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Which one of the following is false
- a)There are exactly two abelian groups of order 6 (upto isomorphism).
- b)Every group of order less than 6 is abelian A group of order 439 is non abelian.
- c)Equation x2 = e can not have more than two solution in some group G with identity e.
- d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.
Correct answer is option 'A,B,C'. Can you explain this answer?
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Verified Answer
Which one of the following is falsea)There are exactly two abelian gro...
Option -(A) Order 6 (2 groups: 1 abelian, 1 nonabelian)
Option -(B) We Know that every group of order 2, 3, 4 and 5 are abelian group but 6 order group is abelian not necessary, and If 0(G) = p (prime)
then G must be cyclic and abelian also. Here 0(G) = 439 (prime) then it is an abelian group,
Option - (B) False statement
Option -(C) If G be a group and x∈G then G has more than two solution such that x2 = e where e is identity element.
Most Upvoted Answer
Which one of the following is falsea)There are exactly two abelian gro...
False Statements:
a) There are exactly two abelian groups of order 6 (up to isomorphism).
This statement is false. There are actually three abelian groups of order 6 (up to isomorphism). The three abelian groups of order 6 are:
1. The cyclic group of order 6, denoted by C6.
2. The direct product of two cyclic groups of order 3, denoted by C3 x C3.
3. The direct product of cyclic groups of order 2 and 3, denoted by C2 x C3.
b) Every group of order less than 6 is abelian.
This statement is false. There are non-abelian groups of order less than 6. For example, the symmetric group S3, which is the group of all permutations of three elements, is non-abelian and has order 6.
c) The equation x^2 = e cannot have more than two solutions in some group G with identity e.
This statement is false. In fact, the equation x^2 = e can have more than two solutions in a group G. For example, consider the Klein four-group V, which is the group with four elements {e, a, b, c} such that e is the identity element, and every element is its own inverse. In this group, every element satisfies x^2 = e, so there are four solutions to the equation.
True Statement:
d) A group of order 59 is cyclic, and the smallest prime integer n such that there are two non-isomorphic groups of order n is 4.
This statement is true. A group of prime order is always cyclic, so a group of order 59 is cyclic. Additionally, the smallest prime integer n such that there are two non-isomorphic groups of order n is indeed 4. The two non-isomorphic groups of order 4 are the cyclic group C4 and the Klein four-group V.
a) There are exactly two abelian groups of order 6 (up to isomorphism).
This statement is false. There are actually three abelian groups of order 6 (up to isomorphism). The three abelian groups of order 6 are:
1. The cyclic group of order 6, denoted by C6.
2. The direct product of two cyclic groups of order 3, denoted by C3 x C3.
3. The direct product of cyclic groups of order 2 and 3, denoted by C2 x C3.
b) Every group of order less than 6 is abelian.
This statement is false. There are non-abelian groups of order less than 6. For example, the symmetric group S3, which is the group of all permutations of three elements, is non-abelian and has order 6.
c) The equation x^2 = e cannot have more than two solutions in some group G with identity e.
This statement is false. In fact, the equation x^2 = e can have more than two solutions in a group G. For example, consider the Klein four-group V, which is the group with four elements {e, a, b, c} such that e is the identity element, and every element is its own inverse. In this group, every element satisfies x^2 = e, so there are four solutions to the equation.
True Statement:
d) A group of order 59 is cyclic, and the smallest prime integer n such that there are two non-isomorphic groups of order n is 4.
This statement is true. A group of prime order is always cyclic, so a group of order 59 is cyclic. Additionally, the smallest prime integer n such that there are two non-isomorphic groups of order n is indeed 4. The two non-isomorphic groups of order 4 are the cyclic group C4 and the Klein four-group V.
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Community Answer
Which one of the following is falsea)There are exactly two abelian gro...
Total no. of abelian group upto isomorphism is equal to p(1 ).p(1)
Option -(B) We Know that every group of order 2, 3, 4 and 5 are abelian group but 6 order group is abelian not necessary, and If 0(G) = p (prime)
then G must be cyclic and abelian also. Here 0(G) = 439 (prime) then it is an abelian group,
Option - (B) False statement
Option -(C) If G be a group and x∈G then G has more than two solution such that x2 = e where e is identity element.
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Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer?
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Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. 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 Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer?.
Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. 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 Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer?.
Solutions for Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
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Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer?, a detailed solution for Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer? has been provided alongside types of Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Which one of the following is falsea)There are exactly two abelian groups of order 6 (upto isomorphism).b)Every group of order less than 6 is abelian A group of order 439 is non abelian.c)Equation x2 = e can not have more than two solution in some group G with identity e.d)A group of order 59 is cyclic and the smallest +ve integer n s.t. there are two non - isomorphic groups of order n is 4.Correct answer is option 'A,B,C'. Can you explain this answer? tests, examples and also practice Mathematics tests.
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