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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.