Abstract Algebra 2023 – BSc Computer Science Part 2
Total No. of Questions : 9] [Total No. of Printed Pages: 4
Paper code: 13511
1511
B.Sc. (Computer Science) (Part 2)
Examination-2023
Paper No. 1.2
ABSTRACT ALGEBRA
Time: Three Hours] [Maximum Marks: 50
Note: Attempt five questions in all selection at least one question from each section. All questions carry equal marks.
Section-A
1. (a) Prove that the set of all cosets of a normal subgroup is a group with respect to multiplication of complexes as composition.
(b) If H is a normal subgroup of a solvable group G, then the quotient group G/H is solvable.
2. (a) State and prove Cayley’s theorem.
(b) Prove that every subgroup of a cyclic group in cyclic.
3. (a) State and prove fundamental theorem of homomorphism of groups.
(b) Show that every permutation can be ex-pressed as product of disjoint cycles.
Section-B
4. (a) Show that the set of matrices is a subring of the ring of 2 x 2 matrices with integral elements.
(b) Prove that the ring of integers is a principle ideal ring.
5. (a) Prove that a field has no proper ideals.
(b) Show that every finite integral domain is a field.
6. (a) State and prove unique factorization theorem.
(b) Show that the set R={0, 1, 2, 3, 4, 5} is commutative ring with respect to ‘+6‘ and ‘X6‘ as two ring compositions.
7. (a) Prove that if two vectors are linearly dependent one of them is a scalar multiple of the other.
(b) Prove that the necessary and sufficient condition that the non-zero element a in the Euclidean ring R is a unit is that d(a) = d(1).
8. (a) Show that the three vectors (1, 1, -1), (2, -3, 5) and (-2, 1, 4) of R3 are linearly independent.
(b) If W be a subspace of a finite dimensional vector space, then show that :
9. (a) Define basis of a vector space and show that there exists a basis for each finite dimensional vector space.
(b) Define direct sum of two subspaces and show that the necessary and sufficient conditions for a vector space V(F) to be a direct sum of its two subspaces W1 and W2 are that-
1. v = w1 + w2
2. w1 ∩ w2 = {0}
……..End……..
Thank you 🙂