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Abstract Algebra

Exercise 4.3.23 (The number of nonidentity elements of $G$ that are contained in conjugates of $M$ is at most $(|M| - 1) |G : M|$)

Recall (cf. Exercise 16, Section 2.4) that a proper subgroup M of G is called maximal if whenever M ≤ H ≤ G , either H = M or H = G . Prove that if M is maximal subgroup of G then either N G ( M ) = M or N G ( M ) = G . Deduce that if M is a maximal subgroup of G that is not normal in G then the number of nonidentity elements of G that are contained in conjugates of M is at most ( | M | − 1 ) | G : M | .

Answers

Proof. Since M ≤ N G ( M ) ≤ G , where M is a maximal subgroup of G , then

N G ( M ) = M or N G ( M ) = G .

If M is a maximal subgroup of G that is not normal in G then N G ( M ) ≠ G , thus

N G ( M ) = M .

The orbit-stabilizer formula shows that the number k of distinct conjugate subgroups 𝑔𝑀 g − 1 of M satisfies

k = | G : N G ( M ) | = | G : M | .

Let g 1 M g 1 − 1 , g 2 M g 2 − 1 , … , g k M g k − 1 be the k distinct conjugate subgroups of M . All this subgroups contains the element 1 , and | M | − 1 non identity elements. Using the general inequality

| ⋃ i = 1 r A i | ≤ ∑ i = 1 r | A i | ,

we obtain

| ( ⋃ g ∈ G 𝑔𝑀 g − 1 ) − { 1 } | = | ⋃ g ∈ G ( 𝑔𝑀 g − 1 − { 1 } ) | = | ⋃ i = 1 k ( g i M g i − 1 − { 1 } ) | ≤ ∑ i = 1 k ( | M | − 1 ) = k ( | M | − 1 ) = ( | M | − 1 ) | G : M | .

If M is a maximal subgroup of G that is not normal in G then the number of nonidentity elements of G that are contained in conjugates of M is at most ( | M | − 1 ) | G : M | . □

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2026-02-09 11:47
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