Mathematics Exam > Mathematics Questions > Which of the following is correct?a)Every sub...
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Which of the following is correct?
- a)Every subgroup of an abelian group is normal
- b)Every subgroup of a cyclic group is normal
- c)Intersection of any two normal subgroup is a normal subgroup
- d)If N is a normal subgroup of G and H is any subgroup of G then NH is normal subgroup of G.
Correct answer is option 'A,B,C'. Can you explain this answer?
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Verified Answer
Which of the following is correct?a)Every subgroup of an abelian group...
Every subgroup of an abelian group is normal as for every x ∈ G and h ∈ H
Every subgroup of cyclic group is normal and intersection of two normal subgroups is normal subgroup again.
If N is a normal subgroup of G and H is any subgroup of G then NH is a subgroup of G but NH is necessarily not normal.
If N is a normal subgroup of G and H is any subgroup of G then NH is a subgroup of G but NH is necessarily not normal.
Most Upvoted Answer
Which of the following is correct?a)Every subgroup of an abelian group...
Every subgroup of an abelian group is normal (Option A):
- An abelian group is a group in which the group operation is commutative. This means that for any elements a and b in the group, ab=ba.
- A subgroup H of a group G is normal if and only if for every element g in G, gHg^(-1) is a subset of H, where g^(-1) represents the inverse of g.
- In an abelian group, the group operation is commutative, so for any element g in G and any subgroup H of G, gHg^(-1) = gg^(-1)H = H. Therefore, every subgroup of an abelian group is normal.
Every subgroup of a cyclic group is normal (Option B):
- A cyclic group is a group that is generated by a single element. This means that every element in the group can be written as a power of the generator.
- Let G be a cyclic group generated by the element a, and let H be a subgroup of G. Since G is cyclic, every element in G can be written as a^n, where n is an integer. Similarly, every element in H can be written as a^m, where m is an integer.
- Now, consider an element g in G and an element h in H. We want to show that ghg^(-1) is in H. Since G is cyclic, g can be written as a^k for some integer k. Therefore, ghg^(-1) = a^ka^ma^(-k) = a^ma^k(a^(-k)) = (a^ma^(-m))^k = (a^m)^k = (a^k)^m.
- Since H is a subgroup of G, (a^k)^m is in H. Therefore, ghg^(-1) is in H, and H is normal in G.
Intersection of any two normal subgroups is a normal subgroup (Option C):
- Let N1 and N2 be normal subgroups of a group G. We want to show that the intersection of N1 and N2, denoted by N1 ∩ N2, is also a normal subgroup of G.
- Consider an element g in G and an element n in N1 ∩ N2. Since n is in N1 ∩ N2, it is in both N1 and N2. Since N1 and N2 are normal subgroups, we have gng^(-1) is in N1 and gng^(-1) is in N2.
- Therefore, gng^(-1) is in both N1 and N2, so it is in the intersection N1 ∩ N2. This shows that N1 ∩ N2 is closed under the group operation.
- Additionally, since N1 and N2 are both normal subgroups, we have g(N1 ∩ N2)g^(-1) = (gN1g^(-1)) ∩ (gN2g^(-1)) = N1 ∩ N2. This shows that N1 ∩ N2 is normal in G.
Therefore, options A, B, and C are all correct.
- An abelian group is a group in which the group operation is commutative. This means that for any elements a and b in the group, ab=ba.
- A subgroup H of a group G is normal if and only if for every element g in G, gHg^(-1) is a subset of H, where g^(-1) represents the inverse of g.
- In an abelian group, the group operation is commutative, so for any element g in G and any subgroup H of G, gHg^(-1) = gg^(-1)H = H. Therefore, every subgroup of an abelian group is normal.
Every subgroup of a cyclic group is normal (Option B):
- A cyclic group is a group that is generated by a single element. This means that every element in the group can be written as a power of the generator.
- Let G be a cyclic group generated by the element a, and let H be a subgroup of G. Since G is cyclic, every element in G can be written as a^n, where n is an integer. Similarly, every element in H can be written as a^m, where m is an integer.
- Now, consider an element g in G and an element h in H. We want to show that ghg^(-1) is in H. Since G is cyclic, g can be written as a^k for some integer k. Therefore, ghg^(-1) = a^ka^ma^(-k) = a^ma^k(a^(-k)) = (a^ma^(-m))^k = (a^m)^k = (a^k)^m.
- Since H is a subgroup of G, (a^k)^m is in H. Therefore, ghg^(-1) is in H, and H is normal in G.
Intersection of any two normal subgroups is a normal subgroup (Option C):
- Let N1 and N2 be normal subgroups of a group G. We want to show that the intersection of N1 and N2, denoted by N1 ∩ N2, is also a normal subgroup of G.
- Consider an element g in G and an element n in N1 ∩ N2. Since n is in N1 ∩ N2, it is in both N1 and N2. Since N1 and N2 are normal subgroups, we have gng^(-1) is in N1 and gng^(-1) is in N2.
- Therefore, gng^(-1) is in both N1 and N2, so it is in the intersection N1 ∩ N2. This shows that N1 ∩ N2 is closed under the group operation.
- Additionally, since N1 and N2 are both normal subgroups, we have g(N1 ∩ N2)g^(-1) = (gN1g^(-1)) ∩ (gN2g^(-1)) = N1 ∩ N2. This shows that N1 ∩ N2 is normal in G.
Therefore, options A, B, and C are all correct.
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Community Answer
Which of the following is correct?a)Every subgroup of an abelian group...
Every subgroup of an abelian group is normal as for every x ∈ G and h ∈ H
Every subgroup of cyclic group is normal and intersection of two normal subgroups is normal subgroup again.
If N is a normal subgroup of G and H is any subgroup of G then NH is a subgroup of G but NH is necessarily not normal.
If N is a normal subgroup of G and H is any subgroup of G then NH is a subgroup of G but NH is necessarily not normal.
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Which of the following is correct?a)Every subgroup of an abelian group is normalb)Every subgroup of a cyclic group is normalc)Intersection of any two normal subgroup is a normal subgroupd)If N is a normal subgroup of G and H is any subgroup of G then NH is normal subgroup of G.Correct answer is option 'A,B,C'. Can you explain this answer?
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Which of the following is correct?a)Every subgroup of an abelian group is normalb)Every subgroup of a cyclic group is normalc)Intersection of any two normal subgroup is a normal subgroupd)If N is a normal subgroup of G and H is any subgroup of G then NH is normal subgroup of G.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 of the following is correct?a)Every subgroup of an abelian group is normalb)Every subgroup of a cyclic group is normalc)Intersection of any two normal subgroup is a normal subgroupd)If N is a normal subgroup of G and H is any subgroup of G then NH is normal subgroup of G.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 of the following is correct?a)Every subgroup of an abelian group is normalb)Every subgroup of a cyclic group is normalc)Intersection of any two normal subgroup is a normal subgroupd)If N is a normal subgroup of G and H is any subgroup of G then NH is normal subgroup of G.Correct answer is option 'A,B,C'. Can you explain this answer?.
Solutions for Which of the following is correct?a)Every subgroup of an abelian group is normalb)Every subgroup of a cyclic group is normalc)Intersection of any two normal subgroup is a normal subgroupd)If N is a normal subgroup of G and H is any subgroup of G then NH is normal subgroup of G.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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