Mathematics Exam > Mathematics Questions > Let H and N be subgroup and normal subgroup o...
Start Learning for Free
Let H and N be subgroup and normal subgroup of a group G respectively then
- a)N is normal subgroup of HN.
- b)N is normal subgroup of HN/G.
- c)N is cyclic subgroup.
- d)N is quotient subgroup.
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Let H and N be subgroup and normal subgroup of a group G respectively ...
Suppose e∈ H and for any arbitrary x ∈ N.
Now HN is a subgroup of G, N is a subgroup of G and N ≤ HN therefore N is a subgroup of HN let hx be an arbitary element of HN st.
again x1 be any arbitary element of N then.
..(1)Since N Δ G therefore
Most Upvoted Answer
Let H and N be subgroup and normal subgroup of a group G respectively ...
Explanation:
To prove that option 'A' is correct, we need to show that N is a normal subgroup of HN.
Definition of a normal subgroup:
A subgroup N of a group G is called a normal subgroup if and only if for every element h in H and every element n in N, the product hnh^(-1) is also in N.
Proof:
We need to show that for every element h in H and every element n in N, the product hnh^(-1) is also in N.
Let's take an arbitrary element nh in HN, where n is an element of N and h is an element of H.
Now, let's consider the conjugate of nh by an element h' in H: (h')^(-1)(nh)(h').
Since N is a normal subgroup of G, we know that for every element n' in N and every element h'' in H, the product h''n'h''^(-1) is also in N.
Therefore, in our case, (h')^(-1)(nh)(h') is also in N.
Now, let's consider the product (h')^(-1)(nh)(h')n^(-1).
Since N is a subgroup of G, it is closed under the group operation. Therefore, the product nh and n^(-1) is also in N.
So, (h')^(-1)(nh)(h')n^(-1) is in N.
Now, let's consider the product h'(h')^(-1)(nh)(h')n^(-1).
Since G is a group, it is closed under the group operation. Therefore, the product h'(h')^(-1) is the identity element e.
So, h'(h')^(-1)(nh)(h')n^(-1) = enh'n^(-1) is in N.
Therefore, we have shown that for every element h in H and every element n in N, the product hnh^(-1) is also in N.
Hence, option 'A' is correct.
To prove that option 'A' is correct, we need to show that N is a normal subgroup of HN.
Definition of a normal subgroup:
A subgroup N of a group G is called a normal subgroup if and only if for every element h in H and every element n in N, the product hnh^(-1) is also in N.
Proof:
We need to show that for every element h in H and every element n in N, the product hnh^(-1) is also in N.
Let's take an arbitrary element nh in HN, where n is an element of N and h is an element of H.
Now, let's consider the conjugate of nh by an element h' in H: (h')^(-1)(nh)(h').
Since N is a normal subgroup of G, we know that for every element n' in N and every element h'' in H, the product h''n'h''^(-1) is also in N.
Therefore, in our case, (h')^(-1)(nh)(h') is also in N.
Now, let's consider the product (h')^(-1)(nh)(h')n^(-1).
Since N is a subgroup of G, it is closed under the group operation. Therefore, the product nh and n^(-1) is also in N.
So, (h')^(-1)(nh)(h')n^(-1) is in N.
Now, let's consider the product h'(h')^(-1)(nh)(h')n^(-1).
Since G is a group, it is closed under the group operation. Therefore, the product h'(h')^(-1) is the identity element e.
So, h'(h')^(-1)(nh)(h')n^(-1) = enh'n^(-1) is in N.
Therefore, we have shown that for every element h in H and every element n in N, the product hnh^(-1) is also in N.
Hence, option 'A' is correct.
|
Explore Courses for Mathematics exam
|
|
Similar Mathematics Doubts
Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer?
Question Description
Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. 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 Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer?.
Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. 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 Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer?.
Solutions for Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.
Here you can find the meaning of Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let H and N be subgroup and normal subgroup of a group G respectively thena)N is normal subgroup of HN.b)N is normal subgroup of HN/G.c)N is cyclic subgroup.d)N is quotient subgroup.Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Mathematics tests.
|
|
Explore Courses for Mathematics exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup