Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian

ISI SOLUTION PAPER & TIFR SOLUTION PAPER ON ABSTRACT ALGEBRA

SOLVED PROBLEMS ON ABSTRACT ALGEBRA 

Q1. Which of the statement is FALSE
A. Any abelian group of order 16 is cyclic      
B.Any abelian group of order 12 is cyclic
C.Any abelian group of order 35 is cyclic        
D.Any abelian group of order 49 is cyclic
Ans:(B) Any abelian group is said to be cyclic if the order of the abelian group is prime or power of prime or product of two prime. Here 49=7^2,7 is prime;    16=2^4,2 is prime;     35=5*7,5,here 5,7 are prime . . . .So any abelian group of order 12 is not cyclic. 

Q2. A Cyclic group of order 60 has no of generators
A.12              B.15               C.16                D.20
Ans:(C) Use the concept of phi function.  phi(60)=60(1-1/2)(1-1/3)(1-1/5)=16.
Here phi(n)=n(1-1/a)(1-1/b) . . . (1-1/p),where a,b, . . .,p are primes.60=2^2 *3*5

Q3. In the ring Z/8Z, the equation x^2=1 has exactly exactly solutions
A.2        B.4          C.1             D.none
Ans:(B) x={0,1,3,,4,5,6,7}
Here x=1,3,5,7 satisfies the equation x^2 -1 is divisible by 8. So there are 4 solutions.

Q4. There groups Z9 & Z3*Z3 are
A. isomorphic       B. abelian      C. non-abelian      D. cyclic
Ans: (A) The order of  Z9 & Z3*Z3 is 9. It may corresponds each member of  Z3*Z3 to any member of Z9. So, they are isomorphic.

Q5. T/F : The symmetric group F5 consisting of permutations on 5 symbols has an element of order 6.
Ans : TRUE   O(F5)=5!=120 So, it could be possible for any a€F5 s.t. a^6=e ,so any element of order 6 will be in F5.

Q6. Let G be a group of order 30 & A,B be the subgroup of G of order 2 & 5 respectively. Then O(G/AB) is
A.1        B.3       C.5         D.none
Ans : (B) O(A)=2,O(B)=5,since 2 & 5 are relatively prime so O(AB)=O(A)O(B)=10, then O(G/AB)=30/10=3.

Q7.Let G={ Z € C : |Z| = 1 }. Then under multiplication of complex numbers G is a
A.finite group         B.infinite group         C.cyclic group           D.none
Ans : (C)  G is closed under multiplication & there exists inverse of every element in G. Hence G is multiplicative  group, i.e. Cyclic.

Q8. T/F : If p is an odd prime then a group of order 2p can't be cyclic.
Ans : FALSE   The group of order 6 is cyclic where p=3 is a prime.

Q9.The no of subgroups of a cyclic group of order 100 is
A.9        B.10        C.2           D.none
Ans : (A) Tool kit: If n =p1^a +p2^b + . . . +pk^k  is the prime factorisation then the total no of subgroups=(a+1)(b+1) . . .(k+1),where pi 's are primes.
Here 100=2^2*5^2,so ans is=(2+1)(2+1)=9.

Q10. T/F : S7 has a subgroup of order 11.
Ans : FALSE   Let H be a subgroup of order 11,then 11 divides |S7|=7! ,but this is a contradiction.So,there is no subgroup of order 11.

Q11. The irreducible polynomials in C[x] are the polynomials of degree
A.0        B.1        C.2         D.none
Ans : (B)  The polynomial of degree 0 are the invertible elements of C[x] . By the fundamental theorem of algebra, any polynomial of positive degree has a root in C & hence a linear factor. Therefore any polynomials of degree more than 1 are reducible & those of degree 1 are irreducible.

Q12. Any group G of order 2p,where p is a prime number has a normal subgroup of order p,then the index of subgroup H in G is
A.p         B.2        C.1         D.none
Ans : (B) Given that order of G=2p. By Cauchy's theorem G has an element of order p(since p is prime),then the cyclic group H={a,a^2, . . . ,a^p} is a subgroup of order p. The index of H in G is=O(G)/O(H)=2p/p=2.

Q13. If f : G->H be a group homomorphism from a group G into a group H with kernel k. If O(G)=75,  O(H)=45,  O(K)=15, then the order of the image  f(G) is
A.3        B.5        C.15       D.45
Ans : (B)O{f(G)}=O(G/K)=O(G)/O(K)=75/15=5.

Q14.T/F : Set S={-2,-1,1,2} forms a group w.r.t. multiplication.
Ans : FALSE  Since (-2)(2)=-4 does not belong to S. So,closure property is not satisfied. Hence,S is nt a group.

Q15. T/F: Let G be a finite group of order n & let a€H then a^n=e where H is a subgroup of G.
Ans : TRUE ,By Lagrange's theorem,the order of each subgroup of a finite group is a divisor of the order of the group. Implying O(H)=m then m|n.
For each a€H, a^m=e implying a^n=e.

Notations have usual meaning followed in algebra:
"Z" denotes set of integers & "C" denotes set of complex numbers. O(G) means order of the group G.