2. In a group G, (ab)³ = a³b² for all a, bEG, and that 3 does not divide order of G. Prove that every element in the group is cube of some element of G, i.e., every element xEG is cube of some element yEG (Thus, x = y³). [Hint: It may be helpful to show that if a³ = b' then a=b.)

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter4: More On Groups
Section4.8: Some Results On Finite Abelian Groups (optional)
Problem 15E: 15. Assume that can be written as the direct sum , where is a cyclic group of order . Prove that...
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2. In a group G, (ab) ³ = a³b³ for all a, bEG, and that 3 does not divide order of
G. Prove that every element in the group is cube of some element of G, i.e.,
every element xEG is cube of some element yEG (Thus, x = y³).
[Hint: It may be helpful to show that if a = b then a=b.)
Transcribed Image Text:2. In a group G, (ab) ³ = a³b³ for all a, bEG, and that 3 does not divide order of G. Prove that every element in the group is cube of some element of G, i.e., every element xEG is cube of some element yEG (Thus, x = y³). [Hint: It may be helpful to show that if a = b then a=b.)
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