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Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?
- a)G is always cyclic, but H may not be cyclic.
- b)
Both G and H are always cyclic. - c)G may not be cyclic, but H is always cyclic.
- d)Both G and H may not be cyclic.
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Let G be a group order 6, and H be a subgroup of G such that 1 < |H...
Based on the given information, we know that the order of group G is 6. This means that G has 6 elements.
We also know that H is a subgroup of G. Since H is a subgroup, it must also be a group.
Now, let's consider the possible orders of H. Since H is a subgroup of G, the order of H must divide the order of G. In this case, the order of G is 6.
The possible orders of H are therefore 1, 2, 3, or 6.
If the order of H is 1, then H contains only the identity element of G. However, H cannot be the trivial subgroup since the given condition states that H is nontrivial. Therefore, the order of H cannot be 1.
If the order of H is 2, then H contains 2 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain one more element. However, this is not possible since the order of G is 6, and H can have at most 2 elements. Therefore, the order of H cannot be 2.
If the order of H is 3, then H contains 3 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain two more elements. However, this is not possible since the order of G is 6, and H can have at most 3 elements. Therefore, the order of H cannot be 3.
If the order of H is 6, then H contains 6 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain five more elements. However, this is not possible since the order of G is 6, and H can have at most 6 elements. Therefore, the order of H cannot be 6.
Since none of the possible orders of H satisfy the given condition, it is not possible to have a subgroup H of G such that the order of H is 1.
We also know that H is a subgroup of G. Since H is a subgroup, it must also be a group.
Now, let's consider the possible orders of H. Since H is a subgroup of G, the order of H must divide the order of G. In this case, the order of G is 6.
The possible orders of H are therefore 1, 2, 3, or 6.
If the order of H is 1, then H contains only the identity element of G. However, H cannot be the trivial subgroup since the given condition states that H is nontrivial. Therefore, the order of H cannot be 1.
If the order of H is 2, then H contains 2 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain one more element. However, this is not possible since the order of G is 6, and H can have at most 2 elements. Therefore, the order of H cannot be 2.
If the order of H is 3, then H contains 3 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain two more elements. However, this is not possible since the order of G is 6, and H can have at most 3 elements. Therefore, the order of H cannot be 3.
If the order of H is 6, then H contains 6 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain five more elements. However, this is not possible since the order of G is 6, and H can have at most 6 elements. Therefore, the order of H cannot be 6.
Since none of the possible orders of H satisfy the given condition, it is not possible to have a subgroup H of G such that the order of H is 1.
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Let G be a group order 6, and H be a subgroup of G such that 1 < |H...
Consider S6 to be that group now it's proper normal subgroup is A6 and cardiniality of it is 360
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Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?a)G is always cyclic, but H may not be cyclic.b)Both G and H are always cyclic.c)G may not be cyclic, but H is always cyclic.d)Both G and H may not be cyclic.Correct answer is option 'C'. Can you explain this answer? for Mathematics 2025 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?a)G is always cyclic, but H may not be cyclic.b)Both G and H are always cyclic.c)G may not be cyclic, but H is always cyclic.d)Both G and H may not be cyclic.Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?a)G is always cyclic, but H may not be cyclic.b)Both G and H are always cyclic.c)G may not be cyclic, but H is always cyclic.d)Both G and H may not be cyclic.Correct answer is option 'C'. Can you explain this answer?.
Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?a)G is always cyclic, but H may not be cyclic.b)Both G and H are always cyclic.c)G may not be cyclic, but H is always cyclic.d)Both G and H may not be cyclic.Correct answer is option 'C'. Can you explain this answer? for Mathematics 2025 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?a)G is always cyclic, but H may not be cyclic.b)Both G and H are always cyclic.c)G may not be cyclic, but H is always cyclic.d)Both G and H may not be cyclic.Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?a)G is always cyclic, but H may not be cyclic.b)Both G and H are always cyclic.c)G may not be cyclic, but H is always cyclic.d)Both G and H may not be cyclic.Correct answer is option 'C'. Can you explain this answer?.
Solutions for Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?a)G is always cyclic, but H may not be cyclic.b)Both G and H are always cyclic.c)G may not be cyclic, but H is always cyclic.d)Both G and H may not be cyclic.Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
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ample number of questions to practice Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?a)G is always cyclic, but H may not be cyclic.b)Both G and H are always cyclic.c)G may not be cyclic, but H is always cyclic.d)Both G and H may not be cyclic.Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Mathematics tests.
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