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Exercise 6.3. Let G be a finite group satisfying that for every positive integer m dividing |G|, there exists exactly one subgroup H of G so that |H| = m. Prove or disprove that G is cyclic.

Exercise 6.3. Let G be a finite group satisfying that for every positive integer m dividing |G|, there exists exactly one subgroup H of G so that |H| = m. Prove or disprove that G is cyclic.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter4: More On Groups
Section4.4: Cosets Of A Subgroup
Problem 29E: Let be a group of order , where and are distinct prime integers. If has only one subgroup of...
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Exercise 6.3. Let G be a finite group satisfying that for every positive integer m dividing |G|, there exists exactly one subgroup H of G so that |H| = m. Prove or disprove that G is cyclic.
Transcribed Image Text:Exercise 6.3. Let G be a finite group satisfying that for every positive integer m dividing |G|, there exists exactly one subgroup H of G so that |H| = m. Prove or disprove that G is cyclic.
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