Division of Marks TOTAL [3 +2] Part -A Part-B Part-C
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1 GROUPS
2 Division of Marks MARK MARKS MARKS [3 +2] TOTAL Part -A Part-B Part-C
3 Set of Integers I or Z= Set Of Non negative integers ={0,1,2,3,4, ,,,, } Set Of Rationals= Q=
4 Set Of Reals=R=
5 Q. [1M]Define a binary operation Ans: Let S be a nonempty set.* is said to be binary operation on S if
6 Q. (1m)a * b = Where a and b are the elements of R Is * a binary operation on R? Ans: If a=1, b=2 then a *b = Therefore * is not a binary operation
7 Q. a * b =, Where a and b are the elements of R. Is * a binary operation on R? Ans:Let a = 2, b = -7 then * is not a binary operation
8 Q. a * b =, Where a and b are the elements of C. Is * a binary operation on C? Ans: Let a=1, b= i then * is not a binary operation
9 Q. In the set of all integers I, Ifa*b=a a b,prove that * is not a binary operation. Ans: Let a = 2 and b = -3 then Therefore * is not a binary operation
10 Q. In the set of non negative integers * is defined by a*b=a. b Verify whether * is a binary operation or not. Ans: Let a = 0 and b = 0 then which is undefined Therefore * is not a binary operation
11 Q. On I, prove that * is not a b.o Ans: Let a = 1, b = 2 then
12 Q. State Closure law. Ans: Closure law A b.o * on a non empty set G is said to satisfy closure law if
13 Q. State Commuatative law. Ans: Commuative law A b.o * on a non empty set G is said to satisfy commutative law if
14 Q. On Z(orI) I), a * b = a + b +2, find the Identity element. Ans: Let e be the identity By definition a*e=e*a=a ea a Now a*e=a a+e+2 +2 =a e=-2 Z
15 Q. On the set of all rationals, * is defined by Show that * is not associative.
16 a*(b*c)= a*x where x = b*c= = --- (1)
17 (a*b)*c=y*c where y=a*b= = ---(2) From (1) and (2) * is not associative
18 Q. If * is defined by, a, b R, Find the identity element in R under * Ans:
19 Q. If * is defined by, find the identity if it exists. Ans: a*e = a e-1 and e*a = e a-1 i.e a*e e*a Therefore identity does not exist
20 Q. On Q, a *b =, Find the identity and inverse of 8
21 Ans: Let e be the identity By defn. a*e=a Let a -1 be the inverse of a a *a -1 =e, a G
22 Q. On Q {-1}, define a * b = a +b + ab Find the identity and the inverse of 2 Ans: a*e= a
23
24 Q. Define a group. Give an example.
25 Ans: A non empty set G together th with a binary operation * is said to be a group if * satisfies closure associative,identity and inverse laws. Ex: (I, +)
26 Q. Define a semi group. Give an example of a semigroup which is not a group.
27 Ans: A non empty set G together th with a binary operation * is said to be a semigroup if * satisfies closure and associative laws. Ex: (N, +)
28 Q. Define a subgroup. Give an example of it. Ans: A non empty set His said to be a subgroup of a group (G,*) if i) H G ii) (H,*) is itself a group. Ex: {1,-1} is a subgroup of {1,-1,i,-i} under multiplication.
29 Q. Prove that the identity element is unique in a group.
30 Ans: Let (G,*) be a group and a G Suppose e 1 and e 2 are two identities. Now e 1 is the identity Now e 2 is the identity From (1) and (2) The identity is unique
31 Q. Prove that in a group the inverse of an element is unique.
32 Ans: Let (G,*) be a group and a G Suppose b and c are two inverses of a Now b is the inverse of a a * b = b * a = e (1) Now c is the inverse of a a * c=c* c a=e (2) From (1) and (2) a*b=a*c a * b=c [ By Left cancellation law] The inverse is unique
33 Q. Show that the set (Z,*), where ee Z is the set of all integers and * is defined by a*b=a+b+1 a,b Z, is an abelian group. Solution: Closure law: Let a,b Z. Clearly, a+b+1 is again an element of Z. Thus a,b Z, a*b=a+b+1 a b Z
34 Associative law: Let a,b,c Z. Consider, a*(b*c) = a*x where x=b*c=b+c+1 =a+x+1=a+(b+c+1)+1 = a+b+c+2
35 And (a*b )*c = y*c where y=a*b=a+b+1 =y+c+1=(a+b+1)+c+1 = a+b+c+2 a*(b*c)=(a*b)*c ( )
36 Identity ty law: Let e be the identity a*e=a ae=a a+e+1=a e=-1 1 Inverse law: 1 Let a -1 be the inverse of a Then a*a -1 =e a+a -1 +1=-1 a -1 =-2-a Z
37 Commutative law: a,b Z, we have a*b=a+b+1=b+a+1=b*a Hence (Z,*) is an abelian group.
38 Q. Prove that the set of fourth roots of unity ie, G = {1, -1, i, -i} forms finite abelian group under multiplication Sol: G={1,-1,i,-i} 1i i} 1-1 i -i i -i i i i i -i i -i i 1-1
39 Closure law: All the entries in the table are in G Closure is valid Associative law: 1,-1 and i G 1 (-1 i)=- i (1-1) i=- i Associative is valid
40 Identity law: The row headed d by 1 is same as the top most row, 1 is the identity Inverse law: From the table 1-1 =1, (-1) -1 =-1 i -1 =-i and (-i) -1 =i
41 Commutative law: The table is symmetric about principal diagonal Hence (G, ) is an abelian group.
42 Q. Show that the set is a group under matrix multiplication.
43 Ans: Closure law. Let A= B = Then AB= closure law is valid in M
44 Associative law: We know that the matrix multiplication is associative. Associative law is valid in M. Identity law: Clearly I = is the identity and A I=I A=A
45 Inverse law: Let A= clearly hence A is non singular A -1 exists A -1 = Hence M is a group under multiplication.
46 Q.Prove that a non empty subset H of a group G is a subgroup of G if and only if (i) a, b H a b H 1 (ii) a H a -1 H
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