Exercise 2.8 (Solutions)

Notes (Solutions) of Exercise 2.8: Textbook of Algebra and Trigonometry Class XI (Mathematics FSc Part 1 or HSSC-I), Punjab Textbook Board (PTB) Lahore.

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,

01=0, 11=3, 21=2, 31=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 0Q, 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- 0Q is an identity element.

d- If aQ then additive inverse aQ 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,bQ+, abQ+ thus closure law holds.

b- For a,b,cQ, a(bc)=(ab)c, thus associative law holds.

c- Since 1Q+ such that for aQ+, a×1=1×a=a. Hence 1 is the identity element.

d- For aQ+, 1aQ+ 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,bZ, a+bZ.

b- Since a+(b+c)=(a+b)+c, thus associative law holds in Z.

c- Since 0Z such that for aZ, a+0=0+a=a. Thus 0 an identity element.

d- For aZ, aZ 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 aZ the multiplicative inverse of a is 1aZ. 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,bG, find the solutions of the equations: ax=b, xa=b

Solution Given that G is a group under the operation and a,bG such that ax=b

As aG and G is group so a1G such that

a1(ax)=a1b(a1a)x=a1bby associative lawex=a1bby inverse law.x=a1bby identity law.

And for

xa=b(xa)a1=ba1 for aG,a1Gx(aa1)=ba1by associative lawxe=ba1by inverse law.x=ba1by 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,bQ}.

(i) Let a+3b, c+3dG, where a, b, c & d are rationals.

(a+3b)+(c+3d)=(a+c)+3(b+d)=a+3bG,

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+3fG, (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+30G as 0 is a rational such that for any a+3bG,

(a+3b)+(0+30)=(a+0)+3(b+0)=a+3b and (0+30)+(a+3b)=(0+a)+3(0+b)=a+3b.

Thus 0+30 is an identity element in G.

(iv) For a+3bG where a & b are rational, there exit rationals a & b such that

(a+3b)+((a)+3(b))=(a+(a))+3(b+(b))=0+30 and ((a)+3(b))+(a+3b)=((a)+a)+3((b)+b)=0+30.

Thus, inverse of a+3b is (a)+3(b), exists in G.

(v) For a+3b, c+3dG

(a+3b)+(c+3d)=(a+c)+3(b+d)=(c+a)+3(d+b), as commutative law hold in Q=(c+d3)+(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,BP(S), where A & B are subsets of S.

As intersection of two subsets of S is subset of S.

Therefore AB=ABP(S). Thus, closure law holds in P(S).

For A,B,CP(S),

A(BC)=A(BC)=(AB)C=(AB)C

Thus, associative law holds in P(S).

Hence (P(S),) is a semi-group.

For AP(S) where A is a subset of S we have SP(S) such that AS=SA=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

(aa)a=a(aa)ca=acx1=b.

Now again by associative law

(aa)b=a(ab)cb=aax2=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,BG, then A2×2×B2×2=C2×2G.

Thus, closure law holds in G under multiplication.

(ii) Associative law in matrices of same order under multiplication holds.

therefore for A,B,CG, 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 AG there exist A1G such that AA1=A1A=I.

v) As we know for any two matrices A,BG, ABBA, 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.

Book:
Exercise 2.8 (Page 78): Textbook of Algebra and Trigonometry Class XI.
Punjab Textbook Board, Lahore - PAKISTAN.
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