Algebra - Group Theory

1. Groups

• Definition 1.1: A non empty set $G$ together with a binary operation $*$ is said to be a semi-group if the following holds
(1.) Closure law: $\forall$ $a,b\in G \Rightarrow a*b \in G$.
(2.) Associative law: $a*(b*c)=(a*b)*c$ $\forall$ $a,b,c \in G$.

• Definition 1.2: A non empty set $G$ together with a binary operation $*$ is said to be a monoid if the following holds
(1.) Closure law: $\forall$ $a,b\in G \Rightarrow a*b \in G$.
(2.) Associative law: $a*(b*c)=(a*b)*c$ $\forall$ $a,b,c \in G$.
(3.) Existence of identity element: $\exists$ $e \in G$ such that $a*e=e*a=a$ $\forall$ $a \in G$.

• Definition 1.3: A non empty set $G$ together with a binary operation $*$ is said to be a group if the following holds
(1.) Closure law: $\forall$ $a,b\in G \Rightarrow a*b \in G$.
(2.) Associative law: $a*(b*c)=(a*b)*c$ $\forall$ $a,b,c \in G$.
(3.) Existence of identity element: $\exists$ $e \in G$ such that $a*e=e*a=a$ $\forall$ $a \in G$.
(4.) Existence of inverse element: For each $a \in G$, $\exists$ $a^{-1} \in G$ such that $a*a^{-1}=a^{-1}*a=e$.

• $a^{-1}$ is called the inverse element of $a$ in $G$. Note that $a^{-1}\neq \frac{1}{a}$ We see $a^{-1}$ as an element in $G$.

• Notation: $a\times b=a\cdot b=ab$
A group $G$ with binary operation $*$ is denoted by $(G,*)$.
Generally we write the group $(G,*)$ as $G$. Note that binary opeation $*$ is defined on $G$ if we write simply as $G$ also.

• Let $(G,+)$ be a group. Then if $a \in G$ $\Rightarrow$ $a^{-1} \in G$. $a^{-1}$ will be of the form $-a$
Let $(G,\times)$ be a group. Then if $a \in G$ $\Rightarrow$ $a^{-1} \in G$. $a^{-1}$ will be of the form $\frac{1}{a}$

• Definition 1.4: A non empty set $G$ together with a binary operation $*$ is said to be a abelian group or commutative group if the following holds
(1.) Closure law: $\forall$ $a,b\in G \Rightarrow a*b \in G$.
(2.) Associative law: $a*(b*c)=(a*b)*c$ $\forall$ $a,b,c \in G$.
(3.) Existence of identity element: $\exists$ $e \in G$ such that $a*e=e*a=a$ $\forall$ $a \in G$.
(4.) Existence of inverse element: For each $a \in G$, $\exists$ $a^{-1} \in G$ such that $a*a^{-1}=a^{-1}*a=e$.
(5.) Commutative law: $a*b=b*a$ $\forall$ $a,b \in G$.

• Definition 1.5: If a group $G$ contains finite number of elements, then it is called a finite group.

• Definition 1.6: If a group $G$ contains infinite number of elements, then it is called a infinite group.

• Definition 1.7: Let $G$ be a finite group. The number of elements of $G$ is called the order of a group denoted by $O(G)$. If a group $G$ consists of $n$ elements, then $O(G)=n$.

• Every abelian group is a group, every group is a monoid, every monoid is a semi-group. i.e., abelian group $\Rightarrow$ group $\Rightarrow$ monoid $\Rightarrow$ semi-group. Converse may not be true.

• Examples:
(1.) $(\mathbb{N},+)$ is a semi group but not a monoid.
(2.) $(\mathbb{Z},\times)$ is a monoid but not a group.
(3.) Set of $2 \times 2$ non singular real matrices under matrix multiplication is a group but not abelian.
(4.) $(\mathbb{Z},+)$ is an abelian group.
(5.) $(\mathbb{N},+)$, $(\mathbb{N},\times)$, $(\mathbb{Z},\times)$, $(\mathbb{Z}^*,+)$ are not groups.
(6.) $(\mathbb{Z},+)$, $(\mathbb{Q},+)$, $(\mathbb{Q},\times)$, $(\mathbb{R},+)$, $(\mathbb{R},\times)$, $(\mathbb{C},+)$, $(\mathbb{C},\times)$ are infinite abelian groups.

• Integral power of an element
(1.) Let $G$ be a group with binary operation $+$, then for $a \in G$ we have
$a=a$
$a+a=2a$
$a+a+a=3a$
$\vdots$
Adding $a$, $n$ times we get $\smash{\underbrace{a+a+a+.....+a}_\text{$n$times}}=0$

(2.) Let $G$ be a group with binary operation $\times$, then for $a \in G$ we have
$a=a$
$a\times a=a^2$
$a\times a\times a=a^3$
$\vdots$
Multiplying $a$, $n$ times we get $\smash{\underbrace{a\times a\times a\times....\times a}_\text{$n$times}}=a^n$

(3.) Let $G$ be a group with binary operation $\times$, then for $a \in G$ $\Rightarrow$ $a^{-1} \in G$ we have $(a^{-1})^n=(a^n)^{-1}=a^{-n}$

• Theorem 1.1: Identity element of a group $G$ is unique.
Proof: Suppose, if possible let $e_1$ and $e_2$ be two distinct identity elements of a group $(G,*)$.
Since $e_1$ is identity, $e_2*e_1=e_1*e_2=e_2$
Since $e_2$ is identity, $e_1*e_2=e_2*e_1=e_1$
Hence, it follows that $e_1=e_2$
$\Rightarrow$ Identity is unique. $\blacksquare$

• Theorem 1.2: Every element $a$ in a group $G$ has a unique inverse.
Proof: Let $(G,*)$ be a group and $a \in G$. Suppose if $a$ has two distinct inverse $a^{-1}_1$ and $a^{-1}_2$ in $G$.
Since $a^{-1}_1$ is the inverse of $a$, $a*a^{-1}_1=a^{-1}_1*a=e$
Since $a^{-1}_2$ is the inverse of $a$, $a*a^{-1}_2=a^{-1}_2*a=e$
Consider, $a^{-1}_1=a^{-1}_1*e=a^{-1}_1*(a*a^{-1}_2)=(a^{-1}_1*a)*a^{-1}_2=e*a^{-1}_2=a^{-1}_2$
$\therefore$ $a^{-1}_1=a^{-1}_2$
$\Rightarrow$ For each $a \in G$ $\exists$ unique inverse $a^{-1} \in G$ $\blacksquare$

• Theorem 1.3: Let $(G,*)$ be a group and $a,b,c \in G$ then
(i) $a*c=b*c \Rightarrow a=b$. (Right cancellation law)
(ii) $c*a=c*b \Rightarrow a=b$. (Left cancellation law)
Proof (i): Consider, $a*c=b*c$
$\Rightarrow$ $(a*c)*c^{-1}=(b*c)*c^{-1}$ [Post multiply $c^{-1}$ on both sides]
$\Rightarrow$ $a*(c*c^{-1})=b*(c*c^{-1})$ [Associative law in $G$]
$\Rightarrow$ $a*e=b*e$ [Inverse law in $G$]
$\Rightarrow$ $a=b$
Proof (ii): Similar arguements as in above. $\square$

• Theorem 1.4: If $(G,*)$ is a group then
(i) $(a^{-1})^{-1}=a$ $\forall$ $a \in G$.
(ii) $(a*b)^{-1}=b^{-1}*a^{-1}$ $\forall$ $a,b \in G$.
Proof (i): Let $a\in G$ $\Rightarrow$ $a^{-1} \in G$ such that $a*a^{-1}=a^{-1}*a=e$ where $e$ is the identity of $G$. Since $a^{-1} \in G$ $\Rightarrow$ $(a^{-1})^{-1} \in G$ such that $a^{-1}*(a^{-1})^{-1}=(a^{-1})^{-1}*a^{-1}=e$
Consider $a*a^{-1}=e=(a^{-1})^{-1}*a^{-1}$
$\Rightarrow$ $a*a^{-1}=(a^{-1})^{-1}*a^{-1}$
$\Rightarrow$ $a=(a^{-1})^{-1}$ [By right cancellation]
Proof (ii): Let $a,b \in G$ $\Rightarrow$ $a^{-1},b^{-1} \in G$
Consider $(a*b)*(b^{-1}*a^{-1})$
$=a*(b*b^{-1})*a^{-1}$
$=a*e*a^{-1}$
$=a*a^{-1}=e$
$\therefore$ $(a*b)*(b^{-1}*a^{-1})=e$
Similarly we have $(b^{-1}*a^{-1})*(a*b)=e$
Thus $(a*b)*(b^{-1}*a^{-1})=e=(b^{-1}*a^{-1})*(a*b)$
$\Rightarrow$ $(a*b)^{-1}=b^{-1}*a^{-1}$ $\blacksquare$

• Theorem 1.5: In a group $(G,*)$, the equation $a*x=b$ and $y*a=b$ where $a,b \in G$ have unique solutions in $G$.
Proof: Consider $a*x=b$
$\Rightarrow$ $a^{-1}*(a*x)=a^{-1}*b$ [Pre multiplying $a^{-1}$ on both sides]
$\Rightarrow$ $(a^{-1}*a)*x=a^{-1}*b$ [Associative law in $G$]
$\Rightarrow$ $e*x=a^{-1}*b$ [Identity law in $G$]
$\Rightarrow$ $x=a^{-1}*b$ $\in G$
$\therefore$ $a*x=b$ has a solution $x=a^{-1}*b \in G$.
Now consider $y*a=b$
$\Rightarrow$ $(y*a)*a^{-1}=b*a^{-1}$
$\Rightarrow$ $y*(a*a^{-1})=b*a^{-1}$
$\Rightarrow$ $y*e=b*a^{-1}$
$\Rightarrow$ $y=b*a^{-1} \in G$
$\therefore$ $y*a=b$ has a solution $y=b*a^{-1} \in G$.
Uniqueness: Suppose, if possible $x_1$ and $x_2$ be two solutions of the equation $a*x=b$
Then, $a*x_1=b$ and $a*x_2=b$.
$\Rightarrow$ $x_1=x_2$.
$\therefore$ $a*x=b$ has a unique solution $x=a^{-1}*b \in G$. Similarly $y*a=b$ has a unique solution $y=b*a^{-1} \in G$. $\blacksquare$

• Theorem 1.6: The left identity is also the right identity in a group $G$.
Proof: Let $e$ be the left identity of a group $G$ and let $a \in G$, then $e*a=a$.
To prove: $e$ is also the right identity i.e., $a*e=a$.
Let $a^{-1}$ be the inverse element of $a$ in $G$, then $a^{-1}*a=e$
Now consider, $a^{-1}*(a*e)=(a^{-1}*a)*e=e*e=e=a^{-1}*a$
$\Rightarrow$ $a^{-1}*(a*e)=a^{-1}*a$
$\Rightarrow$ $a*e=a$ [By left cancellation law in $G$] $\blacksquare$

• Theorem 1.7: The left inverse of an element in a group $G$ is also the right inverse.
Proof: Let $a^{-1}$ be the left inverse of $a \in G$. i.e., $a^{-1}*a=e$.
To prove: $a^{-1}$ is also a right inverse of $a$ i.e., $a*a^{-1}=e$
Consider $a^{-1}*(a*a^{-1})=(a^{-1}*a)*a^{-1}$ \hfill[Associative law in $G$]
$\Rightarrow$ $a^{-1}*(a*a^{-1})=e*a^{-1}$
$\Rightarrow$ $a^{-1}*(a*a^{-1})=a^{-1}$
$\Rightarrow$ $a^{-1}*(a*a^{-1})=a^{-1}*e$
$\Rightarrow$ $a*a^{-1}=e$ [By left cancellation law in $G$] $\blacksquare$

• Theorem 1.8: A finite group $G$ with $*$ which is associative is a group if and only if cancellation holds.
Proof: Let $G$ be a group, then cancellation law holds in $G$ [By theorem 1.3]
Conversely: Suppose that cancellation law holds in $G$. To prove: $G$ is a group.
Consider a set $G=\{a_1,a_2,...,a_n\}$ of $n$ elements.
Now consider a set $B=\{a*a_1,a*a_2,...,a*a_n\}$ where $a \in G$
Claim: Elements of $B$ are distinct. For if $a*a_i=a*a_j$ $\Rightarrow$ $a_i=a_j$ which is a contradiction to the fact that $G$ has $n$ elements.
Hence, $G=\{a*a_1,a*a_2,....,a*a_n\}$
Let $b\in G$ $\Rightarrow$ $b=a*a_i$ for some $i$. Hence the equation $b=a*x$ where $x=a_i$ has a unique solution in $G$. \hfill[By theorem 1.5]
Similarly the equation $y*a=b$ has a unique soluiton in $G$. Hence, $G$ is a group. $\blacksquare$

• Problem 1.1: Show that $\mathbb{Z}$ is an infinite abelian group under addition.
Solution: Consider $\mathbb{Z}=\{0,\pm 1, \pm 2, \pm 3,.....\}$ Let $a,b,c \in \mathbb{Z}$.
(1.) Closure law: $\forall$ $a,b\in \mathbb{Z} \Rightarrow a+b \in \mathbb{Z}$.
(2.) Associative law: $a+(b+c)=(a+b)+c$ $\forall$ $a,b,c \in \mathbb{Z}$.
(3.) Existence of identity element: $\exists$ $0 \in \mathbb{Z}$ such that $a+0=0+a=a$ $\forall$ $a \in \mathbb{Z}$.
(4.) Existence of inverse element: For each $a \in \mathbb{Z}$, $\exists$ $-a \in \mathbb{Z}$ such that $a+(-a)=(-a)+a=0$.
(5.) Commutative law: $a+b=b+a$ $\forall$ $a,b \in \mathbb{Z}$.
$\therefore$ $(\mathbb{Z},+)$ is an abelian group. Since $\mathbb{Z}$ is infinite, it follows that $\mathbb{Z}$ is an infinite abelian group under addition. $\spadesuit$

• Problem 1.2: Show that $\mathbb{Z}$ is not an group under multiplication.
Solution: Consider $\mathbb{Z}=\{0,\pm 1, \pm 2, \pm 3,.....\}$. Let $a,b,c \in \mathbb{Z}$.
(1.) Closure law: $\forall$ $a,b\in \mathbb{Z} \Rightarrow ab \in \mathbb{Z}$.
(2.) Associative law: $a(bc)=(ab)c$ $\forall$ $a,b,c \in \mathbb{Z}$.
(3.) Existence of identity element: $\exists$ $1 \in \mathbb{Z}$ such that $a\times 1=1\times a=a$ $\forall$ $a \in \mathbb{Z}$.
(4.) Existence of inverse element: For each $a \in \mathbb{Z}$, $\nexists$ $\frac{1}{a} \in \mathbb{Z}$
$\therefore$ $(\mathbb{Z},\times)$ is a monoid but not a group. $\spadesuit$

• Problem 1.3: Show that $\mathbb{Q}$ is an infinite abelian group under addition.
Solution: Consider $\mathbb{Q}=\{\frac{x}{y}$ $/ x,y \in \mathbb{Z}$ and $y \neq 0 \}$. Let $a,b,c \in \mathbb{Q}$.
(1.) Closure law: $\forall$ $a,b\in \mathbb{Q} \Rightarrow a+b \in \mathbb{Q}$.
(2.) Associative law: $a+(b+c)=(a+b)+c$ $\forall$ $a,b,c \in \mathbb{Q}$.
(3.) Existence of identity element: $\exists$ $0=\frac{0}{1} \in \mathbb{Q}$ such that $a+0=0+a=a$ $\forall$ $a \in \mathbb{Q}$.
(4.) Existence of inverse element: For each $a \in \mathbb{Q}$, $\exists$ $-a \in \mathbb{Q}$ such that $a+(-a)=(-a)+a=0$.
(5.) Commutative law: $a+b=b+a$ $\forall$ $a,b \in \mathbb{Q}$.
$\therefore$ $(\mathbb{Q},+)$ is an abelian group. Since $\mathbb{Q}$ is infinite, it follows that $\mathbb{Q}$ is an infinite abelian group under addition. $\spadesuit$

• Problem 1.4: Show that $\mathbb{Q}$ is an infinite abelian group under multiplication.
Solution: Consider $\mathbb{Q}=\{\frac{x}{y}$ $/ x,y \in \mathbb{Z}$ and $y \neq 0 \}$. Let $a,b,c \in \mathbb{Q}$.
(1.) Closure law: $\forall$ $a,b\in \mathbb{Q} \Rightarrow ab \in \mathbb{Q}$.
(2.) Associative law: $a(bc)=(ab)c$ $\forall$ $a,b,c \in \mathbb{Q}$.
(3.) Existence of identity element: $\exists$ $1=\frac{1}{1} \in \mathbb{Q}$ such that $a\times 1=1\times a=a$ $\forall$ $a \in \mathbb{Q}$.
(4.) Existence of inverse element: For each $a \in \mathbb{Q}$, $\exists$ $\frac{1}{a} \in \mathbb{Q}$ such that $a\times \frac{1}{a}=\frac{1}{a} \times a=1$.
(5.) Commutative law: $ab=ba$ $\forall$ $a,b \in \mathbb{Q}$.
$\therefore$ $(\mathbb{Q},\times)$ is an abelian group. Since $\mathbb{Q}$ is infinite, it follows that $\mathbb{Q}$ is an infinite abelian group under multiplication. $\spadesuit$

• Problem 1.5: Show that $\mathbb{R}$ is an infinite abelian group under addition.
Solution: Let $a,b,c \in \mathbb{R}$.
(1.) Closure law: $\forall$ $a,b\in \mathbb{R} \Rightarrow a+b \in \mathbb{R}$.
(2.) Associative law: $a+(b+c)=(a+b)+c$ $\forall$ $a,b,c \in \mathbb{R}$.
(3.) Existence of identity element: $\exists$ $0 \in \mathbb{R}$ such that $a+0=0+a=a$ $\forall$ $a \in \mathbb{R}$.
(4.) Existence of inverse element: For each $a \in \mathbb{R}$, $\exists$ $-a \in \mathbb{R}$ such that $a+(-a)=(-a)+a=0$.
(5.) Commutative law: $a+b=b+a$ $\forall$ $a,b \in \mathbb{R}$.
$\therefore$ $(\mathbb{R},+)$ is an abelian group. Since $\mathbb{R}$ is infinite, it follows that $\mathbb{R}$ is an infinite abelian group under addition. $\spadesuit$

• Problem 1.6: Show that $\mathbb{R}$ is an infinite abelian group under multiplication.
Solution: Let $a,b,c \in \mathbb{R}$.
(1.) Closure law: $\forall$ $a,b\in \mathbb{R} \Rightarrow ab \in \mathbb{R}$.
(2.) Associative law: $a(bc)=(ab)c$ $\forall$ $a,b,c \in \mathbb{R}$.
(3.) Existence of identity element: $\exists$ $1 \in \mathbb{R}$ such that $a\times 1=1\times a=a$ $\forall$ $a \in \mathbb{R}$.
(4.) Existence of inverse element: For each $a \in \mathbb{R}$, $\exists$ $\frac{1}{a} \in \mathbb{R}$ such that $a\times \frac{1}{a}=\frac{1}{a} \times a=1$.
(5.) Commutative law: $ab=ba$ $\forall$ $a,b \in \mathbb{R}$.
$\therefore$ $(\mathbb{R},\times)$ is an abelian group. Since $\mathbb{R}$ is infinite, it follows that $\mathbb{R}$ is an infinite abelian group under multiplication. $\spadesuit$

• Problem 1.7: Show that $\mathbb{C}$ is an infinite abelian group under addition.
Solution: Consider, $\mathbb{C}=\{a+ib/ a,b \in \mathbb{R}$ and $i=\sqrt{-1} \}$. Let $x,y,z \in \mathbb{C}$ $\Rightarrow$ $x=a_1+ib_1$, $y=a_2+ib_2$, $z=a_3+ib_3$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{R}$ and $i=\sqrt{-1}$
(1.) Closure law: Consider $x+y$
$=(a_1+ib_1)+(a_2+ib_2)$
$=(a_1+a_2)+i(b_1+b_2) \in$ $\mathbb{C}$
$\therefore$ $\forall$ $x,y \in$ $\mathbb{C}$ $\Rightarrow$ $x+y \in \mathbb{C}$
(2.) Associative law: Consider $(x+y)+z$
$=[(a_1+ib_1)+(a_2+ib_2)]+(a_3+ib_3)$
$=[(a_1+a_2)+i(b_1+b_2)]+(a_3+ib_3)$
$=(a_1+a_2+a_3)+i(b_1+b_2+b_3)$
$=(a_1+ib_1)+[(a_2+a_3)+i(b_2+b_3)]$
$=(a_1+ib_1)+[(a_2+ib_2)]+(a_3+ib_3)]$
$=x+(y+z)$
$\therefore$ $(x+y)+z=x+(y+z)$ $\forall$ $x,y,z \in$ $\mathbb{C}$
(3.) Existence of additive identity: $\exists$ an element $0=0+i0 \in$ $\mathbb{C}$, such that $x+0=0+x=x$ $\forall$ $x \in$ $\mathbb{C}$.
(4.) Existence of additive inverse: For each $x=a_1+ib_1 \in$ $\mathbb{C}$, $\exists$ an element $-x=-(a_1+ib_1)=-a_1-ib_1$ $\in$ $\mathbb{C}$, such that $x+(-x)=(-x)+x=0$.
(5.) Commutative law: Consider $x+y$
$=(a_1+ib_1)+(a_2+ib_2)$
$=(a_1+a_2)+i(b_1+b_2)$
$=(a_2+a_1)+i(b_2+b_1)$
$=(a_2+ib_2)+(a_1+ib_1)$
$=y+x$
$\therefore$ $x+y=y+x$ $\forall$ $x,y \in$ $\mathbb{C}$
$\therefore$ $(\mathbb{C},+)$ is an abelian group. Since $\mathbb{C}$ is infinite, it follows that $\mathbb{C}$ is an infinite abelian group under addition. $\spadesuit$

• Problem 1.8: Show that $\mathbb{C}$ is an infinite abelian group under multiplication.
Solution: Consider $\mathbb{C}=\{a+ib/ a,b \in \mathbb{R}$ and $i=\sqrt{-1} \}$. Let $x,y,z \in \mathbb{C}$ $\Rightarrow$ $x=a_1+ib_1$, $y=a_2+ib_2$, $z=a_3+ib_3$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{R}$ and $i=\sqrt{-1}$
(1.) Closure law: Consider $xy$
$=(a_1+ib_1)(a_2+ib_2)$
$=a_1a_2+ib_1a_2+ia_1b_2+i^2b_1b_2$
$=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1) \in$ $\mathbb{C}$.
$\therefore$ $\forall$ $x,y \in$ $\mathbb{C}$ $\Rightarrow$ $xy \in \mathbb{C}$.
(2.) Associative law: Consider $x(yz)$
$=(a_1+ib_1)[(a_2a_3-b_2b_3)+i(a_2b_3+a_3b_2)]$
$=a_1(a_2a_3)-a_1(b_2b_3)+a_1(ia_2b_3)+a_1(ia_3b_2)+ib_1(a_2a_3)-ib_1(b_2b_3)-b_1(a_2b_3)-b_1(a_3b_2)$
$=(a_1a_2)a_3-(a_1b_2)b_3+(a_1ia_2)b_3+(a_1ib_2)a_3+(ib_1a_2)a_3-(ib_1b_2)b_3-(b_1a_2)b_3-(b_1b_2)a_3$
$=(a_1a_2+ia_1b_2+ib_1a_2-b_1b_2)a_3+(-a_1b_2+a_1a_2i-ib_1b_2-b_1a_2)b_3$
$=(a_1a_2+ia_1b_2+ib_1a_2-b_1b_2)a_3+(ia_1b_2+a_1a_2-b_1b_2+ib_1a_2)ib_3$
$=(a_1a_2+ia_1b_2+ia_2b_1-b_1b_2)(a_3+ib_3)$
$=[(a_1+ib_1)(a_2+ib_2)](a_3+ib_3)$
$=(xy)z$
$\therefore$ $x(yz)=(xy)z$ $\forall$ $x,y,z \in \mathbb{C}$
(3.) Existence of identity element: $\exists$ $1=1+i0 \in \mathbb{C}$ such that $x\times 1=1\times x=x$ $\forall$ $x \in \mathbb{C}$.
(4.) Existence of multiplicative inverse: Consider $\frac{1}{x}$
$=\frac{1}{a_1+ib_1}=\frac{1}{a_1+ib_1}\frac{a_1-ib_1}{a_1-ib_1}=\frac{a_1-ib_1}{(a_1+ib_1)(a_1-ib_1)}=\frac{a_1-ib_1}{a_1^2+b_1^2}=\frac{a_1}{a_1^2+b_1^2}+i\frac{-b_1}{a_1^2+b_1^2} \in \mathbb{C}$
$\therefore$ For each non zero $x \in \mathbb{C}$ $\exists$ $\frac{1}{x} \in \mathbb{C}$ such that $x\times\frac{1}{x}=\frac{1}{x}\times x=1$ $\forall$ $x \in \mathbb{C}$
(5.) Commutative law: Consider, $xy$
$=(a_1+ib_1)(a_2+ib_2)$
$=a_1a_2+ib_1a_2+ia_1b_2-b_1b_2$
$=a_2a_1+ib_2a_1+ia_2b_1-b_2b_1$
$=(a_2+ib_2)(a_1+ib_1)$
$=yx$
$\therefore$ $xy=yx$ $\forall$ $x,y \in \mathbb{C}$
$\therefore$ $(\mathbb{C},\times)$ is an abelian group. Since $\mathbb{C}$ is infinite, it follows that $\mathbb{C}$ is an infinite abelian group under multiplication. $\spadesuit$

• Problem 1.9: Show that the set of Gaussian integers is a group under addition.
Solution: Consider the set of Gaussian integers $\mathbb{Z}[i]=\{a+ib/ a,b \in \mathbb{Z}$ and $i=\sqrt{-1} \}$. Let $x,y,z \in \mathbb{Z}[i]$ $\Rightarrow$ $x=a_1+ib_1$, $y=a_2+ib_2$, $z=a_3+ib_3$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{Z}$ and $i=\sqrt{-1}$
(1.) Closure law: Consider $x+y$
$=(a_1+ib_1)+(a_2+ib_2)$
$=(a_1+a_2)+i(b_1+b_2) \in$ $\mathbb{Z}[i]$
$\therefore$ $\forall$ $x,y \in$ $\mathbb{Z}[i]$ $\Rightarrow$ $x+y \in \mathbb{Z}[i]$
(2.) Associative law: Consider $(x+y)+z$
$=[(a_1+ib_1)+(a_2+ib_2)]+(a_3+ib_3)$
$=[(a_1+a_2)+i(b_1+b_2)]+(a_3+ib_3)$
$=(a_1+a_2+a_3)+i(b_1+b_2+b_3)$
$=(a_1+ib_1)+[(a_2+a_3)+i(b_2+b_3)]$
$=(a_1+ib_1)+[(a_2+ib_2)]+(a_3+ib_3)]$
$=x+(y+z)$
$\therefore$ $(x+y)+z=x+(y+z)$ $\forall$ $x,y,z \in$ $\mathbb{Z}[i]$
(3.) Existence of additive identity: $\exists$ an element $0=0+i0 \in$ $\mathbb{Z}[i]$, such that $x+0=0+x=x$ $\forall$ $x \in$ $\mathbb{Z}[i]$.
(4.) Existence of additive inverse: For each $x=a_1+ib_1 \in$ $\mathbb{Z}[i]$, $\exists$ an element $-x=-(a_1+ib_1)=-a_1-ib_1$ $\in$ $\mathbb{Z}[i]$, such that $x+(-x)=(-x)+x=0$.
(5.) Commutative law: Consider $x+y$
$=(a_1+ib_1)+(a_2+ib_2)$
$=(a_1+a_2)+i(b_1+b_2)$
$=(a_2+a_1)+i(b_2+b_1)$
$=(a_2+ib_2)+(a_1+ib_1)$
$=y+x$
$\therefore$ $x+y=y+x$ $\forall$ $x,y \in$ $\mathbb{Z}[i]$
$\therefore$ $(\mathbb{Z}[i],+)$ is an abelian group. Since $\mathbb{Z}$ is infinite, it follows that $\mathbb{Z}[i]$ is an infinite abelian group under addition. $\spadesuit$

• Problem 1.10: Show that the set consisting of elements of the form $a+b\sqrt{2}$ where $a$ and $b$ are rationals is an abelian group under addittion.
Solution: Set of numbers of the form $a+b\sqrt{2}$, where $a$ and $b$ are rationals, is denoted by $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}$ /$a,b \in \mathbb{Q} \}$ Let $x,y,z \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $x=a_1+b_1\sqrt{2}$, $y=a_2+b_2\sqrt{2}$ and $z=a_3+b_3\sqrt{2}$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{Q}$
(1.) Closure law: Consider $x+y$
$=(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})$
$=(a_1+a_2)+(b_1+b_2)\sqrt{2} \in$ $\mathbb{Q}[\sqrt{2}]$
$\therefore$ $\forall$ $x,y \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $x+y \in \mathbb{Q}[\sqrt{2}]$.
(2.) Associative law: Consider, $(x+y)+z$
$=[(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})]+(a_3+b_3\sqrt{2})$
$=[(a_1+a_2)+(b_1+b_2)\sqrt{2}]+(a_3+b_3\sqrt{2})$
$=(a_1+a_2+a_3)+(b_1+b_2+b_3)\sqrt{2}$
$=(a_1+b_1)\sqrt{2}+[(a_2+a_3)+(b_2+b_3)\sqrt{2}]$
$=(a_1+b_1\sqrt{2})+[(a_2+b_2\sqrt{2})]+(a_3+b_3\sqrt{2})]$
$=x+(y+z)$
$\therefore$ $(x+y)+z=x+(y+z)$ $\forall$ $\in \mathbb{Q}[\sqrt{2}]$
(3.) Existence of identity element: $\exists$ an element $0=0+0\sqrt{2} \in$ $\mathbb{Q}[\sqrt{2}]$, such that $x+0=0+x=x$ $\forall$ $x \in \mathbb{Q}[\sqrt{2}]$.
(4.) Existence of inverse element: For each $x=(a_1+b_1\sqrt{2}) \in$ $\mathbb{Q}[\sqrt{2}]$, $\exists$ an element $-x=-(a_1+b_1\sqrt{2})=-a_1-b_1\sqrt{2}$ $\in$ $\mathbb{Q}[\sqrt{2}]$, such that $x+(-x)=(-x)+x=0$.
(5.) Commutative law: Consider $x+y$
$=(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})$
$=(a_1+a_2)+(b_1+b_2)\sqrt{2}$
$=(a_2+a_1)+(b_2+b_1)\sqrt{2}$
$=(a_2+b_2\sqrt{2})+(a_1+b_1\sqrt{2})$
$=y+x$
$\therefore$ $x+y=y+x$ $\forall$ $x,y \in \mathbb{Q}[\sqrt{2}]$
$\therefore$ $(\mathbb{Q}[\sqrt{2}],+)$ is an abelian group. $\spadesuit$

• Problem 1.11: Show that the set consisting of elements of the form $a+b\sqrt{2}$ where $a$ and $b$ are rationals is an abelian group under multiplication.
Solution: Set of numbers of the form $a+b\sqrt{2}$, where $a$ and $b$ are rationals, is denoted by $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}$ /$a,b \in \mathbb{Q} \}$. Let $x,y,z \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $x=a_1+b_1\sqrt{2}$, $y=a_2+b_2\sqrt{2}$ and $z=a_3+b_3\sqrt{2}$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{Q}$
(1.) Closure law: Consider $xy$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2$
$=(a_1a_2+2b_1b_2)+(a_1b_2+a_2b_1)\sqrt{2} \in$ $\mathbb{Q}[\sqrt{2}]$
$\therefore$ $\forall$ $x,y \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $xy \in \mathbb{Q}[\sqrt{2}]$
(2.) Associative law: Consider $x(yz)$
$=(a_1+b_1\sqrt{2})[(a_2+b_2\sqrt{2})(a_3+b_3\sqrt{2})]$
$=(a_1+b_1\sqrt{2})[(a_2a_3+2b_2b_3)+(a_2b_3+a_3b_2)\sqrt{2}]$
$=a_1a_2a_3+a_1a_2b_3\sqrt{2}+a_1a_3b_2\sqrt{2}+2a_1b_2b_3+b_1a_2a_3\sqrt{2}+2a_2b_1b_3+2a_3b_1b_2+2\sqrt{2}b_1b_2b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)a_3+(a_1a_2\sqrt{2}+2a_1b_2+a_2b_1\sqrt{2}+2\sqrt{2}b_1b_2)b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)a_3+(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)\sqrt{2}b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)(a_3+b_3\sqrt{2})$
$=[(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})](a_3+b_3\sqrt{2})$
$=(xy)z$
$\therefore$ $x(yz)=(xy)z$ $\forall$ $x,y,z \in \mathbb{Q}[\sqrt{2}]$
(3.) Existence of identity element: $\exists$ $1=1+0\sqrt{2}$ $\in$ $\mathbb{Q}[\sqrt{2}]$ such that, $x \times 1=1 \times x=x$ $\forall$ $x \in \mathbb{Q}[\sqrt{2}]$
(4.) Existence of inverse law: Consider $\frac{1}{x}$
$=\frac{1}{a_1+b_1\sqrt{2}}=\frac{1}{a_1+b_1\sqrt{2}}\frac{a_1-b_1\sqrt{2}}{a_1-b_1\sqrt{2}}=\frac{a_1-b_1\sqrt{2}}{a_1^2-2b_1^2}=\frac{a_1}{a_1^2-2b_1^2}+\frac{-b_1\sqrt{2}}{a_1^2-2b_1^2}$ $\in \mathbb{Q}[\sqrt{2}]$
$\therefore$ For each non zero $x \in \mathbb{Q}[\sqrt{2}]$ $\exists$ $\frac{1}{x} \in \mathbb{Q}[\sqrt{2}]$ such that $x\times\frac{1}{x}=\frac{1}{x}\times x=1$
(5.) Commutative law: $xy$
$=(a_1+b_2\sqrt{2})(a_2+b_2\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2$
$=a_2a_1+b_2a_1\sqrt{2}+a_2b_1\sqrt{2}+2b_2b_1$
$=(a_2+b_2\sqrt{2})(a_1+b_1\sqrt{2})$
$=yx$
$\therefore$ $xy=yx$ $\forall$ $x,y \in \mathbb{Q}[\sqrt{2}]$
$\therefore (\mathbb{Q}[\sqrt{2}],\times)$ is an abelian group. $\spadesuit$

• Problem 1.12: Show that the set consisting of elements of the form $a+b\sqrt{2}$ where $a,b \in \mathbb{Z}$ is not a group under multiplication.
Solution: Set of numbers of the form $a+b\sqrt{2}$, where $a$ and $b$ are integers, is denoted by $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}$ /$a,b \in \mathbb{Z} \}$. Let $x,y,z \in \mathbb{Z}[\sqrt{2}]$ $\Rightarrow$ $x=a_1+b_1\sqrt{2}$, $y=a_2+b_2\sqrt{2}$ and $z=a_3+b_3\sqrt{2}$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{Z}$
(1.) Closure law: Consider $xy$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2$
$=(a_1a_2+2b_1b_2)+(a_1b_2+a_2b_1)\sqrt{2} \in$ $\mathbb{Z}[\sqrt{2}]$
$\therefore$ $\forall$ $x,y \in \mathbb{Z}[\sqrt{2}]$ $\Rightarrow$ $xy \in \mathbb{Z}[\sqrt{2}]$
(2.) Associative law: Consider $x(yz)$
$=(a_1+b_1\sqrt{2})[(a_2a_3+2b_2b_3)+(a_2b_3+a_3b_2)\sqrt{2}]$
$=a_1a_2a_3+a_1a_2b_3\sqrt{2}+a_1a_3b_2\sqrt{2}+2a_1b_2b_3+b_1a_2a_3\sqrt{2}+2a_2b_1b_3+2a_3b_1b_2+2\sqrt{2}b_1b_2b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)a_3+(a_1a_2\sqrt{2}+2a_1b_2+a_2b_1\sqrt{2}+2\sqrt{2}b_1b_2)b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)a_3+(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)\sqrt{2}b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)(a_3+b_3\sqrt{2})$
$=[(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})](a_3+b_3\sqrt{2})$
$=(xy)z$
$\therefore$ $x(yz)=(xy)z$ $\forall$ $x,y,z \in \mathbb{Z}[\sqrt{2}]$
(3.) Existence of identity element: $\exists$ $1=1+0\sqrt{2}$ $\in$ $\mathbb{Z}[\sqrt{2}]$ such that, $x \times 1=1 \times x=x$ $\forall$ $x \in \mathbb{Z}[\sqrt{2}]$
(4.) Existence of inverse law: consider $2+0\sqrt{2} \in \mathbb{Z}[\sqrt{2}]$ but $\frac{1}{2+0\sqrt{2}}=\frac{1}{2}+0\sqrt{2} \notin \mathbb{Z}[\sqrt{2}]$ Hence for each $x \in \mathbb{Z}[\sqrt{2}]$ $\nexists$ $\frac{1}{x} \in \mathbb{Z}[\sqrt{2}]$
$\therefore (\mathbb{Z}[\sqrt{2}],\times)$ is not a group. $\spadesuit$

• Problem 1.13: Show that cube roots of unity is a group under multiplication.
Solution: Let the set of cube roots of unity be $G=\{1,\omega,\omega^2\}$ where $\omega=\frac{-1+i\sqrt{3}}{2}$ and $\omega^2=\frac{-1-i\sqrt{3}}{2}$

(1.) Closure law: Since all possible products belong to $G$, $G$ is closed w.r.t $\times$.
$\therefore$ $\forall$ $a,b\in G \Rightarrow a\times b \in G$.
(2.) Associative law: Clearly, $a\times (b\times c)=(a\times b)\times c$ $\forall$ $a,b,c \in G$.
(3.) Existence of identity element: The row headed by $1$ is same as the top row, So $1 \in G$ is the identity element such that $a \times 1=1 \times a=a$ $\forall$ $a \in G$.
(4.) Existence of inverse element: Inverse of $1$ is $1$, inverse of $\omega$ is $\omega^2$ and inverse of $\omega^2$ is $\omega$.
$\therefore$ Each element in $G$ has an inverse element in $G$.
(5.) Commutative law: Since there is a symmetry in the table we have $a\times b=b\times a$ $\forall$ $a,b \in G$.
$\therefore$ $(G,\times)$ is an abelian group. Since $G$ is finite, it follows that $G$ is a finite abelian group under multiplication of order 3.$\spadesuit$

• Problem 1.14: Show that $\{1,-1,i,-i\}$ is an finite abelian group under multiplication.
Solution: Let $G=\{1,-1,i,-i\}$ where $i=\sqrt{-1}$

(1.) Closure law: Since all possible products belong to $G$, $G$ is closed w.r.t $\times$.
$\therefore$ $\forall$ $a,b\in G \Rightarrow a\times b \in G$.
(2.) Associative law: Clearly, $a\times(b\times c)=(a\times b)\times c$ $\forall$ $a,b,c \in G$.
(3.) Existence of identity element: The row headed by $1$ is same as the top row, So $1 \in G$ is the identity element such that $a \times 1=1 \times a=a$ $\forall$ $a \in G$.
(4.) Existence of inverse element: Inverse of $1$ is $1$, inverse of $-1$ is $-1$, inverse of $i$ is $-i$ and inverse of $-i$ is $i$.
$\therefore$ Each element in $G$ has an inverse element in $G$.
(5.) Commutative law: Since there is a symmetry in the table we have $a\times b=b\times a$ $\forall$ $a,b \in G$.
$\therefore$ $(G,\times)$ is an abelian group. Since $G$ is finite, it follows that $G$ is a finite abelian group under multiplication of order 4. $\spadesuit$

• Problem 1.15: Show that the set of all $n^{th}$ roots of unity forms an finite abelian group under multiplication.
Solution: Consider the set of all $n^{th}$ roots of unity $\{e^{\frac{2\pi ik}{n}}/k=0,1,2,3,...\}$. Let $G=\{e^{\frac{2\pi ik}{n}}/k\in \mathbb{Z}\}=\{e^{\theta k}/k\in \mathbb{Z} \}$ where $\theta=\frac{2\pi i}{n}$
Let $a,b,c \in G \Rightarrow a=e^{\theta k_1},b=e^{\theta k_2},c=e^{\theta k_3}$
(1.) Closure law: Consider, $ab=e^{\theta k_1}e^{\theta k_2}=e^{\theta (k_1+k_2)} \in G$
$\therefore$ $\forall$ $a,b \in G \Rightarrow ab \in G$.
(2.) Associative law: Consider, $a(bc)=e^{\theta k_1}(e^{\theta k_2}e^{\theta k_3})=e^{\theta k_1}e^{\theta (k_2+k_3)}=e^{\theta (k_1+k_2+k_3)}=e^{\theta (k_1+k_2)}e^{\theta k_3}=(e^{\theta k_1}e^{\theta k_2})e^{\theta k_3}=(ab)c$
$\therefore$ $\forall$ $a,b,c \in G$ $\Rightarrow$ $a(bc)=(ab)c$.
(3.) Existence of identity: $\exists$ $1=e^{\theta 0} \in G$ such that $a\times 1=1\times a=a$ $\forall$ $a \in G$.
(4.) Existence of inverse: For each $a=e^{\theta k_1} \in G$ $\exists$ $\frac{1}{a}=\frac{1}{e^{\theta k_1}}=e^{-\theta k_1} \in G$ such that $a\times \frac{1}{a}=\frac{1}{a}\times a=1$.
(5.) Commutative law: Consider, $ab=e^{\theta k_1}e^{\theta k_2}=e^{\theta (k_1+k_2)}=e^{\theta (k_2+k_1)}=e^{\theta k_2}e^{\theta k_1}=ba$
$\therefore$ $\forall$ $a,b \in G$ $\Rightarrow$ $ab=ba$.
$\therefore$ $(G,\times)$ is an finite abelian group of order $n$. $\spadesuit$

• Problem 1.16: Show that the set consisting of elements of the form $e^{i\theta}$ is an abelian group under multiplication.
Solution: Let $G=\{e^{i\theta}/0\leq \theta \leq 2\pi\}$. Let $a,b,c \in G$ $\Rightarrow$ $a=e^{i\theta_1}$ , $b=e^{i\theta_2}$ and $c=e^{i\theta_3}$
(1.) Closure law: Consider, $ab=e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)} \in G$
$\therefore$ $\forall$ $a,b \in G \Rightarrow ab \in G$.
(2.) Associative law: Consider, $a(bc)=e^{i\theta_1}(e^{i\theta_2}e^{i\theta_3})=e^{i\theta_1}e^{i(\theta_1+\theta_2)}=e^{i(\theta_1+\theta_2+\theta_3)}=e^{i(\theta_1+\theta_2)}e^{i\theta_3}=(e^{i\theta_1}e^{i\theta_2})e^{i\theta_3}=(ab)c$
$\therefore$ $a(bc)=(ab)c$ $\forall$ $a,b,c \in G$.
(3.) Existence of identity element: $\exists$ $1=e^{i0} \in G$ such that $e^{i\theta_1}e^{i0}=e^{i0}e^{i\theta_1}=e^{i\theta_1}$
$\therefore$ $a\times 1=1 \times a=a$ $\forall$ $a \in G$.
(4.) Existence of inverse law: For each $a=e^{i\theta_1} \in G$ $\exists$ $\frac{1}{a}=e^{-i\theta_1}$ such that $a\times \frac{1}{a}=\frac{1}{a}\times a=1$.
(5.) Commuatative law: Consider, $ab=e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}=e^{i(\theta_2+\theta_1)}=e^{i\theta_2}e^{i\theta_1}=ba$
$\therefore$ $ab=ba$ $\forall$ $a,b \in G$.
$\therefore$ $(G,\times)$ is an abelian group. $\spadesuit$

• Problem 1.17: Show that the set $\{5^n/n \in \mathbb{Z}\}$ is an abelian group under multiplication.
Solution: Let $G=\{5^n/n \in \mathbb{Z}\}$. Let $a,b,c \in G$ $\Rightarrow$ $a=5^m$ , $b=5^n$ and $c=5^r$ where $m,n,r \in \mathbb{Z}$
(1.) Closure law: Consider, $ab=5^m5^n=5^{m+n} \in G$
$\therefore$ $\forall$ $a,b \in G \Rightarrow ab \in G$.
(2.) Associative law: Consider, $a(bc)=5^m(5^n5^r)=5^m5^{n+r}=5^{m+n+r}=5^{m+n}5^r=(5^m5^n)5^r=(ab)c$
$\therefore$ $a(bc)=(ab)c$ $\forall$ $a,b,c \in G$.
(3.) Existence of identity element: $\exists$ $1=5^0 \in G$ such that $5^n5^0=5^0r^n=5^n$
$\therefore$ $a\times 1=1 \times a=a$ $\forall$ $a \in G$.
(4.) Existence of inverse law: For each $a=5^n \in G$ $\exists$ $\frac{1}{a}=5^{-n} \in G$ such that $a\times \frac{1}{a}=\frac{1}{a}\times a=1$.
(5.) Commuatative law: Consider, $ab=5^m5^n=5^{m+n}=5^{n+m}=5^n5^m=ba$
$\therefore$ $ab=ba$ $\forall$ $a,b \in G$.
$\therefore$ $(G,\times)$ is an abelian group. $\spadesuit$

• Problem 1.18: Show that the set of positive rationals form an infinite abelian group under the binary operation $a*b=\frac{ab}{2}$.
Solution: Consider $\mathbb{Q}^+=\{\frac{x}{y}/ x,y \in \mathbb{Z}^+\}$. Let $a,b,c \in \mathbb{Q}^+$.
(1.) Closure law: Since $a,b \in \mathbb{Q}^+ \Rightarrow \frac{ab}{2} \in \mathbb{Q}^+$, we have $a*b=\frac{ab}{2} \in \mathbb{Q}^+$.
$\therefore$ $\forall$ $a,b\in \mathbb{Q}^+ \Rightarrow a*b \in \mathbb{Q}^+$.
(2.) Associative law: Consider, $a*(b*c)=a*\frac{bc}{2}=\frac{a\frac{bc}{2}}{2}=\frac{abc}{4}...(1)$
Now consider, $(a*b)*c=\frac{ab}{2}*c=\frac{\frac{ab}{2}c}{2}=\frac{abc}{4}...(2)$
$\therefore$ From (1) and (2) we have, $a*(b*c)=(a*b)*c$ $\forall$ $a,b,c \in \mathbb{Q}^+$.
(3.) Existence of identity element: $\exists$ $e \in \mathbb{Q}^+$ such that $a*e=e*a=a$ $\forall$ $a \in \mathbb{Q}^+$.
Consider, $a*e=a \Rightarrow \frac{ae}{2}=a \Rightarrow e=2$
$\therefore$ $e=2$ is the identity element in $\mathbb{Q}^+$ such that $a*e=e*a=a$ $\forall$ $a \in \mathbb{Q}^+$.
(4.) Existence of inverse element: For each $a \in \mathbb{Q}^+$, $\exists$ $a^{-1} \in \mathbb{Q}^+$ such that $a*a^{-1}=a^{-1}*a=2$.
Consider, $a*a^{-1}=2 \Rightarrow \frac{aa^{-1}}{2}=2 \Rightarrow a^{-1}=\frac{4}{a}$.
$\therefore$ For each $a \in \mathbb{Q}^+ \exists$ inverse element $\frac{4}{a}$ such that $a*\frac{4}{a}=\frac{4}{a}*a=2$.
(5.) Commutative law: Consider, $a*b=\frac{ab}{2}=\frac{ba}{2}=b*a$
$\therefore$ $a*b=b*a$ $\forall$ $a,b \in \mathbb{Q}^+$.
$\therefore$ $(\mathbb{Q}^+,*)$ is an abelian group. Since $\mathbb{Q}^+$ is infinite, it follows that $\mathbb{Q}^+$ is an infinite abelian group under $*$. $\spadesuit$

• Problem 1.19: Show that the set $\mathbb{Z}$ is a group under $a*b=a+b+1$ $\forall$ $a,b \in \mathbb{Z}$.
Solution: Consider $\mathbb{Z}=\{0,\pm 1, \pm 2, \pm 3,.....\}$. Let $a,b,c \in \mathbb{Z}$.
(1.) Closure law: Since $\forall$ $a,b\in \mathbb{Z} \Rightarrow a+b \in \mathbb{Z} \Rightarrow a+b+1 \in \mathbb{Z}$
$\therefore$ $\forall$ $a,b\in \mathbb{Z} \Rightarrow a*b=a+b+1 \in \mathbb{Z}$.
(2.) Associative law: Consider, $a*(b*c)=a*(b+c+1)=a+b+c+1+1=a+b+c+2...(1)$
Now consider, $(a*b)*c=(a+b+1)*c=a+b+1+c+1=a+b+c+2...(2)$
From (1) and (2), we have $a*(b*c)=(a*b)*c$ $\forall$ $a,b,c \in \mathbb{Z}$.
(3.) Existence of identity element: $\exists$ $e \in \mathbb{Z}$ such that $a*e=e*a=a$ $\forall$ $a \in \mathbb{Z}$.
Consider, $a*e=a \Rightarrow a+e+1=a \Rightarrow e=-1$
$\therefore$ $e=-1$ is the identity element in $\mathbb{Z}$ such that $a*e=e*a=a$ $\forall$ $a \in \mathbb{Z}$.
(4.) Existence of inverse element: For each $a \in \mathbb{Z}$, $\exists$ $a^{-1} \in \mathbb{Z}$ such that $a*a^{-1}=a^{-1}*a=-1$.
Consider, $a*a^{-1}=-1 \Rightarrow a+a^{-1}+1=-1 \Rightarrow a^{-1}=-2-a$
$\therefore$ For each $a \in \mathbb{Z}$, $\exists$ inverse element $-2-a \in \mathbb{Z}$ such that $a*(-2-a)=(-2-a)*a=-1$.
(5.) Commutative law: Consider, $a*b=a+b+1=b+a+1=b*a$
$\therefore$ $a*b=b*a$ $\forall$ $a,b \in \mathbb{Z}$.
$\therefore$ $(\mathbb{Z},*)$ is an abelian group. $\spadesuit$

• Problem 1.20: Show that the set of all real numbers except -1 from an abelian group under the composition $a*b=a+b+ab$
Solution: Let $G=\mathbb{R}-\{-1\}$. Let $a,b,c \in G$ where $a\neq -1,b\neq -1,c \neq -1$.
(1.) Closure law: For $a,b \in G$ $\Rightarrow$ $a+b+ab \in G$
$\therefore$ $\forall$ $a,b \Rightarrow a*b=a+b+ab \in G$.
(2.) Associative law: Consider, $a*(b*c)=a*(b+c+bc)=a+b+c+bc+a(b+c+bc)=a+b+c+bc+ab+ac+abc$
Now consider, $(a*b)*c=(a+b+ab)*c=a+b+ab+c+(a+b+ab)c=a+b+ab+c+ac+bc+abc$
$\therefore$ $a*(b*c)=(a*b)*c$ $\forall$ $a,b,c \in G$.
(3.) Existence of identity: $\exists$ $0 \in G$ such that $a*0=0*a=a$ $\forall$ $a \in G$.
(4.) Existence of inverse: For each $a \in$ there exists $a^{-1}$ such that $a*a^{-1}=a^{-1}*a=0$
Consider, $a*a^{-1}=0 \Rightarrow a+a^{-1}+aa^{-1}=0 \Rightarrow a^{-1}(1+a)=-a \Rightarrow a^{-1}=\frac{-a}{1+a} \in G$ as $a\neq -1$.
(5.) Commutative law: Consider, $a*b=a+b+ab=b+a+ba=b*a$
$\therefore$ $\forall$ $a,b \in G$ $\Rightarrow$ $a*b=b*a$.
$\therefore (G,*)$ is an abelian group.$\spadesuit$

• Problem 1.21: Show that the set of $2 \times 2$ real matrices is an abelian group under matrix addition.
Solution: Set of $2 \times 2$ real matrices is denoted by $\mathbb{M}=\Bigg\{\begin{bmatrix} a & b\\ c & d \end{bmatrix} \Big/ a,b,c,d \in \mathbb{R} \Bigg\}$.
Let $X,Y,Z \in \mathbb{M}$ $\Rightarrow$ $X=$ $\left[ \begin{array}{cc} a_1 & b_1 \\ c_1 & d_1 \end{array} \right]$, $Y=$ $\left[ \begin{array}{cc} a_2 & b_2 \\ c_2 & d_2 \end{array} \right]$ and $Z=$ $\left[ \begin{array}{cc} a_3 & b_3 \\ c_3 & d_3 \end{array} \right]$ where $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3 \in \mathbb{R}$
(1.) Closure law: Consider $X+Y$
$=\left[\begin{array}{cc}a_1 & b_1 \\c_1 & d_1\end{array}\right]+ \left[\begin{array}{cc}a_2 & b_2 \\c_2 & d_2\end{array}\right]$
$=\left[\begin{array}{cc}a_1+a_2 & b_1+b_2 \\c_1+c_2 & d_1+d_2\end{array}\right]$ $\in \mathbb{M}$
$\therefore$ $\forall$ $X,Y \in \mathbb{M}$ $\Rightarrow$ $X+Y \in \mathbb{M}$
(2.) Associative law: Clearly $X+(Y+Z)=(X+Y)+Z$ $\forall$ $X,Y,Z \in \mathbb{M}$
(3.) Existence of additive identity: $\exists$ an element $0=\left[ \begin{array}{cc}0 & 0 \\0 & 0\end{array}\right]$ $\in$ $\mathbb{M}$, such that $X+0=0+X=X$ $\forall$ $X \in$ $\mathbb{M}$.
(4.) Existence of additive inverse: For each $X= \left[ \begin{array}{cc} a_1 & b_1 \\ c_1 & d_1 \end{array} \right] \in$ $\mathbb{M}$, $\exists$ an element $-X=\left[ \begin{array}{cc} -a_1 & -b_1 \\ -c_1 & -d_1 \end{array} \right]$ $\in$ $\mathbb{M}$, such that $X+(-X)=(-X)+X=0$
(5.) Commutative law: Consider $X+Y$
$=\left[ \begin{array}{cc} a_1+a_2 & b_1+b_2 \\ c_1+c_2 & d_1+d_2 \end{array} \right]$
$=\left[ \begin{array}{cc} a_2+a_1 & b_2+b_1 \\ c_2+c_1 & d_2+d_1 \end{array} \right]$
$=\left[ \begin{array}{cc} a_2 & b_2 \\ c_2 & d_2 \end{array} \right]+\left[ \begin{array}{cc} a_1 & b_1 \\ c_1 & d_1 \end{array} \right]$
$=Y+X$
$\therefore$ $X+Y=Y+X$ $\forall$ $X,Y \in \mathbb{M}$
$\therefore$ $(\mathbb{M},+)$ is an abelian group. $\spadesuit$

• Problem 1.22: Show that the set of $2 \times 2$ non singular real matrices is a non commutative group under matrix multiplication.
Solution: Consider $\mathbb{M}=\Bigg\{ \begin{bmatrix} a & b\\ c & d \end{bmatrix} / a,b,c,d \in \mathbb{R}$ and $ad-bc\neq 0 \Bigg\}$
Let $A,B,C \in \mathbb{M}$ $\Rightarrow$ $A=\begin{bmatrix} a_1 & b_1\\ c_1 & d_1 \end{bmatrix}$, $B=\begin{bmatrix} a_2 & b_2\\ c_2 & d_2 \end{bmatrix}$ and $C=\begin{bmatrix} a_3 & b_3\\ c_3 & d_3 \end{bmatrix}$ where $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3 \in \mathbb{R}$
(1.) Closure law: Consider, $AB$ $=\begin{bmatrix} a_1 & b_1\\ c_1 & d_1 \end{bmatrix} \begin{bmatrix} a_2 & b_2\\ c_2 & d_2 \end{bmatrix}$ $=\begin{bmatrix} a_1a_2+b_1c_2 & a_1b_2+b_1d_2\\ c_1a_2+d_1c_2 & c_1b_2+d_1d_2 \end{bmatrix}$ $\in \mathbb{M}$
$\therefore$ $\forall$ $A,B \in \mathbb{M}$ $\Rightarrow AB \in \mathbb{M}$.
(2.) Associative law: Clearly $A(BC)=(AB)C$ $\forall$ $A,B,C \in \mathbb{M}$
(3.) Existence of identity element: $\exists$ $I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \in \mathbb{M}$ such that $AI=IA=A$ $\forall$ $A \in \mathbb{M}$
(4.) Existence of inverse law: Since each $A \in \mathbb{M}$ is non singular $\exists$ $A^{-1}=\frac{1}{|A|}adjA \in \mathbb{M}$ such that $AA^{-1}=A^{-1}A=I$.
(5.) Commuatative law: Consider $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right],\left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right] \in$ $\mathbb{M}$, we have $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]\left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right]=\left[ \begin{array}{cc} 4 & -1 \\ 2 & -2 \end{array} \right]$ and $\left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right]\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]=\left[ \begin{array}{cc} 2 & 3 \\ 2 & 0 \end{array} \right]$
Hence $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]\left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right] \neq \left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right]\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]$
$\therefore$ $(\mathbb{M},\times)$ is a non abelian group. $\spadesuit$

• Problem 1.23: Show that the set of elements of the form $\begin{bmatrix} x & x \\ x & x \end{bmatrix}$ where $x \in \mathbb{R}$, is a group with respect to matrix multiplication.
Solution: Let $\mathbb{M}=\Bigg\{ \begin{bmatrix} x & x\\ x & x \end{bmatrix} / x \in \mathbb{R} \Bigg\}$
Let $A,B,C \in \mathbb{M}$ $\Rightarrow$ $A=\begin{bmatrix} x_1 & x_1\\ x_1 & x_1 \end{bmatrix}$, $B=\begin{bmatrix} x_2 & x_2\\ x_2 & x_2 \end{bmatrix}$ and $C=\begin{bmatrix} x_3 & x_3\\ x_3 & x_3 \end{bmatrix}$ where $x_1,x_2,x_3 \in \mathbb{R}$
(1.) Closure law: Consider, $AB$ $=\begin{bmatrix} x_1 & x_1\\ x_1 & x_1 \end{bmatrix} \begin{bmatrix} x_2 & x_2\\ x_2 & x_2 \end{bmatrix}$ $= \begin{bmatrix} 2x_1x_2 & 2x_1x_2\\ 2x_1x_2 & 2x_1x_2 \end{bmatrix} \in \mathbb{M}$
$\therefore$ $\forall$ $A,B \in \mathbb{M} \Rightarrow AB \in \mathbb{M}$.
(2.) Associative law: $A(BC)=(AB)C$ $\forall$ $A,B,C \in \mathbb{M}$.
(3.) Existence of identity element: $\exists$ $I=\begin{bmatrix} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \in \mathbb{M}$ such that $AI=IA=A$ $\forall$ $A \in \mathbb{M}$.
(4.) Existence of inverse law: For $A=\begin{bmatrix} x_1 & x_1\\ x_1 & x_1 \end{bmatrix}$ $\Rightarrow$ $|A|$ does not exist $\Rightarrow$ $A^{-1}$ does not exist.
$\therefore$ For each $A \in \mathbb{M}$ $\nexists$ $A^{-1} \in \mathbb{M}$.
$\therefore$ $(\mathbb{M},\times)$ is not a group. $\spadesuit$

• Problem 1.24: Show that the set of elements of the form $\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ where $\alpha \in \mathbb{R}$, is a group with respect to matrix multiplication.
Solution: Let $\mathbb{M}=\Bigg\{ \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} / \alpha \in \mathbb{R} \Bigg\}$
Let $A,B,C \in \mathbb{M}$ $\Rightarrow$ $A=\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, $B=\begin{bmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{bmatrix}$ and $C=\begin{bmatrix} \cos\gamma & -\sin\gamma \\ \sin\gamma & \cos\gamma \end{bmatrix}$ where $\alpha,\beta,\gamma \in \mathbb{R}$
(1.) Closure law: Consider $AB$
$=\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{bmatrix}$
$=\begin{bmatrix} \cos\alpha\cos\beta-\sin\alpha\sin\beta & -\cos\alpha\sin\beta-\sin\alpha\cos\beta \\ \sin\alpha\cos\beta+\cos\alpha\sin\beta & -\sin\alpha\sin\beta+\cos\alpha\cos\beta \end{bmatrix}$
$=\begin{bmatrix} \cos(\alpha+\beta) &-\sin(\alpha+\beta)\\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{bmatrix} \in \mathbb{M}$
$\therefore$ $\forall$ $A,B \in \mathbb{M}$ $\Rightarrow$ $AB \in \mathbb{M}$.
(2.) Associative law: Clearly $A(BC)=(AB)C$ $\forall$ $A,B,C \in \mathbb{M}$.
(3.) Existence of identity element: $\exists$ $I=\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=\begin{bmatrix} \cos 0 & -\sin 0 \\ \sin 0 & \cos 0 \end{bmatrix} \in \mathbb{M}$ such that $AI=IA=A$ $\forall$ $A \in \mathbb{M}$
(4.) Existence of inverse element: For each $A=\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \in \mathbb{M}$ $\exists$ $A^{-1}=\frac{1}{|A|}$adj$A=\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}=\begin{bmatrix} \cos(-\alpha) & -\sin(-\alpha) \\ \sin(-\alpha) & \cos(-\alpha) \end{bmatrix}$ $\in$ $\mathbb{M}$ such that $AA^{-1}=A^{-1}A=I$
$\therefore$ $(\mathbb{M},\times)$ is a group. $\spadesuit$

• Exercise 1.1: Show that the set $\{3^n/n \in \mathbb{Z}\}$ is an abelian group under multiplication.

• Exercise 1.2: Show that the set consisting of elements of the form $a+b\sqrt{2}$ where $a,b \in \mathbb{Z}$ is an abelian group under addittion.

• Exercise 1.3: Show that $\mathbb{Q}[\sqrt{3}]$ is an abelian group under multiplication.

• Exercise 1.4: Show that $\mathbb{Z}[\sqrt{5}]$ is an abelian group under addition.

• Exercise 1.5: Show that the set of Gaussian integers is not a group group under multiplication.