Exercise 2.8 (Solutions)
The main topic of this exercise are binary operation, semi-group, monoid, groups and abelian groups. These notes are based on the new Student Learning Outcomes (SLOs). Version: 4.1, Available at MathCity.org
Question # 1 Operation ⊕ performed on the two-member set G={0,1} is shown in the adjoining table. ⊕01011110 Answers the questions:
(i) Name the identity element if it exists?
(ii) What is the inverse of 1?
(iii) Is the set G, under the given operation a group? Abelian or non-abelian?
Solutions
(i) From the given table we have 0+0=0 and 0+1=1.
This show that 0 is the identity element.
(ii) Since 1+1=0 (identity element) so the inverse of 1 is 1.
(iii) It is clear from table that element of the given set satisfies closure law, associative law, identity law and inverse law thus given set is group under ⊕.
Also it satisfies commutative law so it is an abelian group.
Question # 2 The operation ⊕ as performed on the set {0,1,2,3} is shown in the adjoining table, show that the set is an Abelian group?
⊕012300123112302230103012
Solution
Suppose G={0,1,2,3}
i) The given table show that each element of the table is a member of G thus closure law holds.
ii) ⊕ is associative in G.
iii) Table show that 0 is identity element w.r.t. ⊕.
iv) Since 0+0=0, 1+3=0, 2+2=0, 3+1=0,
0−1=0, 1−1=3, 2−1=2, 3−1=1.
v) As the table is symmetric w.r.t. to the principal diagonal. Hence commutative law holds.
Question # 3 (i) Determine whether or not the set of rational number form groups with respect to ×.
Solution
As 0∈Q, multiplicative inverse of 0 in not in set Q. Therefore the set of rational number is not a group w.r.t to “×”.
Question # 3 (ii) Determine whether or not the set of rational number form groups with respect to +.
Solution
a- Closure property holds in Q under + because sum of two rational number is also rational.
b- Associative property holds in Q under addition.
c- 0∈Q is an identity element.
d- If a∈Q then additive inverse −a∈Q such that a+(−a)=(−a)+a=0.
Therefore the set of rational number is group under addition.
Question # 3 (iii) Determine whether or not the set of positive rational number form groups with respect to ×.
Solution
a- Since for a,b∈Q+, ab∈Q+ thus closure law holds.
b- For a,b,c∈Q, a(bc)=(ab)c, thus associative law holds.
c- Since 1∈Q+ such that for a∈Q+, a×1=1×a=a. Hence 1 is the identity element.
d- For a∈Q+, 1a∈Q+ such that a×1a=1a×a=1. Thus inverse of a is 1a.
Hence Q+ is group under addition.
Question # 3 (iv) Determine whether or not the set of integers form groups with respect to +.
Solution
Given Z={0,±1,±2,±3,...}.
a- Since sum of integers is an integer therefore for a,b∈Z, a+b∈Z.
b- Since a+(b+c)=(a+b)+c, thus associative law holds in Z.
c- Since 0∈Z such that for a∈Z, a+0=0+a=a. Thus 0 an identity element.
d- For a∈Z, −a∈Z such that a+(−a)=(−a)+a=0. Thus inverse of a is −a.
Hence, we conclude that set of integers form groups with respect to +.
Question # 3 (v) Determine whether or not the set of integers form groups with respect to ×.
Solution
Given Z={0,±1,±2,±3,...}.
For any a∈Z the multiplicative inverse of a is 1a∉Z. Hence Z is not a group under multiplication.
Question # 4 Show that the adjoining table represents the sums of the elements of the set {E,O}. What is the identity element of this set? Show that this set is Abelian group.
⊕EOEEOOOE
Solution
As E+E=E, E+O=O, O+O=E
Thus the table represents the sums of the elements of set {E,O}.
The identity element of the set is E because
E+E=E+E=E & E+O=O+E=E.
(i) From the table each element belongs to the set {E,O}.
Hence closure law is satisfied.
(ii) ⊕ is associative in {E,O}
(iii) E is the identity element of w.r.t to ⊕
(iv) As O+O=E and E+E=E, thus inverse of O is O and inverse of E is E.
(v) As the table is symmetric about the principal diagonal therefore ⊕ is commutative.
Hence {E,O} is Abelian group under ⊕.
Question # 5 Show that the set {1,ω,ω2}, when ω3=1 is an abelian group w.r.t. ordinary multiplication.
Solution
Suppose G={1,ω,ω2}.
×1ωω211ωω2ωωω21ω2ω21ω
(i) A table show that all the entries belong to G, therefore colure law holds.
(ii) Associative law holds in G w.r.t. multiplication.
e.g. 1×(ω×ω2)=1×1=1.
(1×ω)×ω2=ω×ω2=1.
(iii) Since 1×1=1, 1×ω=ω×1=ω, 1×ω2=ω2×1=ω2.
Thus 1 is an identity element in G.
(iv) As 1×1=1, ω×ω2=ω2×ω=1, ω2×ω=ω×ω2=1,
therefore, inverse of 1 is 1, inverse of ω is ω2, inverse of ω2 is ω.
(v) As table is symmetric about principal diagonal therefore commutative law holds in G.
Hence G is an abelian group under multiplication.
Question # 6 If G is a group under the operation ∗ and a,b∈G, find the solutions of the equations: a∗x=b, x∗a=b
Solution Given that G is a group under the operation ∗ and a,b∈G such that a∗x=b
As a∈G and G is group so a−1∈G such that
a−1∗(a∗x)=a−1∗b⇒(a−1∗a)∗x=a−1∗bby associative law⇒e∗x=a−1∗bby inverse law.⇒x=a−1∗bby identity law.
And for
x∗a=b⇒(x∗a)∗a−1=b∗a−1 for a∈G,a−1∈G⇒x∗(a∗a−1)=b∗a−1by associative law⇒x∗e=b∗a−1by inverse law.⇒x=b∗a−1by identity law.
Question # 7 Show that the set consisting of elements of the form a+√3b (a,b being rational), is an abelian group w.r.t. addition.
Solution
Consider G={a+√3b|a,b∈Q}.
(i) Let a+√3b, c+√3d∈G, where a, b, c & d are rationals.
(a+√3b)+(c+√3d)=(a+c)+√3(b+d)=a′+√3b′∈G,
where a′=a+c and b′=b+d are rationals as sum of rationals is rational. Thus closure law holds in G under addition.
(ii) For a+√3b, c+√3d, e+√3f∈G, (a+√3b)+((c+√3d)+(e+√3f))=(a+√3b)+((c+e)+√3(d+f))=(a+(c+e))+√3(b+(d+f))=((a+c)+e)+√3((b+d)+f), as associative law holds in Q=((a+c)+√3(b+d))+(e+√3f)=((a+√3b)+(c+√3d))+(e+√3f).
Thus, associative law holds in G under addition.
(iii) 0+√3⋅0∈G as 0 is a rational such that for any a+√3b∈G,
(a+√3b)+(0+√3⋅0)=(a+0)+√3(b+0)=a+√3b and (0+√3⋅0)+(a+√3b)=(0+a)+√3(0+b)=a+√3b.
Thus 0+√3⋅0 is an identity element in G.
(iv) For a+√3b∈G where a & b are rational, there exit rationals –a & −b such that
(a+√3b)+((−a)+√3(−b))=(a+(−a))+√3(b+(−b))=0+√3⋅0 and ((−a)+√3(−b))+(a+√3b)=((−a)+a)+√3((−b)+b)=0+√3⋅0.
Thus, inverse of a+√3b is (−a)+√3(−b), exists in G.
(v) For a+√3b, c+√3d∈G
(a+√3b)+(c+√3d)=(a+c)+√3(b+d)=(c+a)+√3(d+b), as commutative law hold in Q=(c+d√3)+(a+√3b).
Thus, Commutative law holds in G under addition.
Hence G is an abelian group under addition.
Question 8 Determine whether (P(S),∗), where ∗ stands for intersection is a semi group, a monoid or neither. If it is a monoid, specify its identity.
Solution Suppose A,B∈P(S), where A & B are subsets of S.
As intersection of two subsets of S is subset of S.
Therefore A∗B=A∩B∈P(S). Thus, closure law holds in P(S).
For A,B,C∈P(S),
A∗(B∗C)=A∩(B∩C)=(A∩B)∩C=(A∗B)∗C
Thus, associative law holds in P(S).
Hence (P(S),∗) is a semi-group.
For A∈P(S) where A is a subset of S we have S∈P(S) such that A∩S=S∩A=A.
Thus S is an identity element in P(S). Hence (P(S),∗) is a monoid.
Question 9 Complete the following table to obtain a semi-group under ∗
∗abcacabbabccx1x2a
Solution
Let x1 and x2 be the required elements.
By associative law
(a∗a)∗a=a∗(a∗a)⇒c∗a=a∗c⇒x1=b.
Now again by associative law
(a∗a)∗b=a∗(a∗b)⇒c∗b=a∗a⇒x2=c
Question 10 Prove that all 2×2 non-singular matrices over the real field form a non-abelian group under multiplication.
Solution
Let G be the all non-singular 2×2 matrices over the real field.
(i) Suppose A,B∈G, then A2×2×B2×2=C2×2∈G.
Thus, closure law holds in G under multiplication.
(ii) Associative law in matrices of same order under multiplication holds.
therefore for A,B,C∈G, A×(B×C)=(A×B)×C.
(iii) I2×2=(1001) is a non-singular matrix such that A2×2×I2×2=I2×2×A2×2=A2×2. Thus, I2×2 is an identity element in G.
iv) Since inverse of non-singular square matrix exists, for A∈G there exist A−1∈G such that AA−1=A−1A=I.
v) As we know for any two matrices A,B∈G, AB≠BA, in general.
Therefore, commutative law does not hold in G under multiplication.
Hence the set of all 2×2 non-singular matrices over a real field is a non-abelian group under multiplication.
Exercise 2.8 (Page 78): Textbook of Algebra and Trigonometry Class XI.
Punjab Textbook Board, Lahore - PAKISTAN.
Page URL: https://www.mathcity.org/fsc-part1-ptb/sol/unit-02/ex2-8
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