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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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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