Exercise 2.8 (Solutions)

Question # 1 Operation $\oplus$ performed on the two-member set $G=\{0,1\}$ is shown in the adjoining table. \[ \begin{array}{|c|c|c|} \hline \oplus & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array} \] 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 $\oplus$.

Also it satisfies commutative law so it is an abelian group.

Question # 2 The operation $\oplus$ as performed on the set $\{0,1,2,3\}$ is shown in the adjoining table, show that the set is an Abelian group?

\[ \begin{array}{|c|c|c|c|c|} \hline \oplus & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ \hline 1 & 1 & 2 & 3 & 0 \\ \hline 2 & 2 & 3 & 0 & 1 \\ \hline 0 & 3 & 0 & 1 & 2 \\ \hline \end{array} \]

Solution

Suppose $G=\left\{ 0,1,2,3 \right\}$

i) The given table show that each element of the table is a member of $G$ thus closure law holds.

ii) $\oplus$ is associative in $G$.

iii) Table show that $0$ is identity element w.r.t. $\oplus$.

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 $\times$.

Solution

As $0\in \mathbb{Q}$, multiplicative inverse of $0$ in not in set $\mathbb{Q}$. Therefore the set of rational number is not a group w.r.t to “$\times $”.

Question # 3 (ii) Determine whether or not the set of rational number form groups with respect to $+$.

Solution

a- Closure property holds in $\mathbb{Q}$ under $+$ because sum of two rational number is also rational.

b- Associative property holds in $\mathbb{Q}$ under addition.

c- $0\in \mathbb{Q}$ is an identity element.

d- If $a\in \mathbb{Q}$ then additive inverse $-a\in \mathbb{Q}$ 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 $\times$.

Solution

a- Since for $a,b\in {{\mathbb{Q}}^{+}}$, $ab\in {{\mathbb{Q}}^{+}}$ thus closure law holds.

b- For $a,b,c\in \mathbb{Q}$, $a(bc)=(ab)c$, thus associative law holds.

c- Since $1\in {{\mathbb{Q}}^{+}}$ such that for $a\in {{\mathbb{Q}}^{+}}$, $a\times 1=1\times a=a$. Hence 1 is the identity element.

d- For $a\in {{\mathbb{Q}}^{+}}$, $\dfrac{1}{a}\in {{\mathbb{Q}}^{+}}$ such that $a\times \dfrac{1}{a}=\dfrac{1}{a}\times a=1$. Thus inverse of $a$ is $\dfrac{1}{a}$.

Hence ${{\mathbb{Q}}^{+}}$ is group under addition.

Question # 3 (iv) Determine whether or not the set of integers form groups with respect to $+$.

Solution

Given $\mathbb{Z}=\{0,\pm 1,\pm \,2,\pm \,3,...\}$.

a- Since sum of integers is an integer therefore for $a,b\in \mathbb{Z}$, $a+b\in \mathbb{Z}$.

b- Since $a+\left( b+c \right)=\left( a+b \right)+c$, thus associative law holds in $\mathbb{Z}$.

c- Since $0\in \mathbb{Z}$ such that for $a\in \mathbb{Z}$, $a+0=0+a=a$. Thus $0$ an identity element.

d- For $a\in \mathbb{Z}$, $-a\in \mathbb{Z}$ 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 $\times $.

Solution

Given $\mathbb{Z}=\{0,\pm 1,\pm \,2,\pm \,3,...\}$.

For any $a\in \mathbb{Z}$ the multiplicative inverse of $a$ is $\frac{1}{a}\notin \mathbb{Z}$. Hence $\mathbb{Z}$ 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.

\[ \begin{array}{|c|c|c|} \hline \oplus & E & O \\ \hline E & E & O \\ \hline O & O & E \\ \hline \end{array} \]

Solution

As $E + E = E$, $E + O = O$, $O + O = E$

Thus the table represents the sums of the elements of set $\left\{ E,O \right\}$.

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 $\left\{ E,O \right\}$.

Hence closure law is satisfied.

(ii) $\oplus$ is associative in $\left\{ E,O \right\}$

(iii) $E$ is the identity element of w.r.t to $\oplus$

(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 $\oplus$ is commutative.

Hence $\left\{ E,O \right\}$ is Abelian group under $\oplus $.

Question # 5 Show that the set $\left\{ 1,\omega ,{{\omega }^{2}} \right\}$, when ${{\omega }^{3}}=1$ is an abelian group w.r.t. ordinary multiplication.

Solution

Suppose $G=\left\{ 1,\omega ,{{\omega }^{2}} \right\}$.

\[ \begin{array}{|c|c|c|c|} \hline \times & 1 & \omega & \omega^2 \\ \hline 1 & 1 & \omega & \omega^2 \\ \hline \omega & \omega & \omega^2 & 1 \\ \hline \omega^2 & \omega^2 & 1 & \omega \\ \hline \end{array} \]

(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. $1\times (\omega \times {{\omega }^{2}})=1\times 1=1$.

$(1\times \omega )\times {{\omega }^{2}}=\omega \times {{\omega }^{2}}=1$.

(iii) Since $1\times 1=1$, $1\times \omega =\omega \times 1=\omega $, $1\times {{\omega }^{2}}={{\omega }^{2}}\times 1={{\omega }^{2}}$.

Thus $1$ is an identity element in $G$.

(iv) As $1\times 1=1$, $\omega \times {{\omega }^{2}}={{\omega }^{2}}\times \omega =1$, ${{\omega }^{2}}\times \omega =\omega \times {{\omega }^{2}}=1$,

therefore, inverse of $1$ is $1$, inverse of $\omega$ is ${{\omega }^{2}}$, inverse of ${{\omega }^{2}}$ is $\omega $.

(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 $a,b\in G$, find the solutions of the equations: $a*x=b$, $x*a=b$

Solution Given that $G$ is a group under the operation $*$ and $a,b\in G$ such that $a*x=b$

As $a\in G$ and $G$ is group so $a^{-1}\in G$ such that

\begin{align}&{{a}^{-1}}*\left( a*x \right)={{a}^{-1}}*b \\ \Rightarrow \,\,\,\,& \left( {{a}^{-1}}*a \right)*x={{a}^{-1}}*b\,\,\,\, \text{by associative law}\\ \Rightarrow \,\,\,\,& e*x={{a}^{-1}}*b \,\,\,\,\text{by inverse law.}\\ \Rightarrow \,\,\,\,& x={{a}^{-1}}*b\,\,\,\, \text{by identity law.}\end{align}

And for

\begin{align} & x*a=b \\ \Rightarrow \,\,\,\,& \left( x*a \right)*{{a}^{-1}}=b*{{a}^{-1}} \text{ for } a \in G, {{a}^{-1}}\in G \\ \Rightarrow \,\,\,\, & x*\left( a*{{a}^{-1}} \right)=b*{{a}^{-1}}\,\,\,\, \text{by associative law}\\ \Rightarrow \,\,\,\, & x*e=b*{{a}^{-1}}\,\,\,\, \text{by inverse law.}\\ \Rightarrow \,\,\,\, & x=b*{{a}^{-1}} \,\,\,\,\text{by identity law.}\end{align}

Question # 7 Show that the set consisting of elements of the form $a+\sqrt{3}b$ ($a,b$ being rational), is an abelian group w.r.t. addition.

Solution

Consider $G=\left\{ a+\sqrt{3}\,b | a,b\in \mathbb{Q} \right\}$.

(i) Let $a+\sqrt{3}b$, $c+\sqrt{3}d \in G$, where $a$, $b$, $c$ & $d$ are rationals.

\begin{align}& \left( a+\sqrt{3} b \right) +\left( c+\sqrt{3} d\, \right) \\ = & \left( a+c \right)+\sqrt{3}\left( b+d \right) \\ = & {a}'+\sqrt{3}{b}'\in G,\end{align}

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 $a+\sqrt{3}b$, $c+\sqrt{3}d$, $e+\sqrt{3}f \in G$, \begin{align}& (a+\sqrt{3} b) +\left( (c+\sqrt{3} d)+(e+\sqrt{3} f)\right) \\ = & (a+\sqrt{3} b)+\left((c+e)+\sqrt{3}(d+f)\right) \\ = & \left( a+(c+e) \right)+\sqrt{3}\left( b+(d+f) \right) \\ = & \left( (a+c)+e \right) + \sqrt{3}\left( (b+d)+f \right), \\ & \qquad \text{ as associative law holds in }\mathbb{Q} \\ = & \left((a+c)+\sqrt{3}(b+d)\right)+(e+\sqrt{3}f) \\ = & \left( (a+\sqrt{3}b)+(c+\sqrt{3}d)\, \right)+(e+\sqrt{3}f). \end{align}

Thus, associative law holds in $G$ under addition.

(iii) $0+\sqrt{3}\cdot 0\in G$ as $0$ is a rational such that for any $a+\sqrt{3}b \in \,G$,

\begin{align}& (a+\sqrt{3}b) + (0+\sqrt{3} \cdot 0) \\ = & (a+0)+\sqrt{3}(b+0) \\ = & a+\sqrt{3}b \end{align} and \begin{align}& (0+\sqrt{3} \cdot 0) +(a+\sqrt{3}b)\\ = & (0+a)+\sqrt{3}(0+b) \\ = & a+\sqrt{3}b.\end{align}

Thus $0+\sqrt{3}\cdot 0$ is an identity element in $G$.

(iv) For $a+\sqrt{3}b \in G$ where $a$ & $b$ are rational, there exit rationals $– a$ & $-b$ such that

\begin{align} & (a+\sqrt{3}b) +\left((-a)+\sqrt{3}(-b)\right)\\ = & \left( a+(-a)\right)+\sqrt{3}\left( b+(-b)\right) \\ = & 0+\sqrt{3}\cdot 0 \end{align} and \begin{align}& \left( (-a)+\sqrt{3}(-b) \right) + (a+\sqrt{3}b)\\ = & \left( (-a)+a \right)+\sqrt{3}\left( (-b)+b \right) \\ = & 0+\sqrt{3}\cdot 0.\end{align}

Thus, inverse of $a+\sqrt{3}b$ is $(-a)+\sqrt{3}(-b)$, exists in $G$.

(v) For $a+\sqrt{3}b$, $c+\sqrt{3}d \in G$

\begin{align} & \left( a+\sqrt{3}b \right) + \left( c+\sqrt{3}d \right)\\ = & (a+c)+\sqrt{3}(b+d) \\ = & (c+a)+\sqrt{3}(d+b), \\ & \qquad \text{ as commutative law hold in } \mathbb{Q} \\ = & (c+d\sqrt{3})+(a+\sqrt{3}b). \end{align}

Thus, Commutative law holds in $G$ under addition.

Hence $G$ is an abelian group under addition.

Question 8 Determine whether $\left( P(S),* \right)$, where $*$ stands for intersection is a semi group, a monoid or neither. If it is a monoid, specify its identity.

Solution Suppose $A,B\in P(S)$, where $A$ & $B$ are subsets of $S$.

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

Therefore $A*B=A\cap B\in P(S)$. Thus, closure law holds in $P(S)$.

For $A,B,C\in P(S)$,

\begin{eqnarray} & A*(B*C)=A\cap (B\cap C) \\ = & (A\cap B)\cap C=(A*B)*C \end{eqnarray}

Thus, associative law holds in $P(S)$.

Hence $\left( P(S),* \right)$ is a semi-group.

For $A\in P(S)$ where A is a subset of $S$ we have $S\in P(S)$ such that $A\cap S=S\cap A=A$.

Thus $S$ is an identity element in $P(S)$. Hence $\left( P(S),* \right)$ is a monoid.

Question 9 Complete the following table to obtain a semi-group under $*$

\[ \begin{array}{|c|c|c|c|} \hline * & a & b & c \\ \hline a & c & a & b \\ \hline b & a & b & c \\ \hline c & x_1 & x_2 & a \\ \hline \end{array} \]

Solution

Let $x_1$ and $x_2$ be the required elements.

By associative law

\begin{align} & (a*a)*a=a*(a*a) \\ \Rightarrow \,\,\,\,\,\,\, & c*a=a*c\\ \Rightarrow \,\,\,\,\,\, & x_1=b.\end{align}

Now again by associative law

\begin{align} & (a*a)*b=a*(a*b)\\ \Rightarrow \,\,\,\,\,\, & c*b=a*a\\ \Rightarrow \,\,\,\,\,\, & x_2=c\end{align}

Question 10 Prove that all $2\times 2$ non-singular matrices over the real field form a non-abelian group under multiplication.

Solution

Let $G$ be the all non-singular $2\times 2$ matrices over the real field.

(i) Suppose $A,B\in G$, then $A_{2\times 2} \times B_{2\times 2} = C_{2\times 2} \in G$.

Thus, closure law holds in $G$ under multiplication.

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

therefore for $A,B,C\in G$, $A\times (B\times C)=(A\times B)\times C$.

(iii) $I_{2\times 2}= \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$ is a non-singular matrix such that $$A_{2\times 2} \times I_{2\times 2} = I_{2\times 2} \times A_{2\times 2} = A_{2\times 2}.$$ Thus, $I_{2\times 2}$ is an identity element in $G$.

iv) Since inverse of non-singular square matrix exists, for $A\in G$ there exist ${{A}^{-1}}\in G$ such that $A{{A}^{-1}}={{A}^{-1}}A=I$.

v) As we know for any two matrices $A,B\in G$, $AB \ne BA$, in general.

Therefore, commutative law does not hold in G under multiplication.

Hence the set of all $2\times 2$ non-singular matrices over a real field is a non-abelian group under multiplication.