Mathematics Exam > Mathematics Questions > G is a group of order 51. Then which one of t...
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G is a group of order 51. Then which one of the following statements is false?
- a)All proper subgroups of G are cyclic.
- b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.
- c)G must have an element of order 17.
- d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.
Correct answer is option 'D'. Can you explain this answer?
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G is a group of order 51. Then which one of the following statements i...
Given G is a group of order 51.
i.e. o(G) = 51 = 3 x 17
All subgroup of group is cyclic.
The subgroups are of odd order so each element will have its inverse different from itself.
So, If G is abelian then there exists no proper subgroup H of G such that product of all elements of H is identity is false.
i.e. o(G) = 51 = 3 x 17
All subgroup of group is cyclic.
The subgroups are of odd order so each element will have its inverse different from itself.
So, If G is abelian then there exists no proper subgroup H of G such that product of all elements of H is identity is false.
Most Upvoted Answer
G is a group of order 51. Then which one of the following statements i...
Statement: G is a group of order 51.
False Statement: If G is abelian then there exists no subgroup H of G such that the product of all elements of H is the identity.
To prove this false statement, let's consider the given group G of order 51.
Proof:
1. All proper subgroups of G are cyclic:
This statement is true for any group of prime order. Since 51 is not a prime number, G cannot have all proper subgroups as cyclic. Hence, option (a) is true.
2. If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic:
This statement is true according to the Fundamental Theorem of Cyclic Groups. If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic. Therefore, option (b) is true.
3. G must have an element of order 17:
This statement is true according to Cauchy's Theorem. Since 17 is a prime divisor of the order of G, G must have an element of order 17. Hence, option (c) is true.
4. If G is abelian then there exists no subgroup H of G such that the product of all elements of H is the identity:
This statement is false. Let's prove it by contradiction.
Assume that G is an abelian group and there exists a subgroup H of G such that the product of all elements of H is the identity.
Since G is abelian, the product of all elements of H can be written as:
(1) h1 * h2 * h3 * ... * hk = e, where e is the identity element of G.
Since H is a subgroup of G, it is closed under the group operation. Therefore, the product of any two elements in H will also be in H.
Now, consider the element h1 in H. Since H is closed under the group operation, h1 * h1 = h2 is also in H.
Similarly, h2 * h2 = h3, and so on.
Continuing this process, we can deduce that h1 * h1 * h1 * ... * h1 = hk is also in H.
But hk is the identity element of G, which means H contains the identity element.
Therefore, the product of all elements of H is the identity element of G.
Hence, the false statement is option (d).
In conclusion, the false statement is option (d).
False Statement: If G is abelian then there exists no subgroup H of G such that the product of all elements of H is the identity.
To prove this false statement, let's consider the given group G of order 51.
Proof:
1. All proper subgroups of G are cyclic:
This statement is true for any group of prime order. Since 51 is not a prime number, G cannot have all proper subgroups as cyclic. Hence, option (a) is true.
2. If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic:
This statement is true according to the Fundamental Theorem of Cyclic Groups. If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic. Therefore, option (b) is true.
3. G must have an element of order 17:
This statement is true according to Cauchy's Theorem. Since 17 is a prime divisor of the order of G, G must have an element of order 17. Hence, option (c) is true.
4. If G is abelian then there exists no subgroup H of G such that the product of all elements of H is the identity:
This statement is false. Let's prove it by contradiction.
Assume that G is an abelian group and there exists a subgroup H of G such that the product of all elements of H is the identity.
Since G is abelian, the product of all elements of H can be written as:
(1) h1 * h2 * h3 * ... * hk = e, where e is the identity element of G.
Since H is a subgroup of G, it is closed under the group operation. Therefore, the product of any two elements in H will also be in H.
Now, consider the element h1 in H. Since H is closed under the group operation, h1 * h1 = h2 is also in H.
Similarly, h2 * h2 = h3, and so on.
Continuing this process, we can deduce that h1 * h1 * h1 * ... * h1 = hk is also in H.
But hk is the identity element of G, which means H contains the identity element.
Therefore, the product of all elements of H is the identity element of G.
Hence, the false statement is option (d).
In conclusion, the false statement is option (d).
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G is a group of order 51. Then which one of the following statements is false?a)All proper subgroups of G are cyclic.b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.c)G must have an element of order 17.d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.Correct answer is option 'D'. Can you explain this answer?
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G is a group of order 51. Then which one of the following statements is false?a)All proper subgroups of G are cyclic.b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.c)G must have an element of order 17.d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.Correct 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 G is a group of order 51. Then which one of the following statements is false?a)All proper subgroups of G are cyclic.b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.c)G must have an element of order 17.d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.Correct 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 G is a group of order 51. Then which one of the following statements is false?a)All proper subgroups of G are cyclic.b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.c)G must have an element of order 17.d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.Correct answer is option 'D'. Can you explain this answer?.
G is a group of order 51. Then which one of the following statements is false?a)All proper subgroups of G are cyclic.b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.c)G must have an element of order 17.d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.Correct 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 G is a group of order 51. Then which one of the following statements is false?a)All proper subgroups of G are cyclic.b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.c)G must have an element of order 17.d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.Correct 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 G is a group of order 51. Then which one of the following statements is false?a)All proper subgroups of G are cyclic.b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.c)G must have an element of order 17.d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.Correct answer is option 'D'. Can you explain this answer?.
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ample number of questions to practice G is a group of order 51. Then which one of the following statements is false?a)All proper subgroups of G are cyclic.b)If G has only one subgroup of order 3 and only one subgroup of order 17, then G is cyclic.c)G must have an element of order 17.d)If G is abelian then there exists no ?? subgroup Hot G such that product of all elements of H is identity.Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Mathematics tests.
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