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Consider the following statements:
1. A group of order 289 is abelian.
2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.
3. Every proper subgroup of S3 is cyclic.
4. (Z30, has 4 subgroups of order 15 .
Correct statements are:
  • a)
    l and 4
  • b)
    1 ,3 and 4
  • c)
    1 and 2
  • d)
    1 and 3
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Consider the following statements:1. A group of order 289 is abelian.2...
every group of order p2 (p -> prime) is always abelian group.
(1) true
In S3 → total no. of elements are 3 of order 2 i.e. x2 = e and identity element also satisfying this condition total elements are 4. but only 3 elements are exists which satisfies the condition y3 = e.
(3) A3 is proper subgroup of S3 which is cyclic.
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Most Upvoted Answer
Consider the following statements:1. A group of order 289 is abelian.2...
every group of order p2 (p -> prime) is always abelian group.
(1) true
In S3 → total no. of elements are 3 of order 2 i.e. x2 = e and identity element also satisfying this condition total elements are 4. but only 3 elements are exists which satisfies the condition y3 = e.
(3) A3 is proper subgroup of S3 which is cyclic.
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Community Answer
Consider the following statements:1. A group of order 289 is abelian.2...
every group of order p2 (p -> prime) is always abelian group.
(1) true
In S3 → total no. of elements are 3 of order 2 i.e. x2 = e and identity element also satisfying this condition total elements are 4. but only 3 elements are exists which satisfies the condition y3 = e.
(3) A3 is proper subgroup of S3 which is cyclic.
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Consider the following statements:1. A group of order 289 is abelian.2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.3. Every proper subgroup of S3 is cyclic.4. (Z30, has 4 subgroups of order 15 .Correct statements are:a)l and 4b)1 ,3 and 4c)1 and 2d)1 and 3Correct answer is option 'D'. Can you explain this answer?
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Consider the following statements:1. A group of order 289 is abelian.2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.3. Every proper subgroup of S3 is cyclic.4. (Z30, has 4 subgroups of order 15 .Correct statements are:a)l and 4b)1 ,3 and 4c)1 and 2d)1 and 3Correct answer is option 'D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Consider the following statements:1. A group of order 289 is abelian.2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.3. Every proper subgroup of S3 is cyclic.4. (Z30, has 4 subgroups of order 15 .Correct statements are:a)l and 4b)1 ,3 and 4c)1 and 2d)1 and 3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the following statements:1. A group of order 289 is abelian.2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.3. Every proper subgroup of S3 is cyclic.4. (Z30, has 4 subgroups of order 15 .Correct statements are:a)l and 4b)1 ,3 and 4c)1 and 2d)1 and 3Correct answer is option 'D'. Can you explain this answer?.
Solutions for Consider the following statements:1. A group of order 289 is abelian.2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.3. Every proper subgroup of S3 is cyclic.4. (Z30, has 4 subgroups of order 15 .Correct statements are:a)l and 4b)1 ,3 and 4c)1 and 2d)1 and 3Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics. Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.
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