in the proof of Theorem 24.2. For each part, provide your
reasoning.
a. 9 = 3 + 3 + 3
b. 21 = 1 + 1 + 3 + 3 + 3 + 3 + 7
c. 10 = 1 + 2 + 2 + 5
d. 18 = 1 + 3 + 6 + 8

Theorem 24.2 p-Groups Have Nontrivial Centers
Let G be a nontrivial finite group whose order is a power of a prime p.
Then Z(G) has more than one element.
PROOF First observe that cl(a) = {a} if and only if a E Z(G) (see
Exercise 4). Thus, by culling out these elements, we may write the class
equation in the form
IGI = IZ(G)I + EIG:C(a)l,
where the sum runs over representatives of all conjugacy classes with
more than one element (this set may be empty). But |G:C(a)| = |G|/|C(a)l,
so each term in EIG:C(a)) has the form pk with k > 1. Hence,
IGI – 2IG:C(a)| = IZ(G)I,
where each term on the left is divisible by p. It follows, then, that p also
divides IZ(G)I, and hence [Z(G)| # 1.
Transcribed Image Text:Theorem 24.2 p-Groups Have Nontrivial Centers Let G be a nontrivial finite group whose order is a power of a prime p. Then Z(G) has more than one element. PROOF First observe that cl(a) = {a} if and only if a E Z(G) (see Exercise 4). Thus, by culling out these elements, we may write the class equation in the form IGI = IZ(G)I + EIG:C(a)l, where the sum runs over representatives of all conjugacy classes with more than one element (this set may be empty). But |G:C(a)| = |G|/|C(a)l, so each term in EIG:C(a)) has the form pk with k > 1. Hence, IGI – 2IG:C(a)| = IZ(G)I, where each term on the left is divisible by p. It follows, then, that p also divides IZ(G)I, and hence [Z(G)| # 1.
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