Explanation Given information: a and b are elements of a finite order in a group such that a b = b a and 〈 a 〉 ∩ 〈 b 〉 = { e } . Formula used: Definition of a cyclic group: G is a cyclic group if there is an element a ∈ G such that G = 〈 a 〉 Proof: Let a and b be elements of finite order in a group G such that a b = b a and 〈 a 〉 ∩ 〈 b 〉 = { e } If | a | = m and | b | = n , then a m = e and a n = e Let | a b | = r then ( a b ) r = e . But ( a b ) r = a r b r ( ∵ a b = b a ) This gives a r b r = e ⇒ a r = b − r a r ∈ 〈 a 〉 ⇒ b − r ∈ 〈 a 〉 and b − r ∈ 〈 b 〉 ⇒ a r ∈ 〈 b 〉 ⇒ a r = b − r ∈ 〈 a 〉 ∩ 〈 b 〉 = { e } ⇒ a r = b − r = e Therefore, | a | divides r , i.e., m | r and | b | divides r , i

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Author: Gilbert, Linda, Jimmie

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Chapter 3.4, Problem 36E

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To prove: Let a and b be elements of finite order in a group such that ab=ba and 〈a〉∩ 〈b〉={e}, then |ab| is the least common multiple of |a| and |b|.

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Chapter 3.4, Problem 35E

Chapter 3.4, Problem 37E

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To prove: Let a and b be elements of finite order in a group such that a b = b a and 〈 a 〉 ∩ 〈 b 〉 = { e } , then | a b | is the least common multiple of | a | and | b | .

Solution Summary: The author explains that a and b are elements of finite order in the group G.

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To prove: Let a and b be elements of finite order in a group such that ab=ba and 〈a〉∩ 〈b〉={e}, then |ab| is the least common multiple of |a| and |b|.

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