MKU University B.Sc., Mathematics SMTJC51 Modern Algebra Chapter 1

Table of Contents

Table of Contents
1
 1.1 Groups
 1.2 Properties of a Group
 1.3 Permutation Groups
 1.4 Subgroups
 1.5 Cyclic Groups
 1.6 Order of an Element

Chapter 1

1.1 Groups

Definition 1.1.1. A group is an ordered pair $(G,\ast )$, where $G$ is a nonempty set and $\ast$ is a binary operation on $G$ such that the following properties hold:

1. For all $a,b,c\in G,a\ast (b\ast c)=(a\ast b)\ast c$ (associative law).
2. There exists $e\in G$ such that for all $a\in G,a\ast e=a=e\ast a$ (existence of an identity).
3. For all $a\in G$, there exists $a^{\prime }\in G$ such that $a\ast a^{\prime }=e=a^{\prime }\ast a$ (existence of an inverse).

Example 1.1.2.

1. $Z,Q$, $R$ and $C$ are groups under usual addition.
2. The set of all $2\times 2$ matrices $\left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right)$ where $a,b,c,d\in R$is a group under matrix addition. $\left(\begin{array}{cc}0\hfill & 0\hfill \\ 0\hfill & 0\hfill \end{array}\right)$ is the identity element and $\left(\begin{array}{cc}-a\hfill & -b\hfill \\ -c\hfill & -d\hfill \end{array}\right)$ is the inverse of $\left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right).$
3. The set of $\mathit{al}12\times 2$ non-singular matrices $\left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right)$ where $a,b,c,d\in R$ is a group under matrix multiplication. $\left(\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right)$ is the identity element. The inverse of $\left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right)$ is $\frac{1}{\left|A\right|}\left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right)$ where $\left|A\right|=\mathit{ad}-\mathit{bc}\ne 0.$
4. $N$ is not a group under usual addition since there is no element $e\in N$ such that $x+e=x.$
5. The set $E$ of all even integers under usual addition is a group.
6. $Q^{\ast }$ and $R^{\ast }$ under usual multiplication are groups. 1 is the identity element and the inverse of a non-zero element $a$ is $1/a.$
7. $Q^{+}$ is a group under usual multiplication. For $a,b\in Q^{+}\Rightarrow \mathit{ab}\in Q^{+}$. Therefore usual multiplication is a binary operation in $Q^{+}.$ $1\in Q^{+}$ is the identity element. If $a\in Q^{+},(1/a)\in Q^{+}$ is the inverse of $a.$
8. $Z$ under the usual multiplication is not a group.
9. Let $A$ be any non-empty set. Let $B\left(A\right)$ be the set of all bijections from $A$ to itself. $B\left(A\right)$ is a group under the composition of functions. We know that $f,g\in B\left(A\right)\Rightarrow f◦g\in B\left(A\right)$. The composition of functions is associative. $i_A:A\to A$ is the identity element. If $f:A\to A$ is a bijection, then $f^{-1}:A\to A$ is also a bijection and $f◦f^{-1}=f^{-1}◦f=i_A$.
10. Let $G=\left\{e\right\}$ and $e\ast e=e$. Obviously $G$ is a group.

11. Let $G=\{1,-1\}.G$ is a group under multiplication. 1 is the identity element. The inverse of each element is itself.
    The Cayley table for this group is

    
12. $\left(\rho \right(S),△)$ is a group. $△$ is associative. Also $A△\Phi =\Phi △A=A$ for all $A\in \rho \left(S\right)$. Hence $\Phi$ is the identity element. $A△A=\Phi$ so that inverse of each element is itself.

13. $G=\{1,i,-1,-i\}$. $G$ is a group under usual multiplication. The identity element is 1. The inverse of 1, $i,-1$ and $-i$ are $1,-i,-1$ and $i$ respectively.

    The Cayley table for this group is given by

    
14. Let $G=\left\{\left(\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right),\left(\begin{array}{cc}-1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right),\left(\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & -1\hfill \end{array}\right),\left(\begin{array}{cc}-1\hfill & 0\hfill \\ 0\hfill & -1\hfill \end{array}\right)\right\}$. $G$ is a group under matrix multiplication.

15. $C^{\ast }$ is a group under usual multiplication given by $(a+\mathit{ib})(c+\mathit{id})=(\mathit{ac}-\mathit{bd})+i(\mathit{ad}+\mathit{bc})$.

    Proof : Let $x,y\in C^{\ast }$ Then $x=a+\mathit{ib}$ where $a$ and $b$ are not simultaneously zero and $y=c+\mathit{id}$ where $c$ and $d$ are simultaneously zero. Now,

    $$\mathit{xy}=(a+\mathit{ib})(c+\mathit{id})=(\mathit{ac}-\mathit{bd})+i(\mathit{ad}+\mathit{bc}).$$

    We shall first prove that $\mathit{ad}-\mathit{bc}$ and $\mathit{ad}+\mathit{bc}$ are not simultaneously zero. Suppose,

    $$\begin{array}{lll}\hfill \mathit{ac}-\mathit{bd}& =0\text{ and }& \hfill \text{(1.1)}\\ \hfill \mathit{ad}+\mathit{bc}& =0& \hfill \text{(1.2)}\end{array}$$

    Multiplying (1.1) by $d$ and (1.2) by $c$ and subtracting, we get

    $$b(d^2+c^2)=0.$$

    Therefore either $b=0$ or $d^2+c^2=0$. Thus either $b=0$ or $(c=0$ and $d=0)$.
    Similarly, either $a=0$ or $(c=0$ and $d=0)$ . Thus $(a=0$ and $b=0)$ or $(c=0$ and $d=0)$ .
    Thus $x=0$ or $y=0$ which is a contradiction.
    Hence $\mathit{xy}\in C^{\ast }$.
    Now, let $x=a+\mathit{ib},y=c+\mathit{id}$ and $z=e+\mathit{if}$. Then

    $$\begin{array}{llll}\hfill x\left(\mathit{yz}\right)& =(a+\mathit{ib})\left[\right(\mathit{ce}-\mathit{df})+i(\mathit{de}+\mathit{cf}\left)\right]& \hfill & \\ \hfill & =(\mathit{ace}-\mathit{adf}-\mathit{bde}-\mathit{bcf})+(\mathit{bce}+\mathit{bdf}+\mathit{ade}+\mathit{acf})& \hfill & \end{array}$$

    Similarly,

    $$\begin{array}{llll}\hfill \left(\mathit{xy}\right)z& =(\mathit{ace}-\mathit{adf}-\mathit{bde}-\mathit{bcf})+(\mathit{bce}+\mathit{bdf}+\mathit{ade}+\mathit{acf})& \hfill & \end{array}$$

    Hence $x\left(\mathit{yz}\right)=\left(\mathit{xy}\right)z$.
    Clearly $1+i0$ is the identity element. Also

    $$\begin{array}{llll}\hfill \frac{1}{x}& =\frac{1}{a+\mathit{ib}}& \hfill & \\ \hfill & =\frac{a-\mathit{ib}}{(a+\mathit{ib})(a-\mathit{ib})}& \hfill & \\ \hfill & =\frac{a-\mathit{ib}}{a^2+b^2}& \hfill & \\ \hfill & =\left(\frac{a}{a^2+b^2})-i(\frac{b}{a^2+b^2}\right)& \hfill & \end{array}$$

    Since $a^2+b^2\ne 0,1/x\in C^{\ast }$ and is the inverse of $x$.
    Hence $C^{\ast }$ is a group under usual multiplication. □

16. Let $G=\{z:z\in C$ and $\left|z\right|=1\}$. Then $G$ is a under usual multiplication.

    Proof : Let $z_1,z_2\in G$. Then

    $$\left|z_1\right|=\left|z_2\right|=1,\left|z_1{z}_2\right|=\left|z_1\right|\left|z_2\right|=1$$

    and so $z_1,z_2\in G$.
    We know that usual multiplication of complex numbers is associative.
    Also $1=1+i0\in G$ and is the identity element.
    Now, let $z\in G$. Then $\left|z\right|=1$.
    Hence $|1/z|=1/\left|z\right|=1$ and so $1/z\in G$ and is the inverse of $z$.
    Hence $G$ is a group. □

17. The set of all $n^{\mathit{th}}$ roots of unity with usual multiplication is a group.

    Proof : Let $\omega =\mathit{cos}(2\pi /n)+\mathit{isin}(2\pi /n)$.
    Then the $n^{\mathit{th}}$ roots of unity are given by 1, $\omega ,{\omega }^2,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\omega }^{n-1}$.
    Let $G=\{1,\omega ,{\omega }^2,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\omega }^{n-1}\}$.
    We know that ${\omega }^n=1,{\omega }^{n+1}=\omega$ etc.
    Let ${\omega }^r,{\omega }^s\in G$.
    Let $r+s=\mathit{qn}+t$ where $0\le t<n$. Then

    $$\begin{array}{llll}\hfill {\omega }^r{\omega }^s& ={\omega }^{r+s}& \hfill & \\ \hfill & ={\omega }^{\mathit{qn}+t}& \hfill & \\ \hfill & ={\left({\omega }^n\right)}^q{\omega }^t& \hfill & \\ \hfill & ={\omega }^t\in G& \hfill & \end{array}$$

    We know that usual multiplication of complex number is associative.
    Clearly $1\in G$ is the identity element and the inverse of ${\omega }^r$ is ${\omega }^{n-r}$. Hence $G$ is a group. □

18. Let $G=\{a+b\sqrt{2}:a,b\in Z\}$. Then $G$ is a group under addition.

    Proof : Let $a+b\sqrt{2}$ and $c+d\sqrt{2}\in G$. Then

    $$(a+b\sqrt{2})+(c+d\sqrt{2})=(a+c)+(b+d)\sqrt{2}\in G.$$

    We know that usual addition is associative.
    Clearly $0=0+0\sqrt{2}\in G$ is the identity element and $-a-b\sqrt{2}$ is the inverse of $a+b\sqrt{2}$. Hence $G$ is a group. □

19. Let $G$ be the set of all real numbers except $-1$. Define$\ast$on $G$ by $a\ast b=a+b+\mathit{ab}.$ Then $(G,\ast )$ is a group.

    Proof : Let $a,b\in G$. Then $a\ne -1$ and $b\ne -1$.
    We claim that $a\ast b\ne -1$.
    Suppose $a\ast b=-1$. Then $a+b+\mathit{ab}=-1$ so that $a+b+\mathit{ab}+1=0$,
    i.e., $(a+1)(b+1)=0$ so that either $a=-1$ or $b=-1$ which is a contradiction.
    Hence $a\ast b\ne -1$ and thus$\ast$is a binary operation on $G$. Now

    $$\begin{array}{llll}\hfill a\ast (b\ast c)& =a\ast (b+c+\mathit{bc})& \hfill & \\ \hfill & =a+(b+c+\mathit{bc})+a(b+c+\mathit{bc})& \hfill & \\ \hfill & =a+b+c+\mathit{bc}+\mathit{ab}+\mathit{ac}+\mathit{abc}& \hfill & \end{array}$$

    Also

    $$\begin{array}{llll}\hfill (a\ast b)\ast c& =(a+b+\mathit{ab})\ast c& \hfill & \\ \hfill & =a+b+\mathit{ab}+c+(a+b+\mathit{bc})c& \hfill & \\ \hfill & =a+b+c+\mathit{ab}+\mathit{ac}+\mathit{bc}+\mathit{abc}& \hfill & \end{array}$$

    Hence $a\ast (b\ast c)=(a\ast b)\ast c$.
    Also $0$ is the identity, for $a\ast 0=a+O+\mathit{Oa}=a$ and $0\ast a=0+a+0a=a$.
    Now, let $a^{\prime }$ be such that $a\ast a^{\prime }=0.$
    Then $a+a^{\prime }+a{a}^{\prime }=0$ so that $a^{\prime }=-a/(1+a)$.
    Since $a\ne -1$, we have $a^{\prime }\in R-\{-1\}.$ Also

    $$\begin{array}{llll}\hfill a^{\prime }\ast a& =\frac{-a}{1+a}\ast a& \hfill & \\ \hfill & =\frac{-a}{1+a}+a+\frac{-a^2}{1+a}=0& \hfill & \end{array}$$

    Hence $a^{\prime }$ is the inverse of $a$ and so $G$ is a group. □

20. In $R^{\ast }$ we define $a\ast b=(1/2)\mathit{ab}$. Then $(R^{\ast },\ast )$ is a group.

    Proof : Obviously $\ast$ is a binary operation in $R^{\ast }$.
    Let $a,b,c\in R^{\ast }$. Then

    $$\begin{array}{llll}\hfill (a\ast b)\ast c& =\left[\right(1/2\left)\mathit{ab}\right]\ast c& \hfill & \\ \hfill & =(1/4)\mathit{abc}& \hfill & \\ \hfill & =a\ast (b\ast c)& \hfill & \end{array}$$

    Hence $\ast$ is associative.
    Let $e\in R^{\ast }$ be such that $a\ast e=a$.
    Therefore $(1/2)\mathit{ae}=a$ and hence $e=2$.
    Let $a\in R^{\ast }$. Let $b\in R^{\ast }$ be such that $a\ast b=2$.
    Then $a\in R^{\ast }$ be such that $a\ast b=2$. Then $(1/2)\mathit{ab}=2$, (i.e) $b=4/a.$
    Thus $a\ast (4/a)=1/2(4/a)a=2$ i.e., $(4/a)$ is the inverse of $a$. Thus $(R^{\ast },\ast )$ is a group. □

21. Let $f_a:R\to R$ be the function defined by $f_a\left(x\right)=x+a$. Then $G=\{f_a:a\in R\}$ is a group under composition of functions.

    Proof : Let $f_a,f_b\in G$. Then

    $$\begin{array}{llll}\hfill (f_a◦f_b)\left(x\right)& =\left(f_a\right(f_b\left(x\right)\left)\right)& \hfill & \\ \hfill & =f_a(x+b)& \hfill & \\ \hfill & =x+b+a& \hfill & \\ \hfill & =f_{b+a}\left(x\right)& \hfill & \end{array}$$

    Hence

    $$f_a◦f_b=f_{b+a}\in G.$$

    We know that composition of mappings is associative.
    Also

    $$f_a◦f_0=f_a=f_0◦f_a$$

    and so $f_0$ is the identity. Also

    $$f_a◦f_{-a}=f_0=f_{-a}◦f_a.$$

    Hence $f_{-a}$ is the inverse of $f_a$. Hence $G$ is a group. □

Definition 1.1.3. Let $Z_n=\{0,1,2,\dots <!--\; FUNCTION\; APPLICATION\; -->,n-1\}$. Let $a,b\in Z_n$. Then $a+b=\mathit{qn}+r$ where $0\le r<n$. We define $a\oplus b=r$. Let $\mathit{ab}=q^{\prime }n+s$ where $0\le s<n$. We define $a\odot b=s$. The binary operations $\oplus$ and $\odot$ are called addition modulo $n$ and multiplication modulo $n$ respectively.

Show that $(Z_n,\oplus )$ is a group.

Proof : Clearly $\oplus \mathit{is}$ a binary operation in $Z_n$. Let $a,b\in Z_n$. Then

$$\begin{array}{lll}\hfill a+b& =q_{1}n+r_1\text{ where }0\le r_1<n& \hfill \text{(1.3)}\\ \hfill b+c& =q_{2}n+r_2\text{ where }0\le r_2<n& \hfill \text{(1.4)}\\ \hfill r_1+c& =q_{3}n+r_3\text{ where }0\le r_3<n& \hfill \text{(1.5)}\end{array}$$

and so

$$\begin{array}{llll}\hfill a+b+c& =(q_1+q_3)n+r_3\text{ and }& \hfill & \\ \hfill a+q_{2}n+r_2& =(q_1+q_3)n+r_3& \hfill & \end{array}$$

Hence $a+r_2=q_{4}n+r_3$ where $q_4=q_1+q_3-q_2$.
Now

$$(a\oplus b)\oplus c=r_1\oplus c=r_3.$$

Also

$$a\oplus (b\oplus c)=a\oplus r_2=r_3.$$

Hence $\oplus$ is associative.
Clearly the identity element is $0$ and the inverse of $a\in Z_n$ is $n-a$.
Hence $(Z_n,\oplus )$ is a group. □

Let $n$ be a prime. Then $Z_n-\left\{0\right\}$ is a group under multiplication modulo $n.$

Proof : Let $a,b\in Z_n-\left\{0\right\}$. Then $a\ne 0$ and $b\ne 0$.
Now, by definition $a\odot b\in Z_n.$
We claim that $a\odot b\ne 0$.
Suppose $a\odot b=0$.
Then $n|\mathit{ab}$. Since $n$ is prime, $n|a$ or $n|b$ and so $a=0$ or $b=0$ which is a contradiction.
Hence $a\odot b\in Z_n-\left\{0\right\}$.
Now, let $a,b,c\in Z_n-\left\{0\right\}$. Clearly

$$\begin{array}{lll}\hfill \mathit{ab}& =q_{1}n+r_1\text{ where }0\le r_1<n& \hfill \text{(1.6)}\\ \hfill \mathit{bc}& =q_{2}n+r_2\text{ where }0\le r_2<n& \hfill \text{(1.7)}\\ \hfill r_{1}c& =q_{3}n+r_3\text{ where }0\le r_3<n& \hfill \text{(1.8)}\end{array}$$

Thus

$$\begin{array}{lllllll}\hfill \mathit{abc}& =q_1\mathit{nc}+r_{1}c\text{ and so }a(q_{2}n+r_2)& \hfill =q_1\mathit{cn}+q_{3}n+r_3& & \hfill & & \hfill \end{array}$$

Hence

$$\begin{array}{llll}\hfill a{r}_2& =q_{4}n+r_3,& \hfill & \\ \hfill \text{ where }{}_4& =q_{1}c+q_3-a{q}_2& \hfill & \end{array}$$

Now,

$$(a·b)·c=r_1·c=r_3.$$

Also,

$$a·(b·c)=a·r_2=r_3.$$

Thus

$$a\odot b)\odot c=a\odot (b\odot c)$$

and hence $\odot$ is associative.
Clearly $1\in Z_n-\left\{0\right\}$ is the identity element.
Let $a\in Z_n-\left\{0\right\}$. Since $n$ is prime $(a,n)=1$. Hence the linear congruence $\mathit{ax}\equiv 1$ ( mod $n$) has a unique solution, say, $b\in Z_n-\left\{0\right\}$. Clearly $a·b=b·a=1$. Thus $b$ is the inverse of $a$. Hence $Z_n-\left\{0\right\}$ is a group. □

The set of all positive integers less than $n$ and prime to it is a group under multiplication modulo $n.$

Proof : Let $G=\{m:m<n$ and $(m,n)=1\}$.
Let $p,q\in G$. Obviously $\mathit{pq}\ne n$ and $(\mathit{pq},n)=1$.
Now let $\mathit{pq}=\mathit{sn}+r,0<r<n$. Then $p\odot q=r$.
We claim that $(r,n)=1$.
Suppose $(r,n)=a>1$.
Then $a|r$ and $a|n$. Hence $a\left|\right(r+\mathit{sn})$ i.e., $a|\mathit{pq}$. Also $a|n$.
Hence $(\mathit{pq},n)\ne 1$ which is a contradiction.
Thus $r\in G$ and so $G$ is closed under $\odot$.
We know that multiplication modulo $n$ is associative.
Clearly $1\in G$ is the identity element.
Let $a\in G$. Then $(a,n)=1$. Hence the linear congruence $\mathit{ax}\equiv 1$ ( mod $n$) has a unique solution for $x$ say $b$.
Clearly $\mathit{ab}\equiv 1$ ( mod $n$) and so $a\odot b=1$.
Now we have to prove that $b\in G$.
Suppose $(b,n)=c$. Since $\mathit{ab}\equiv 1$ ( mod $n$), $\mathit{ab}=\mathit{qn}+1$. Now $c|b$ and $c|n$

$$\begin{array}{llll}\hfill \Rightarrow & c\left|\right(\mathit{ab}-\mathit{qn})& \hfill & \\ \hfill \Rightarrow & c|1& \hfill & \\ \hfill \Rightarrow & c=1& \hfill & \end{array}$$

Thus $(b,n)=1$ and $b\in G$ and is the inverse of $a$. Thus $G$ is a group. □

Let $G$ denotes the set of all matrices of the form $\left(\begin{array}{cc}x\hfill & x\hfill \\ x\hfill & x\hfill \end{array}\right)$ where $x\in R^{\ast }$ Then $G$ is a group under multiplication.

Proof : Let $A,B\in G$. Let $A=\left(\begin{array}{cc}x\hfill & x\hfill \\ x\hfill & x\hfill \end{array}\right)$ and $B=\left(\begin{array}{cc}y\hfill & y\hfill \\ y\hfill & y\hfill \end{array}\right).$
Then $\mathit{AB}=\left(\begin{array}{cc}2\mathit{xy}\hfill & 2\mathit{xy}\hfill \\ 2\mathit{xy}\hfill & 2\mathit{xy}\hfill \end{array}\right)\in G$.
We know that matrix multiplication is associative.
Let $E=\left(\begin{array}{cc}e\hfill & e\hfill \\ e\hfill & e\hfill \end{array}\right)$ be such that $\mathit{AE}=A$. Then $\left(\begin{array}{cc}x\hfill & x\hfill \\ x\hfill & x\hfill \end{array}\right)\left(\begin{array}{cc}e\hfill & e\hfill \\ e\hfill & e\hfill \end{array}\right)=\left(\begin{array}{cc}x\hfill & x\hfill \\ x\hfill & x\hfill \end{array}\right)$ and so $\left(\begin{array}{cc}2\mathit{xe}\hfill & 2\mathit{xe}\hfill \\ 2\mathit{xe}\hfill & 2\mathit{xe}\hfill \end{array}\right)=\left(\begin{array}{cc}x\hfill & x\hfill \\ x\hfill & x\hfill \end{array}\right)$. Therefore $2\mathit{xe}=x$ and $e=1/2$.
Hence $E=\left(\begin{array}{cc}1/2\hfill & 1/2\hfill \\ 1/2\hfill & 1/2\hfill \end{array}\right)$ is the identity element of $G$.
Let $\left(\begin{array}{cc}y\hfill & y\hfill \\ y\hfill & y\hfill \end{array}\right)$ be the inverse of $\left(\begin{array}{cc}x\hfill & x\hfill \\ x\hfill & x\hfill \end{array}\right)$. Then $\left(\begin{array}{cc}x\hfill & x\hfill \\ x\hfill & x\hfill \end{array}\right)\left(\begin{array}{cc}y\hfill & y\hfill \\ y\hfill & y\hfill \end{array}\right)=\left(\begin{array}{cc}1/2\hfill & 1/2\hfill \\ 1/2\hfill & 1/2\hfill \end{array}\right)$ and so $\left(\begin{array}{cc}2\mathit{xy}\hfill & 2\mathit{xy}\hfill \\ 2\mathit{xy}\hfill & 2\mathit{xy}\hfill \end{array}\right)=\left(\begin{array}{cc}1/2\hfill & 1/2\hfill \\ 1/2\hfill & 1/2\hfill \end{array}\right)$. Thus $2\mathit{xy}=1/2$ and $y=x/4$.
Inverse of $\left(\begin{array}{cc}x\hfill & x\hfill \\ x\hfill & x\hfill \end{array}\right)$ is $\left(\begin{array}{cc}x/4\hfill & x/4\hfill \\ x/4\hfill & x/4\hfill \end{array}\right)$.
Hence $G$ is a group. □

In $N$ we define $a\ast b=a$. Then $(N,\ast )$ is not a group.

Proof : Clearly $\ast$is an associative binary operation on $N$. However, there is no element $e\in N$ such that $e\ast a=a$ for all $a\in N$. Hence there is no identity element in $(N,\ast )$. Hence $(N,\ast )$ is not a group. □

Definition 1.1.4. A group $G$ is said to be abelian if $\mathit{ab}=\mathit{ba}$ for all $a,b\in G$. A group which is not abelian is called a non-abelian group.

Example 1.1.5.

1. . $Z\text{'}\left\{0\right\},Q\text{'}\left\{0\right\},R\text{'}\left\{0\right\}$ and $C\text{'}\left\{0\right\}$ under usual multiplication are abelian groups.
2. $\left(\rho \right(S),△)$ is an abelian group, since $A△B=B△A$ for all $A,B\in \rho \left(S\right)$.
3. $(Z_n,\oplus )$ is an abelian group.

1.2 Properties of a Group

Theorem 1.2.1. Let $G$ be a group. Then

1. There exists a unique identity element $e\in G$ such that $e\ast a=a=a\ast e$ for all $a\in G.$
2. For all $a\in G$, there exists a unique inverse $a^{\prime }\in G$ such that $a\ast a^{\prime }=e=a^{\prime }\ast a.$

Proof :

1. Now $G$ is group. Therefore, by Definition of group, there exists $e\in G$ such that

   $$e\ast a=a=a\ast e\text{ for all }a\in G.$$

   Suppose, let $e$ and $e^{\prime }$ be two identity elements of $G$. Then $e{e}^{\prime }=e^{\prime }$ (since $e$ is an identity element). Also $e{e}^{\prime }=e$( since e is an identity element). Hence $e=e^{\prime }.$

2. Let $a\in G$. By Definition of group, there exists $a^{\prime }\in G$ such that $a\ast a^{\prime }=e=a^{\prime }\ast a$. Suppose there exists $a^{″<!--\; FUNCTION\; APPLICATION\; -->}\in G$ such that $a\ast a^{″<!--\; FUNCTION\; APPLICATION\; -->}=e=a^{″<!--\; FUNCTION\; APPLICATION\; -->}\ast a$. We show that $a^{\prime }=a^{″<!--\; FUNCTION\; APPLICATION\; -->}$ Now

   $$\begin{array}{llll}\hfill a^{\prime }& =a^{\prime }\ast e=a^{\prime }\ast (a\ast a^{″<!--\; FUNCTION\; APPLICATION\; -->})\text{ (substituting }e=a\ast a^{″<!--\; FUNCTION\; APPLICATION\; -->}& \hfill & \\ \hfill & =(a^{\prime }\ast a)\ast a^{″<!--\; FUNCTION\; APPLICATION\; -->}=e\ast a^{″<!--\; FUNCTION\; APPLICATION\; -->}\text{ (because }{a}^{\prime }\ast a=e\text{) }=a^{″<!--\; FUNCTION\; APPLICATION\; -->}& \hfill & \end{array}$$

   Thus, $a^{\prime }$ is unique.

□

We denote the inverse of $a$ by $a^{-1}$

Theorem 1.2.2. In a group, the left and right cancellation laws hold $(i.e,)\mathit{ab}=\mathit{ac}\Rightarrow b=c$ and $\mathit{ba}=\mathit{ca}\Rightarrow b=c.$

Proof : Suppose

$$\begin{array}{llll}\hfill \mathit{ab}& =\mathit{ac}& \hfill & \\ \hfill \Rightarrow a^{-1}\left(\mathit{ab}\right)& =a^{-1}\left(\mathit{ac}\right)& \hfill & \\ \hfill \Rightarrow \left(a^{-1}a\right)b& =\left(a^{-1}a\right)c& \hfill & \\ \hfill \Rightarrow \mathit{eb}& =\mathit{ec}& \hfill & \\ \hfill \Rightarrow b& =c& \hfill & \end{array}$$

Similarly, we can prove that $\mathit{ba}=\mathit{ca}\Rightarrow b=c.$ □

Theorem 1.2.3. Let $G$ be a group and $a,b\in G$. Then the equation $\mathit{ax}=b$ and $\mathit{ya}=b$ have unique solutions for $x$ and $y$ in $G.$

Proof : Consider $a^{-1}b\in G$. Then $a\left(a^{-1}b\right)=\left(a{a}^{-1}\right)b=\mathit{eb}=b$.
Hence $a^{-1}b$ is a solution of $\mathit{ax}=b$.
Now, to prove the uniqueness, let $x_1$ and $x_2$ be two solutions of $\mathit{ax}=b$.
Then $a{x}_1=b$ and $a{x}_2=b$.
Therefore $a{x}_1=a{x}_2$ which implies $x_1=x_2$.
Hence $x=a^{-1}b$ is the unique solution for $\mathit{ax}=b$.
Similarly we can prove that $y=b{a}^{-1}$ is the solution of the equation $\mathit{ya}=b.$ □

Theorem 1.2.4. Let $G$ be a group. Let $a,b\in G$. Then ${\left(\mathit{ab}\right)}^{-1}=b^{-1}{a}^{-1}$ and ${\left(a^{-1}\right)}^{-1}=a.$

Proof : Now

$$\begin{array}{llll}\hfill \left(\mathit{ab}\right)\left(b^{-1}{a}^{-1}\right)& =a\left(b{b}^{-1}\right)a^{-1}& \hfill & \\ \hfill & =\mathit{ae}{a}^{-1}& \hfill & \\ \hfill & =a{a}^{-1}& \hfill & \\ \hfill & =e& \hfill & \end{array}$$

Similarly $\left(b^{-1}{a}^{-1}\right)\left(\mathit{ab}\right)=e$.
Hence ${\left(\mathit{ab}\right)}^{-1}=b^{-1}{a}^{-1}$.
Proof of the second part is obvious. □

Corollary 1.2.5. If $a_1,a_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,a_n\in G$ then

$${(a_1{a}_2\cdots a_n)}^{-1}=a_n^{-1}{a}_{n-1}^{-1}\cdots a_1^{-1}$$

Definition 1.2.6. Let $G$ be a group and $a\in G$. For any positive integer $n$, we define $a^n=\mathit{aa}\cdots a$ ( $a$ written $n$ times). Clearly

$$\begin{array}{llll}\hfill {\left(a^n\right)}^{-1}& ={(\mathit{aa}\cdots a)}^{-1}& \hfill & \\ \hfill & =(a^{-1}{a}^{-1}\cdots a^{-1})& \hfill & \\ \hfill & ={\left(a^n\right)}^{-1}& \hfill & \end{array}$$

Now we define

$$a^{-n}={\left(a^{-1}\right)}^n={\left(a^n\right)}^{-1}$$

Finally we define $a^0=e$. Thus $a^n$ is defined for all integers $n.$

Note 1.2.7. When the binary operation on $G$ is $“+$”, we denote $a+a+\cdots +a$ ($a$ written $n$ times) as $\mathit{na}.$

Theorem 1.2.8. Let $G$ be a group and $a\in G$. Then

1. $a^m{a}^n=a^{m+n},m,n\in Z$.
2. ${\left(a^m\right)}^n=a^{\mathit{mn}},m,n\in Z$.

Proof :

1. When $n=0$ the result follows directly from the definition.
   Now let $n>0.$ We prove the result by induction $n$.
   When $m\ge 0,a^{m+1}=a^m{a}^1$ (by definition).
   When $m=-1,a^{m+1}=a^0=e$ and $a^m{a}^1=a^{-1}a=e$.
   Hence $a^{m+1}=a^m{a}^1$.
   When $m\le -2$, let $m=-p$, where $p\ge 2.$
   

   $$\begin{array}{llll}\hfill \left(a^m\right)a& =\left(a^{-p}\right)a={\left(a^{-1}\right)}^{p}a& \hfill & \\ \hfill & ={\left(a^{-1}\right)}^{p-1}{a}^{-1}a& \hfill & \\ \hfill & ={\left(a^{-1}\right)}^{p-1}& \hfill & \\ \hfill & =a^{-p+1}& \hfill & \\ \hfill & =a^{m+1}& \hfill & \end{array}$$

   Hence $a^{m+1}=a^m{a}^1$, for all $m\in Z$ and so the result is true for $n=1$. Suppose now that the theorem is valid for $n=k>1$. Then

   $$\begin{array}{llll}\hfill a^m{a}^k& =a^{m+k}& \hfill & \\ \hfill a^m{a}^{k+1}& =a^m\left(a^{k}a\right)& \hfill & \\ \hfill & =\left(a^m{a}^k\right)a& \hfill & \\ \hfill & =a^{m+k}a\text{ (hypothesis) }& \hfill & \\ \hfill & =a^{m+k+1}\text{ (by definition)}& \hfill & \end{array}$$ Thus is follows that the theorem is valid for $n=k+1$. Hence by induction the theorem holds for all positive integers $n$. Finally if $n<0$, we can prove the result by induction on $-n.$
2. Obvious.

□

Solved Problems

Problem 1.2.9. Show that, in a group $G,x^2=x$ if and only if $x=e.$

Solution. Clearly $e^2=\mathit{ee}=e.$
Conversely, let $x^2=x$.
Then $\mathit{xx}=\mathit{xe}$. Hence by cancellation law $x=e.$

Note 1.2.10. An element $a\in G$ is called idempotent if $a^2=a$. Thus we have shown that in a group G. the identity element is the only idempotent element.

Problem 1.2.11. In an abelian group, ${\left(\mathit{ab}\right)}^2=a^2{b}^2$.

Solution. Clearly

$$\begin{array}{llll}\hfill {\left(\mathit{ab}\right)}^2& =\left(\mathit{ab}\right)\left(\mathit{ab}\right)& \hfill & \\ \hfill & =a\left(\mathit{ba}\right)b& \hfill & \\ \hfill & =a\left(\mathit{ab}\right)b& \hfill & \\ \hfill & =\left(\mathit{aa}\right)\left(\mathit{bb}\right)& \hfill & \\ \hfill & =a^2{b}^2& \hfill & \end{array}$$

Note 1.2.12. In general for any positive integer $n,{\left(\mathit{ab}\right)}^n=a^n{b}^n$ (prove by using induction)

Problem 1.2.13. Let $G$ be a group such that $a^2=e$ for all $a\in G$. Then $G$ is abelian.

Solution. Since $a^2=e,\mathit{aa}=e\Rightarrow a=a^{-1}$
Now, $\mathit{ab}=($ ab ${)}^{-1}=b^{-1}{a}^{-1}=\mathit{ba}$.
Hence $G$ is abelian.

Problem 1.2.14. Let $G$ be a group in which ${\left(\mathit{ab}\right)}^m=a^m{b}^m$ for three consecutive integers and for all $a,b\in G$. Then $G$ is abelian.

Solution. Let $a,b\in G$. Then by hypothesis,

$$\begin{array}{llll}\hfill {\left(\mathit{ab}\right)}^m& =a^m{b}^m& \hfill & \\ \hfill {\left(\mathit{ab}\right)}^{m+1}& =a^{m+1}{b}^{m+1}\text{ and }& \hfill & \\ \hfill {\left(\mathit{ab}\right)}^{m+2}& =a^{m+2}{b}^{m+2}& \hfill & \end{array}$$

Now,

$$\begin{array}{llll}\hfill {\left(\mathit{ab}\right)}^{m+1}& =a^{m+1}{b}^{m+1}& \hfill & \\ \hfill \Rightarrow {\left(\mathit{ab}\right)}^m\left(\mathit{ab}\right)& =\left(a^{m}a\right)\left(b^{m}b\right)& \hfill & \\ \hfill \Rightarrow \left(a^m{b}^m\right)\left(\mathit{ab}\right)& =\left(a^{m}a\right)\left(b^{m}b\right)& \hfill & \\ \hfill b^{m}a& =a{b}^m\text{ (by cancellation law).}& \hfill & \end{array}$$

Similarly

$$\begin{array}{llll}\hfill {\left(\mathit{ab}\right)}^{m+2}& =a^{m+2}{b}^{m+2}& \hfill & \\ \hfill \Rightarrow b^{m+1}a& =a{b}^{m+1}& \hfill & \\ \hfill \Rightarrow b^m\mathit{ba}& =a{b}^{m}b& \hfill & \\ \hfill \Rightarrow b^m\mathit{ba}& =b^m\mathit{ab}\text{ and so }& \hfill & \\ \hfill \mathit{ba}& =\mathit{ab}.& \hfill & \end{array}$$

Hence $G$ is abelian.

Problem 1.2.15. Let $(H,·)$ and $(K,\ast )$ be groups. We define a binary operation $\square$ on $H\times K$ by

$$(h_1,k_1)\square (h_2,k_2)=(h_1{h}_2,k_1\ast k_2).$$

Then $H\times K$ is a group.
$(H\times K$ is called the direct product of $H$ and $K$).

Solution. First we shall prove that $\square$ is associative.
Let $(h_1,k_1),(h_2,k_2),(h_3,k_3)\in H\times K.$

$$\begin{array}{llll}\hfill \left[\right(h_1,k_1)\square (h_2,k_2\left)\right]\square (h_3,k_3)& =(h_1{h}_2,k_1\ast k_2)\square (h_3,k_3)& \hfill & \\ \hfill & =\left(\right(h_1{h}_2)h_3,(k_1\ast k_2)\ast k_3)& \hfill & \\ \hfill & =\left(\right(h_1\left(h_2{h}_3\right),k_1\ast (k_2\ast k_3))& \hfill & \\ \hfill & =(h_1,k_1)\square (h_2{h}_3,k_2\ast k_3)& \hfill & \\ \hfill & =(h_1,k_1)\square \left[\right(h_2,k_2)\square (h_3,k_3\left)\right].& \hfill & \end{array}$$

Let $e,e_1$ be the identities of the groups $H$ and $K$ respectively. Clearly $(e,e_1)$ is the identity element in $H\times K$.
Also $(h^{-1},k^{-1})$ is the inverse of $(h,k)$.
Hence $H\times K$ is a group.

1.3 Permutation Groups

Definition 1.3.1. Let $A$ be a finite set. A bijection from $A$ to itself is called a permutation of $A.$

Example 1.3.2. If $A=\{1,2,3,4\}f:A\to A$ given by $f\left(1\right)=2,f\left(2\right)=1,f\left(3\right)=4$ and $f\left(4\right)=3$ is a permutation of $A$. We shall write this permutation as

$$\left(\begin{array}{cccc}1\hfill & 2\hfill & 3\hfill & 4\hfill \\ 2\hfill & 1\hfill & 4\hfill & 3\hfill \end{array}\right).$$

An element in the bottom row is the image of the element just above it in the upper row.

Definition 1.3.3. Let $A$ be a finite set containing $n$ elements. The set of all permutations of $A$ is clearly a group under the composition of functions. This group is called the symmetric group of degree $n$ and is denoted by $S_n.$

Example 1.3.4. Let $A=\{1,2,3\}$. Then $S_3$ consists of

$$\begin{array}{llll}\hfill e& =\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 1\hfill & 2\hfill & 3\hfill \end{array}\right);p_1=\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 2\hfill & 3\hfill & 1\hfill \end{array}\right);p_2=\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 3\hfill & 1\hfill & 2\hfill \end{array}\right);& \hfill & \\ \hfill p_3& =\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 1\hfill & 3\hfill & 2\hfill \end{array}\right);p_4=\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 3\hfill & 2\hfill & 1\hfill \end{array}\right);p_5=\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 2\hfill & 1\hfill & 3\hfill \end{array}\right)& \hfill & \end{array}$$

In this group, $e$ is the identity element.
We now compute the product $p_1{p}_2.$

	1	2	3
$p_1$	$\downarrow$	$\downarrow$	$\downarrow$		1	2	3
	2	3	1	Hence $p_1{p}_2$	$\downarrow$	$\downarrow$	$\downarrow$
$p_2$	$\downarrow$	$\downarrow$	$\downarrow$		1	2	3
	1	2	3

So that $p_1{p}_2=e$. Now,

$$p_1{p}_4=\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 2\hfill & 3\hfill & 1\hfill \end{array}\right)\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 3\hfill & 2\hfill & 1\hfill \end{array}\right)=\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 2\hfill & 1\hfill & 3\hfill \end{array}\right)=p_5.$$

Similarly we can compute all other products and Cayley table for this group is given by

Thus $S_3$ is a group containing $3!=6$ elements.

Remark 1.3.5.

1. In $S_3$, $p_1{p}_2=p_2{p}_1=e$ so that the inverse of $p_1$ is $p_2$. In general the inverse of a permutation can be obtained by interchanging the rows of the permutation.

   Example 1.3.6. If $p=\left(\begin{array}{ccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill \\ 3\hfill & 4\hfill & 2\hfill & 5\hfill & 1\hfill \end{array}\right)$ then the inverse of $p$ is the permutation given by

   $$p^{-1}=\left(\begin{array}{ccccc}3\hfill & 4\hfill & 2\hfill & 5\hfill & 1\hfill \\ 1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill \end{array}\right)=\left(\begin{array}{ccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill \\ 5\hfill & 3\hfill & 1\hfill & 2\hfill & 4\hfill \end{array}\right).$$

2. In $S_3,p_1{p}_4=p_5$ and $p_4{p}_1=p_3$. Hence $p_1{p}_4\ne p_4{p}_1$ so that $S_3$ is non-abelian.
3. The symmetric group $S_n$ containing $n!$ elements, for, let $A=\{1,2,\dots <!--\; FUNCTION\; APPLICATION\; -->,n\}$. Any permutation on $A$ is given by specifying the image of each element. The image of 1 can be chosen in $n$ different ways. Since the image of two is different from the image of 1, it can be chosen in $(n-1)$ different ways and so on. Hence the number of permutations of $A$ is $n(n-1)\cdots 2·1=n!$ so that the number of elements in $S_n$ is $n!.$

Definition 1.3.7. Let $G$ be a finite group. Then the number of elements in $G$ is called the order of $G$ and is denoted by $\left|G\right|$ or $o\left(G\right)$.

Definition 1.3.8. Let $p$ be a permutation on $A=\{1,2,\dots <!--\; FUNCTION\; APPLICATION\; -->,n\}.p$ is called a cycle of length $r$ if there exist distinct symbols $a_1,a_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,a_r$ such that

$$p\left(a_1\right)=a_2,p\left(a_2\right)=a_3,\dots <!--\; FUNCTION\; APPLICATION\; -->,p\left(a_{r-1}\right)=a_r,\text{ and }p\left(a_r\right)=a_1,$$

and $p\left(b\right)=b$ for all

$$b\in A-\{a_1,a_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,a_r\}.$$

This cycle is represented by the symbol $(a_1,a_2,\cdots ,a_r)$.

Thus under the cycle $(a_1,a_2,\cdots ,a_r)$ each symbol is mapped onto the following symbol except the last one which is mapped onto the first symbol and all the other symbols not in the cycle are fixed.

Example 1.3.9. Let $A=\{1,2,3,4,5\}$. Consider the cycle of length 4 given by $p=\left(2451\right)$. hen

$$p=\left(\begin{array}{ccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill \\ 2\hfill & 4\hfill & 3\hfill & 5\hfill & 1\hfill \end{array}\right)$$

and so $\left(2451\right)=\left(4521\right)=\left(5124\right)=\left(1245\right)$.

Note 1.3.10. Since cycles are special types of permutations, they can be multiplied in the usual way. The product of cycles need not be a cycle.

Example 1.3.11. Let $p_1=\left(234\right)$ and $p_2=(1,5)$. Then

$$p_1{p}_2=\left(\begin{array}{ccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill \\ 1\hfill & 3\hfill & 4\hfill & 2\hfill & 5\hfill \end{array}\right)\left(\begin{array}{ccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill \\ 5\hfill & 2\hfill & 3\hfill & 4\hfill & 1\hfill \end{array}\right)=\left(\begin{array}{ccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill \\ 5\hfill & 3\hfill & 4\hfill & 2\hfill & 1\hfill \end{array}\right)$$

which is not a cycle.

Definition 1.3.12. Two cycles are said to be disjoint if they have any no symbols in common.

Example 1.3.13. (215) and (34) are disjoint cycles.

Remark 1.3.14. If $p_1$ and $p_2$ are disjoint cycles the symbols which are moved by $p_1$ are fixed by $p_2$ and vice versa. Hence multiplication of disjoint cycles is commutative.

Example 1.3.15.

1. Consider the permutation $\left(\begin{array}{ccccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill & 6\hfill & 7\hfill \\ 2\hfill & 1\hfill & 3\hfill & 5\hfill & 6\hfill & 7\hfill & 4\hfill \end{array}\right).$
   We shall write this permutation as a product of disjoint cycles. First of all 1 is moved to 2 and then 2 is moved to 1 thus giving the cycle $\left(12\right)$. The element 3 is left fixed. Again starting with 4, 4 is moved to 5, 5 is moved to 6, 6 is moved to 7 and 7 is moved to 4, thus giving the cycle $\left(4567\right)$. Thus

   $$\begin{array}{llll}\hfill \left(\begin{array}{ccccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill & 6\hfill & 7\hfill \\ 2\hfill & 1\hfill & 3\hfill & 5\hfill & 6\hfill & 7\hfill & 4\hfill \end{array}\right)& =\left(12\right)\left(4567\right)& \hfill & \\ \hfill & =\left(4567\right)\left(12\right)& \hfill & \end{array}$$

2. Consider the permutation

   $$\alpha =\left(\begin{array}{ccccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill & 6\hfill & 7\hfill \\ 2\hfill & 3\hfill & 7\hfill & 5\hfill & 4\hfill & 1\hfill & 6\hfill \end{array}\right)\in S_7.$$

   Starting with 1 we get the cycle $\left(12376\right)$ . The elements 4,5 do not appear in it. Starting with 4 we get the cycle $\left(45\right)$. Each element of the set $\{$1, 2, $\dots <!--\; FUNCTION\; APPLICATION\; -->,7\}$ occurs in one of the two cycles. Thus $\alpha =\left(12376\right)\left(45\right)$.

3. Consider the permutation $\alpha =\left(\begin{array}{cccccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill \\ \hfill 4\hfill & \hfill 6\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 2\hfill & \hfill 5\hfill \end{array}\right)$. Clearly $\alpha =\left(143\right)\left(265\right)$ .

Theorem 1.3.16. Any permutation can be expressed as a product of disjoint cycles.

Proof : Let $p$ be a given permutation of the set $S=\{1,2,\dots <!--\; FUNCTION\; APPLICATION\; -->,n\}$.
Let us start with any symbol $a_1\in S$.
Let $p\left(a_1\right)=a_2,p\left(a_2\right)=a_3,\dots <!--\; FUNCTION\; APPLICATION\; -->$.
Since $S$ is finite, these symbols cannot all the distinct and hence there exists a least positive integer $r$ such that $1\le r\le n$ and $p\left(a_r\right)=a_1.$
Let $c=(a_1,a_2,\cdots ,a_r)$.
If $r=n$ then $p=c$ so that $p$ is cycle.
If $r<n$, let $b_1$ be a symbol in $S$ such that $b_1\notin (a_1,a_2,\cdots ,a_r)$.
Starting with $b_1$ we can construct the cycles $d=(b_1,b_2,\cdots ,b_s)$ as before.
Clearly the cycles $c$ and $d$ are disjoint.
If $r+s=n$ then $p=\mathit{cd}$.
If $r+s<n$ then we repeat this process to obtain more cycles until all symbols appear in one of the cycles.
Thus we get a decomposition of $p$ into disjoint cycles. □

Note 1.3.17. The decomposition of a permutation into disjoint cycles is unique except for the order of the factors.

Definition 1.3.18. A cycle of length two is called a transposition Thus a transposition $\left(a_1{a}_2\right)$ interchanges the symbols $a_1$ and $a_2$ and leaves all the other elements fixed.

Theorem 1.3.19. Any permutation can be expressed as a product of transpositions.

Proof : Since any permutation is a product of disjoint cycles it is enough to prove that each cycle is a product of transpositions. Let $c=(a_1{a}_2\cdots a_1)$ be a cycle. Then

$$(a_1{a}_2\cdots a_1)=\left(a_1{a}_2\right)\left(a_2{a}_3\right)\cdots \left(a_1{a}_r\right).$$

This proves the theorem. □

Example 1.3.20.

1. Let $\left(\begin{array}{ccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill \\ 2\hfill & 4\hfill & 3\hfill & 5\hfill & 1\hfill \end{array}\right)=\left(1245\right)=\left(12\right)\left(14\right)\left(15\right)$. Then $\left(1245\right)=\left(2451\right)=\left(24\right)\left(25\right)\left(21\right)$ and so the representation of a permutation as a product of transpositions is unique.
2. Clearly $\left(1345\right)\left(26\right)=\left(13\right)\left(14\right)\left(15\right)\left(26\right)=\left(13\right)\left(12\right)\left(12\right)\left(14\right)\left(15\right)\left(26\right)$ . Thus in the representation of a permutation as a product of transpositions one can always insert $\left(\mathit{ab}\right)\left(\mathit{ab}\right)$ in any place since $\left(\mathit{ab}\right)\left(\mathit{ab}\right)$ is the identity permutation.

Theorem 1.3.21. If a permutation $p\in S_n$ is a product of $r$ transpositions and also a product of $s$ transpositions then either $r$ and $s$ are both even or both odd.

Proof : Let $p=t_1{t}_2\cdots t_r=t_1^{\prime }{t}_2^{\prime }\cdots t_s^{\prime }$ where $t_i,t_i^{\prime }$ are transpositions. Now consider the polynomial in $n$ variables $x_1,x_2,\cdots ,x_n$ given by

$$\begin{array}{llll}\hfill △& =(x_1-x_2)(x_1-x_3)\cdots (x_1-x_n)\times (x_2-x_3)(x_2-x_4)\cdots (x_2-x_n)\times \cdots \cdots \times (x_{n-1}-x_n)& \hfill & \\ \hfill & =\prod _{i<j}(x_i-x_j)& \hfill & \end{array}$$

For any permutation $p\in S_n$ we define

$$p(△)=\prod _{i<j}(x_{p\left(i\right)}-x_{p\left(j\right)}).$$

Consider the transposition $t=\left(\mathit{ij}\right)$ . Then the factor $x_i-x_j$ in $△$ becomes $x_j-x_i.$ Any factor of $△$ in which neither $i$ nor $j$ is equal to $k$ or $l$ is unchanged. All other factors of $△$ can be paired to form products of the form $\pm (x_i-x_k)(x_k-x_j)$ , the sign being determined by the relative magnitudes of $i,j$ and $k$. Since $t$ interchanges $x_i$ and $x_j$ any such product is unchanged. Hence the effect of the transposition $t$ on $△$ is just to change the sign of $△$ i.e, $t(△)=-△$. Therefore

$$p(△)=(t_1{t}_2\cdots t_r)(△)={(-1)}^r△.$$

Also

$$p(△)=(t_1^{\prime }{t}_2^{\prime }\cdots t_r^{\prime })(△)={(-1)}^s△$$

Therefore ${(-1)}^r={(-1)}^s\Rightarrow r$ and $s$ are both even or both odd. □

Definition 1.3.22. A permutation $p\in S_n$ is called even or odd according as $p$ can be expressed as a product of an even number of transpositions or an odd number of transpositions respectively.

Example 1.3.23.

1. Consider the permutation

   $$\begin{array}{llll}\hfill p& =\left(\begin{array}{ccccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill & 6\hfill & 7\hfill \\ 3\hfill & 6\hfill & 4\hfill & 1\hfill & 7\hfill & 2\hfill & 5\hfill \end{array}\right)\text{ and }& \hfill & \\ \hfill p& =\left(134\right)\left(26\right)\left(57\right)& \hfill & \\ \hfill & =\left(13\right)\left(14\right)\left(26\right)\left(57\right)& \hfill & \end{array}$$

   Therefore $p$ is a product of 4 transposition. Hence $p$ is an even permutation.

2. Consider the permutation

   $$\begin{array}{llll}\hfill p& =\left(\begin{array}{ccccccccc}1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill & 6\hfill & 7\hfill & 8\hfill & 9\hfill \\ 2\hfill & 5\hfill & 4\hfill & 3\hfill & 6\hfill & 1\hfill & 7\hfill & 9\hfill & 8\hfill \end{array}\right)& \hfill & \\ \hfill p& =\left(1256\right)\left(34\right)\left(89\right)& \hfill & \\ \hfill & =\left(12\right)\left(15\right)\left(16\right)\left(34\right)\left(89\right)& \hfill & \end{array}$$

   Therefore $p$ is a product of 5 transposition and so $p$ is an odd permutation.

Theorem 1.3.24.

1. The product of two even permutations is an even permutation.
2. The product of two odd permutations is an even permutation.
3. The product of an even permutation and an odd permutation is an odd permutation.
4. The inverse of an even permutation is an even permutation.
5. The inverse of an odd permutation is an odd permutation.
6. The identity permutation $e$ is an even permutation.

Proof : Let $p_1,p_2$ be two permutations. If $p_1$ is a product of $r$ transpositions and $p_2$ is a product of $s$ transposition, then $p_1{p}_2$ is a product of $r+s$ transpositions. Hence (1), (2) and (3) follows.

Now suppose that a permutation $p$ is a product of $r$ transpositions, say, $p=t_1,t_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,t_r$. Then

$$\begin{array}{llll}\hfill p^{-1}& ={(t_1,t_2,\cdots ,t_r)}^{-1}& \hfill & \\ \hfill & =t_r^{-1}\cdots t_2^{-1}{t}_1^{-1}& \hfill & \\ \hfill & =t_r\cdots t_2{t}_1& \hfill & \end{array}$$

and so $p^{-1}$ is also a product of $r$ transpositions. This proves (4) and (5).

Now, $e=\left(12\right)\left(12\right)$ and hence $e$ is an even permutation which proves (6). □

Theorem 1.3.25. Let $A_n$ be the set of all even permutations in $S_n$. Then $A_n$ is a group containing $\frac{n!}{2}$ permutations.

Proof : From (1),(6) and (4) of Theorem 1.3.24, we see that $A_n$ is a group.
Now let $B_n$ be the set of all permutations in $S_n$.
Define $f$ : $A_n\to B_n$ by $f\left(p\right)=\left(12\right)p.$ Suppose $f\left(p_1\right)=f\left(p_2\right)\Rightarrow \left(12\right)p_1=\left(12\right)p_2\Rightarrow p_1=p_2$.
Hence $f$ is 1-1. If $\alpha \in B_n$, then $\left(12\right)\alpha \in A_n$ and $f\left[\right(12\left)\alpha \right]=\left(12\right)\left(12\right)\alpha =\alpha$.
Hence $f$ is onto. Thus $f$ is a bijection and hence the number of odd permutations in $S_n$= the number of even permutations in $S_n.$
Since $S_n$ contains $n!$ permutations, $A_n$ has $\frac{n!}{2}$ elements. □

Definition 1.3.26. The group $A_n$ of all even permutations in $S_n$ is called the alternating group on $n$ symbols.

1.4 Subgroups

Definition 1.4.1. Let $G$ be a set with binary operation $\ast$defined on it. Let $S\subseteq G.$ If for each $a,b\in S,a\ast b$ is in $S$, we say that $S$ is closed with respect to the binary operation $\ast .$

Example 1.4.2.

1. $(Z,+)$ is a group. The set $E$ of all even integers is closed under $+$ and further $(E,+)$ is itself a group.
2. The set of $G$ of all non-singular $2\times 2$ matrices form a group under matrix multiplication. Let $H$ be the set of all matrices of the form $\left(\begin{array}{cc}\cos <!--\; FUNCTION\; APPLICATION\; -->\theta \hfill & -\sin <!--\; FUNCTION\; APPLICATION\; -->\theta \hfill \\ \sin <!--\; FUNCTION\; APPLICATION\; -->\theta \hfill & \cos <!--\; FUNCTION\; APPLICATION\; -->\theta \hfill \end{array}\right)$. Then $H$ is subset of $G$ and $H$ itself a group under matrix multiplication.

Definition 1.4.3. A subset $H$ of group $G$ is called subgroup of $G$ if $H$ forms a group with respect to the binary operation in $G.$

Example 1.4.4.

1. Let $G$ be any group. Then $\left\{e\right\}$ and $G$ are trivial subgroups of $G$. They are called improper subgroups of $G.$
2. $(Q,+)$ is a subgroup of $(R,+)$ and $(R,+)$ is a subgroup of $(C,+)$.

3. In $(Z_8,\oplus )$ , let $H_1=\{0,4\}$ and $H_2=\{0,2,4,6\}$. The Cayley tables for $H_1$ and $H_2$ are given by

   

   It is easily seen that $H_1$ and $H_2$ are closed under $\oplus \mathit{and}(H_1,\oplus )$ and $(H_2,\oplus )$ are groups. Hence $H_1$ and $H_2$ are subgroups of $Z_8.$

4. $\{1,-1\}$ is a subgroup of $(R^{\ast },·)$ .
5. $\{1,i,-1,-i\}$ is a subgroup of $(C^{\ast },·)$.
6. For any integer $n$ we define $\mathit{nZ}=\{\mathit{nx}:x\in Z\}$. Then $(\mathit{nZ},+)$ is a subgroup of $(Z,+)$.
   For, let $a,b\in \mathit{nZ}$. Then $a=\mathit{nx}$ and $b=\mathit{ny}$ where $x,y\in Z$.
   Hence $a+b=n(x+y)\in \mathit{nZ}$ and so $\mathit{nZ}$ is closed under $+$.
   Clearly $0\in \mathit{nZ}$ is the identity element.
   Inverse of $\mathit{nx}$ is $-\mathit{nx}=n(-x)\in \mathit{nZ}$. Hence $(\mathit{nZ},+)$ is a group.

7. In the symmetric group $S_3$,

   $$\begin{array}{llll}\hfill H_1& =\{e,p_1,p_2\};& \hfill & \\ \hfill H_2& =\{e,p_3\};& \hfill & \\ \hfill H_3& =\{e,p_4\};\text{ and }& \hfill & \\ \hfill H_4& =\{e,p_5\}& \hfill & \end{array}$$

   are subgroups.

8. $A_n$ is a subgroup of $S_n.$
9. The set of permutations $H=\{e,p_1,p_2,p_3\}$, where $e=\left(\begin{array}{cccc}1\hfill & 2\hfill & 3\hfill & 4\hfill \\ 1\hfill & 2\hfill & 3\hfill & 4\hfill \end{array}\right)$ ; $p_1=\left(\begin{array}{cccc}1\hfill & 2\hfill & 3\hfill & 4\hfill \\ 2\hfill & 1\hfill & 4\hfill & 3\hfill \end{array}\right)$ ; $p_2=\left(\begin{array}{cccc}1\hfill & 2\hfill & 3\hfill & 4\hfill \\ 3\hfill & 4\hfill & 1\hfill & 2\hfill \end{array}\right)$ ; $p_3=\left(\begin{array}{cccc}1\hfill & 2\hfill & 3\hfill & 4\hfill \\ 4\hfill & 3\hfill & 2\hfill & 1\hfill \end{array}\right)$, is a subgroup of $S_4.$

Note 1.4.5. In all the above examples we see that the identity element in the subgroup is the same as the identity element of the group.

Theorem 1.4.6. Let $H$ be a subgroup of $G$. Then

1. the identity element of $H$ is the same as that of $G.$
2. for each $a\in H$ the inverse of $a$ in $H$ is the same as the inverse of $a$ in $G.$

Proof :

1. Let $e$ and $e^{\prime }$ be the identity of $G$ and $H$ respectively. Let $a\in H$. Now,

   $$\begin{array}{llll}\hfill e^{\prime }a& =a\text{( sincee’ is the identity of }H\text{)}& \hfill & \\ \hfill & =\mathit{ea}\text{( sincee’ is the identity of }G\text{ and }a\in G\text{)}& \hfill & \\ \hfill e^{\prime }a& =\mathit{ea}& \hfill & \\ \hfill \Rightarrow e^{\prime }& =a\text{( by cancellation law)}& \hfill & \end{array}$$
2. Let $a^{\prime }$ and $a^{″<!--\; FUNCTION\; APPLICATION\; -->}$ be the inverse of $a$ in $G$ and $H$ respectively. Since by (1), $G$ and $H$ have the same identity element $e$, we have $a^{\prime }a=e=a^{″<!--\; FUNCTION\; APPLICATION\; -->}a$. Hence by cancellation law, $a^{\prime }=a^{″<!--\; FUNCTION\; APPLICATION\; -->}$.

□

Theorem 1.4.7. A subset $H$ of a group $G$ is a subgroup of $G$ if and only if

1. it is closed under the binary operation in $G.$
2. The identity $e$ of $G$ is in $H$.
3. $a\in H\Rightarrow a^{-1}\in H.$

Proof : Let $H$ be subgroup of $G$. The result follows immediately from Theorem 1.4.6.

Conversely, let $H$ be a subset of $G$ satisfying conditions (1), (2) and (3). Then, obviously $H$ itself a group with respect to the binary operation in $G$. Therefore $H$ is a subgroup of $G.$ □

Theorem 1.4.8. A non-empty subset $H$ of a group $G$ is a subgroup of $G$ if and only if $a,b\in H\Rightarrow a{b}^{-1}\in H.$

Proof : Let $H$ be a subgroup of $G$. Then

$$a,b\in H\Rightarrow a,b^{-1}\in H\Rightarrow a{b}^{-1}\in H.$$

Conversely, suppose $H$ is a non-empty subset of $G$ such that $a,b\in H\Rightarrow a{b}^{-1}\in H.$
Since $H\ne \varnothing$, there exists $a\in H$. Hence $a,a^{-1}\in H$.
Therefore, $e=a{a}^{-1}\in H$, i.e., $H$ contains the identity element $e$.
Also, since $a,b\in H.e{a}^{-1}\in H$. Hence $a^{-1}\in H$.
Now, let $a,b\in H$. Then $a,b^{-1}\in H$. Hence $a{\left(b^{-1}\right)}^{-1}=\mathit{ab}\in H$ and so $H$ is closed under the binary operation in $G$. Hence by Theorem 1.4.7, $H$ is a subgroup of $G.$ □

Note 1.4.9. If the operation is $+$ then $H$ is a subgroup of $G$ if and only if $a,b\in H\Rightarrow a-b\in H.$

Theorem 1.4.10. Let $H$ be a non-empty finite subset subset of $G$. If $H$ is closed under the operation in $G$ then $H$ is a subgroup of $G.$

Proof : Let $a\in H$. Then $a,a^2,\dots <!--\; FUNCTION\; APPLICATION\; -->,a^n,\dots <!--\; FUNCTION\; APPLICATION\; -->$ are all elements of $H$. But since $H$ is finite the elements $a,a^2,a^3\dots <!--\; FUNCTION\; APPLICATION\; -->$, cannot all be distinct. Hence let $a^r=a^s,r<s$. Then $a^{s-r}=e\in H$. Now, let $a\in H$. We have proved that $a^n=e$ for some $n$. Hence $a{a}^{n-1}=e$. Hence $a^{-1}=a^{n-1}\in H$. Thus $H$ is a subgroup of $G.$ □

Note 1.4.11. Theorem 1.4.10 is not true if $H$ is infinite.
For example, $N$ is an infinite subset of $(Z,+)$ and $N$ is closed under addition. However $N$ is not a subgroup of $(Z,+)$.

Theorem 1.4.12. If $H$ and $K$ are subgroups of a group $G$ then $H\cap K$ is also a subgroup of $G.$

Proof : Clearly $e\in H\cap K$ and so $H\cap K$ is non-empty.
Now let $a,b\in H\cap K$. Then $a,b\in H$ and $a,b\in K$. Since $H$ and $K$ are subgroups of $G,a{b}^{-1}\in H$ and $a{b}^{-1}\in K.$ Therefore $a{b}^{-1}\in H\cap K$. Hence by Theorem 1.4.10, $H\cap K$ is a subgroup of $G.$ □

It can be similarly proved that the intersection of any number of subgroups of $G$ is again a subgroup of $G.$

Remark 1.4.13. The union of two subgroups of a group need not be a subgroup.
For example, $2Z$ and $3Z$ are subgroups of $(Z,+)$ but $2Z\cup 3Z$ is not a subgroup of $Z$ since $3,2\in 2Z\cup 3Z$ but $3+3=5\notin 2Z\cup 3Z.$

Theorem 1.4.14. The union of two subgroups of a group $G$ is a subgroup if and only if one is contained in the other.

Proof : Let $H$ and $K$ be two subgroups of $G$ such that one is contained in the other. Then either $H\subseteq K$ or $K\subseteq H$.
Therefore $H\cup K=K$ or $H\cup K=H$. Hence $H\cup K$ is a subgroup of $G.$
Conversely, suppose $H$ is not contained in $K$ and $K$ is not contained in $H$. Then there exist elements $a,b$ such that $a\in H,a\notin K,b\in K$, and $b\notin H.$
Clearly $a,b\in H\cup K$. Since $H\cup K$ is a subgroup of $\mathit{Gab}\in H\cup K$. Hence $\mathit{ab}\in H$ or $\mathit{ab}\in K$.
If $\mathit{ab}\in H$, then $a^{-1}\in H$ since $a\in H$.
Hence $a^{-1}\left(\mathit{ab}\right)=b\in H$, a contradiction.
If $\mathit{ab}\in K,b^{-1}\in K$ since $b\in K$. Hence $\left(\mathit{ab}\right)b^{-1}=a\in K$, a contradiction.
Hence our assumption that $H$ is not contained in $K$ and $K$ is not contained in $H$ is false.
Therefore $H\subseteq K$ or $K\subseteq H.$ □

Definition 1.4.15. Let $A$ and $B$ be two subsets of a group G. We define $\mathit{AB}=\{\mathit{ab}:a\in A,b\in B\}.$

Note 1.4.16. If $A$ and $B$ are two subgroups of $G$, then $\mathit{AB}$ need not be a subgroup of $G.$
For example, consider $G=S_3.A=\{e,p_3\}$ and $B=\{e,p_4\}$. Then $A$ and $B$ are subgroups of $S_3$. Also

$$\mathit{AB}=\{\mathit{ee},e{p}_4,e{p}_3,p_3{p}_4\}=\{e,p_4,p_3,p_2\}.$$

Now, $p_4{p}_2=p_5\notin \mathit{AB}.$ Hence $\mathit{AB}$ is not a subgroup of $S_3.$

Theorem 1.4.17. Let $A$ and $B$ be subgroups of a group $G$. Then $\mathit{AB}$ is a subgroup of $G$ if and only if $\mathit{AB}=\mathit{BA}.$

Proof : Suppose $\mathit{AB}$ is a subgroup of $G$.
Let $\mathit{ba}\in \mathit{BA}$, where $a\in A$ and $b\in B.$
Now $a=\mathit{ae}\in \mathit{AB}$ and $b=\mathit{eb}\in \mathit{AB}$. Because $\mathit{AB}$ is a subgroup, it follows that $\mathit{ab}\in \mathit{AB}$. Hence,

$$\mathit{BA}\subseteq \mathit{AB}$$

(1.9)

On the other hand, let $\mathit{ab}\in \mathit{AB}$. Then ${\left(\mathit{ab}\right)}^{-1}\in \mathit{AB},$ so ${\left(\mathit{ab}\right)}^{-1}=a_1{b}_1$ for some $a_1\in A$ and $b_1\in B$. Thus, $\mathit{ab}={\left(a_1{b}_1\right)}^{-1}=b_1^{-1}{a}_1^{-1}\in \mathit{BA}.$ This implies that

$$\mathit{AB}\subseteq \mathit{BA}.$$

(1.10)

From (1.9) and (1.10), we get, $\mathit{AB}=\mathit{BA}$.

Conversely, suppose $\mathit{AB}=\mathit{BA}$. Let $a_1{b}_1,a_2{b}_2\in \mathit{AB}$, where $a_1,a_2\in A$ and $b_1,b_2\in B$.
We show that $\left(a_1{b}_1\right){\left(a_2{b}_2\right)}^{-1}\in \mathit{AB}$.
Now $b_2\in B$ and $a_2\in A$. Therefore, $b_2^{-1}{a}_2^{-1}\in \mathit{BA}=\mathit{AB}$.
This implies that $b_2^{-1}{a}_2^{-1}=a_3{b}_3$ for some $a_3\in A$ and $b_3\in B.$
Similarly, $b_1{a}_3\in \mathit{BA}=\mathit{AB}$, so $b_1{a}_3=a_4{b}_4$ for some $a_4\in A$ and $b_4\in B$. Thus,

$$\begin{array}{llll}\hfill \left(a_1{b}_1\right){\left(a_2{b}_2\right)}^{-1}& =a_1{b}_1{b}_2^{-1}{a}_2^{-1}\text{ (because }{\left(a_2{b}_2\right)}^{-1}=b_2^{-1}{a}_2^{-1}\text{)}& \hfill & \\ \hfill & =a_1{b}_1{a}_3{b}_3\text{ (substitute }{b}_2^{-1}{a}_2^{-1}=a_3{b}_3\text{)}& \hfill & \\ \hfill & =a_1{a}_4{b}_4{b}_3\in \mathit{AB}\text{(substitute }{b}_1{a}_3=a_4{b}_4\text{)}& \hfill & \end{array}$$

Hence, $\mathit{AB}$ is a subgroup of $G$. □

Corollary 1.4.18. If $A$ and $B$ are subgroups of an abelian group $G$, then $\mathit{AB}$ is a subgroup of $G.$

Proof : Let $x\in \mathit{AB}.$ Then $x=\mathit{ab}$ where $a\in A$ and $b\in B$.
Since $G$ is abelian, $\mathit{ab}=\mathit{ba}$ and so $x\in \mathit{BA}$. Hence $\mathit{AB}\subseteq \mathit{BA}$.
Similarly $\mathit{BA}\subseteq \mathit{AB}$ and $\mathit{AB}=\mathit{BA}$. Hence $\mathit{AB}$ is a subgroup of $G.$ □

Solved Problems

Problem 1.4.19. Let $a\in R^{\ast }$ Let $H=\{a^n:n\in Z\}$. Then $H$ is a subgroup of $R^{\ast }$

Solution. Clearly $H$ is non-empty. Now, let $x,y\in H$. Then $x=a^s$ and $y=a^t$ where $s,t\in Z$. Thus

$$x{y}^{-1}=a^s{\left(a^t\right)}^{-1}=a^{s-t}\in H$$

and so $H$ is a subgroup of $R^{\ast }$

Problem 1.4.20. Let $H$ denote the set of all permutations in $S_n$ fixing the symbol l. Then $H$ is a subgroup of $S_n.$

Solution. Clearly $e\in H$ and so $H$ is non-empty. Let $\alpha ,\beta \in H$. Then $\alpha$ and $\beta$ fix the symbol 1. Now $\beta$ fixes the symbol $1\Rightarrow {\beta }^{-1}$ fixes the symbol 1. Hence $\alpha {\beta }^{-1}$ fixes symbol 1. Hence $\alpha {\beta }^{-1}\in H$. Thus $H$ is a subgroup of $S_n.$

Problem 1.4.21. Let $G$ be the set of all $2\times 2$ matrices with entries from $R$. Then $G$ is a group under matrix addition. Let $H=\{\left(\begin{array}{cc}a\hfill & 0\hfill \\ 0\hfill & b\hfill \end{array}\right):a,b\in R\}$. Then $H$ is a subgroup of $G.$

Solution. Let $A,B\in H$. Then $A=\left(\begin{array}{cc}a\hfill & 0\hfill \\ 0\hfill & b\hfill \end{array}\right)$ and $B=\left(\begin{array}{cc}c\hfill & 0\hfill \\ 0\hfill & d\hfill \end{array}\right)$. Now

$$\begin{array}{llll}\hfill A-B& =\left(\begin{array}{cc}a\hfill & 0\hfill \\ 0\hfill & b\hfill \end{array}\right)-\left(\begin{array}{cc}c\hfill & 0\hfill \\ 0\hfill & d\hfill \end{array}\right)=\left(\begin{array}{cc}a-c\hfill & 0\hfill \\ 0\hfill & b-d\hfill \end{array}\right)\in H& \hfill & \end{array}$$

Hence $H$ is a subgroup of $G.$

Problem 1.4.22. Let $G$ be a group. Let $H=\{a:a\in G$ and $\mathit{ax}=\mathit{xa}$ for all $x\in G\}.$
(ie) $H$ is the set of all elements which commute with every other element. Show that $H$ is a subgroup of $G.$

Solution. Clearly $\mathit{ex}=\mathit{xe}=x$ for all $x\in G$.
Hence $e\in H$, so that $H$ is non-empty.
Now, let $a,b\in H$.
Then $\mathit{ax}=\mathit{xa}$ and $\mathit{bx}=\mathit{xb}$ for all $x\in G$. Now,

$$\begin{array}{llll}\hfill \mathit{bx}=\mathit{xb}& \Rightarrow b^{-1}\left(\mathit{bx}\right)b^{-1}=b^{-1}\left(\mathit{xb}\right)b^{-1}& \hfill & \\ \hfill & \Rightarrow \left(b^{-1}b\right)x{b}^{-1}=b^{-1}x\left(b{b}^{-1}\right)& \hfill & \\ \hfill & \Rightarrow \mathit{ex}{b}^{-1}=b^{-1}\mathit{xe}& \hfill & \\ \hfill & \Rightarrow x{b}^{-1}=b^{-1}x.& \hfill & \end{array}$$

Now

$$\begin{array}{llll}\hfill \left(a{b}^{-1}\right)x& =a\left(b^{-1}x\right)& \hfill & \\ \hfill & =a\left(x{b}^{-1}\right)& \hfill & \\ \hfill & =\left(\mathit{ax}\right)b^{-1}& \hfill & \\ \hfill & =\left(\mathit{xa}\right)b^{-1}& \hfill & \\ \hfill & =x\left(a{b}^{-1}\right)& \hfill & \end{array}$$

Thus $a{b}^{-1}$ commutes with every element of $G$ and so $a{b}^{-1}\in H$.
Hence $H$ is a subgroup of $G.$

Note 1.4.23. The above subgroup of $G$ is called the center of $G$ and is denoted by $Z\left(G\right)$.

Problem 1.4.24. Let $G$ be a group and let $a$ be a fixed element of $G$. Let $H_a=\{x:x\in G$ and $\mathit{ax}=\mathit{xa}\}$. Show that $H_a$ is a subgroup of $G.$

Solution. Clearly $\mathit{ea}=\mathit{ae}=a$. Hence $e\in H_a$ so that $H_a$ is non-empty. Then $\mathit{ax}=\mathit{xa}$ and $\mathit{ay}=\mathit{ya}$. Now,

$$\mathit{ay}=\mathit{ya}\Rightarrow y^{-1}a=a{y}^{-1}.$$

Hence

$$\begin{array}{llll}\hfill a\left(x{y}^{-1}\right)& =\left(\mathit{ax}\right)y^{-1}& \hfill & \\ \hfill & =\left(\mathit{xa}\right)y^{-1}& \hfill & \\ \hfill & =x\left(a{y}^{-1}\right)& \hfill & \\ \hfill & =x\left(y^{-1}a\right)& \hfill & \\ \hfill & =\left(x{y}^{-1}\right)a& \hfill & \end{array}$$

Hence $x{y}^{-1}$ commutes with $a,x{y}^{-1}\in H_a$ and so $H_a$ is a subgroup of $G.$

Note 1.4.25. $H_a$ is called the normalizer of $a$ in $G.$

1.5 Cyclic Groups

Definition 1.5.1. Let $G$ be a group. Let $a\in G$. Then $H=\{a^n:n\in Z\}$ is a subgroup of $G.$ $H$ is called the cyclic subgroup of $G$ generated by $a$ and is denoted by $\langle a⟩.$

Example 1.5.2.

1. In $(Z,+),\langle a⟩=2Z$ which is the group of even integers.
2. In the group $G=(Z_{12},\oplus )$ , $\langle 3⟩=\{0,3,6,9\},\langle 5⟩=\{0,5,10,3,8,1,6,11,4,9,2,7\}=Z_{12}.$
3. In the group $G=\{1,i,-1,-i\},\langle i⟩=\{i,i^2,i^3,\cdots \}=\{i,-1,-i,1\}=G.$

Definition 1.5.3. Let $G$ be a group and let $a\in G,a$ is called a generator of $G$ if $\langle a⟩=G.$

A group $G$ is cyclic if there exists an element $a\in G$ such that $\langle a⟩=G.$

Note 1.5.4. If $G$ is cyclic group generated by an element $a$, then every element of $G$ is of the form $a^n$ for some $n\in Z.$

Example 1.5.5.

1. $(Z,+)$ is a cyclic group and $1$ is the generator of this group. Clearly $-1$ is also a generator of this group. Thus a cyclic group can have more than one generator.
2. $(\mathit{nZ},+)$ is a cyclic group and $n$ and $-n$ are generators of this group.
3. $(Z_8,\oplus )$ is a cyclic group and 1, 3, 5, 7 are all generators of this group.
4. $(Z_n,\oplus )$ is a cyclic group for all $n\in N;1$ is a generator of this group. In fact if $m\in Z_n$ and $(m,n)=1$ then $m$ is a generator of this group.
5. $G=\{1,i,-1,-i\}$ is a cyclic group under usual multiplication; $i$ is a generator, $-i$ is also a generator of $G$. However $-1$ is not a generator of $G$ since $\langle -1⟩=\{1,-1\}\ne G.$
6. $G=\{1,\omega ,{\omega }^2\}$ where $\omega \ne 1$ is a cube root of unity is a cyclic group, $\omega$ and ${\omega }^2$ are both generators of this group.
7. In this group $G=(Z_7-\{0\},\odot )$, $3$ and $5$ are both generators. Here 2 is not a generator of $G$ since $\langle 2⟩=\{2,4,1\}\ne G.$
8. Let $A$ be a set containing more than one element. Then $\left(\rho \right(A),△)$ is not cyclic; for let $B\in \rho \left(A\right)$ be any element. Then $B△B=\Phi$ so that $\langle B⟩=\{B,\Phi \}\ne \rho \left(A\right)$.
9. $(R,+)$ is not a cyclic group since for any $x\in R,\langle x⟩=\{\mathit{nx}:n\in Z\}\ne R$

Theorem 1.5.6. Any cyclic group is abelian.

Proof : Let $G=\langle a⟩$ be a cyclic group.
Let $x,y\in G$. Then $x=a^r$ and $y=a^s$ for some $r,s\in Z$. Hence

$$\begin{array}{llll}\hfill \mathit{xy}& =a^r{a}^s=a^{r+s}=a^{s+r}=a^s{a}^r=\mathit{yx}& \hfill & \end{array}$$

Hence $G$ is abelian. □

Theorem 1.5.7. A subgroup of cyclic group is cyclic.

Proof : Let $G$ be a cyclic group generated by $a$ and let $H$ be a subgroup of $G$.
We claim that $H$ is cyclic.
Clearly every element of $H$ is of the form $a^n$ for some integer $n$.
Let $m$ be the smallest positive integer such that $a^m\in H$.
We claim that $a^m$ is the generator of $H$.
Let $b\in H$. Then $b=a^n$ for some $n\in Z$. Then

$$b=a^n=a^{\mathit{mq}+r}=a^{\mathit{mq}}{a}^r={\left(a^m\right)}^q{a}^r.$$

Therefore $a^r={\left(a^m\right)}^{-q}b$.
Now, $a^m\in H$. Since $H$ is a subgroup, ${\left(a^m\right)}^{-q}\in H$. Also, $b\in H$.
Clearly $a^r\in H$ and $0\le r<m$.
But $m$ is the least positive integer such that $a^n\in H$. Therefore $r=0$.
Hence $b=a^n=a^{\mathit{qm}}={\left(a^m\right)}^q.$ Every element of $H$ is a power of $a^m$.
Thus $H=\langle a^m⟩$ and so $H$ is cyclic. □

1.6 Order of an Element

Definition 1.6.1. Let $G$ be a group and let $a\in G$. The least positive integer $n$ if it exists) such that $a^n=e$ is called the order of $a$. If there is no positive integer $n$ such that $a^n=e$, then the order of $a$ is said to be infinite.

Example 1.6.2.

1. Consider the group $S_3$,

   $$\begin{array}{llll}\hfill p_1& =\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 2\hfill & 3\hfill & 1\hfill \end{array}\right),& \hfill & \\ \hfill p_1^2& =\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 2\hfill & 3\hfill & 1\hfill \end{array}\right)\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 2\hfill & 3\hfill & 1\hfill \end{array}\right)=\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 3\hfill & 1\hfill & 2\hfill \end{array}\right)=p_2& \hfill & \\ \hfill p_1^3& =\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 3\hfill & 1\hfill & 2\hfill \end{array}\right)\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 2\hfill & 3\hfill & 1\hfill \end{array}\right)=\left(\begin{array}{ccc}1\hfill & 2\hfill & 3\hfill \\ 1\hfill & 2\hfill & 3\hfill \end{array}\right)=e.& \hfill & \end{array}$$

   In this case, 3 is the least positive integer such that $p_1^3=e$. Thus $p_1$ is of order 3.

2. Consider $(R^{\ast },·)$ , From this sequence of elements 2, $2^2,2^3,\dots <!--\; FUNCTION\; APPLICATION\; -->,2^n,\dots <!--\; FUNCTION\; APPLICATION\; -->$.
   In this case there is no positive integer $n$ such that $2^n=1$ and $\langle 2⟩$ contains infinite numbers of elements.
   Thus the order 2 is infinite.

Theorem 1.6.3. Let $G$ be a group and $a\in G$. Then the order of $a$ is the same as the order of the cyclic group generated by $a.$

Proof : Let $a$ be an element of order $n$. Then $a^n=e$.
We claim that $e,a,a^2,\dots <!--\; FUNCTION\; APPLICATION\; -->,a^{n-1}$ are all distinct.
Suppose $a^r=a^s$ where $0<r<s<n$.
Then $a^{s-r}=e$ and $s-r<n$ which contradicts the definition of the order of $a$.
Hence $e,a,a^2,\dots <!--\; FUNCTION\; APPLICATION\; -->,a^{n-1}$ are $n$ distinct elements and $\langle a⟩=\{e,a,a^2,\dots <!--\; FUNCTION\; APPLICATION\; -->,a^{n-1}\}$ which is of order $n.$
If $a$ is of infinite order, the sequence of elements $e,a,a^2,\dots <!--\; FUNCTION\; APPLICATION\; -->,a^{n-1},\dots <!--\; FUNCTION\; APPLICATION\; -->$ are all distinct and are in $\langle a⟩$. Hence $\langle a⟩$ is an infinite group. □

Theorem 1.6.4. In a finite group every element is of finite order.

Proof : Let $a\in G$. If $a$ is of infinite order, then $\langle a⟩$ is an infinite subgroup of $G$, which is a contradiction since $G$ is finite. Hence the order of $a$ is finite. □

Remark 1.6.5. The converse of the above theorem is not true.
i.e., if $G$ is of group in which every element is of finite order then the group $G$ need not be finite. For example, if $S$ is of infinite set, then $\left(\rho \right(S),△)$ is an infinite group. In this group $A△A=\Phi$ for every $A\in \rho \left(S\right)$ so that the order of every element other than $\varnothing$ is 2.

Theorem 1.6.6. Let $G$ be a group and $a$ be an element of order $n$ in $G$. Then $a^m=e$ if and only if $n$ divides $m.$

Proof : Suppose $n|m$. Then $m=\mathit{nq}$ where $q\in Z$ and

$$a^m=a^{\mathit{nq}}={\left(a^n\right)}^q=e^q=e.$$

Conversely, let $a^m=e$.
Let $m=\mathit{nq}+r$ where $0\le r<n$. Now

$$a^m=a^{\mathit{nq}+r}=a^{\mathit{nq}}{a}^r=e{a}^r=a^r.$$

Thus $a^r=e$ and $0\le r<n$.  Now, since $n$ is the least positive integer such that $a^n=e$, we have $r=0$. Hence $m=\mathit{nq}$ and so $n|m.$ □

Theorem 1.6.7. Let $G$ be a group and $a,b\in G$. Then

1. order of $a=$order of $a^{-1}$
2. order of $a=$order of $b^{-1}\mathit{ab}.$
3. order of $\mathit{ab}=$order of $\mathit{ba}.$

Proof :

1. Let $a$ be an element of order $n$. Then $a^n=e$ and

   $${\left(a^{-1}\right)}^n={\left(a^n\right)}^{-1}=e^{-1}=e.$$

   Now, if possible let $0<m<n$ and ${\left(a^{-1}\right)}^m=e$.
   Therefore ${\left(a^m\right)}^{-1}=e$ and $a^m=e$ which contradicts the definition of the order of $a$.
   Thus $n$ is the least positive integer such that ${\left(a^{-1}\right)}^n=e$. Hence the order of $a^{-1}$ is $n.$

2. We shall first prove that for any positive integer $r$. Now ${\left(b^{-1}\mathit{ab}\right)}^r=b^{-1}{a}^{r}b$ and is trivially true if $r=1$. Now, suppose that it is true for $r=k$ so that ${\left(b^{-1}\mathit{ab}\right)}^k=b^{-1}{a}^{k}b.$ Then

   $$\begin{array}{llll}\hfill {\left(b^{-1}\mathit{ab}\right)}^{k+1}& ={\left(b^{-1}\mathit{ab}\right)}^k\left(b^{-1}\mathit{ab}\right)& \hfill & \\ \hfill & =\left(b^{-1}{a}^{k}b\right)\left(b^{-1}\mathit{ab}\right)& \hfill & \\ \hfill & =b^{-1}{a}^{k+1}b& \hfill & \end{array}$$

   Hence it is true for all positive integers.
   Now, let $a$ be an element of order $n$. Then $a^n=e$. and so

   $${\left(b^{-1}\mathit{ab}\right)}^n=b^{-1}{a}^{n}b=b^{-1}\mathit{eb}=e.$$

   Now, if possible, let $0<m<n$ and ${\left(b^{-1}\mathit{ab}\right)}^m=e$.
   Then $b^{-1}{a}^{m}b=e$ and $a^m=e$ which contradicts the definition of the order of $a$.
   Thus $n$ is the least positive integer such that ${\left(b^{-1}\mathit{ab}\right)}^n=e$.
   Therefore the order of $b^{-1}\mathit{ab}$ is $n.$

3. The order of $\mathit{ab}$= the order of $a^{-1}\left(\mathit{ab}\right)a$= the order of $\mathit{ba}$ by (1).

□

Theorem 1.6.8. Let $G$ be a group and let $a$ be an element of order $n$ in $G$. Then the order of $a^s$, where $0<s<n$, is $n/d$ where $d$ is the g.c.d of $n$ and $s.$

Proof : Let $(n/d)=k$ and $(s/d)=l$ so that $k$ and $l$ are relatively prime. Now,

$${\left(a^s\right)}^k=a^{\mathit{sk}}=a^{\mathit{ldk}}=a^{\mathit{ln}}={\left(a^n\right)}^l=e.$$

Further if $m$ is any positive integer such that ${\left(a^s\right)}^m=e$ then $a^{\mathit{sm}}=e$.
Since order of $a$ is $n$, we have $n|\mathit{sm}$. Therefore $\mathit{kd}|\mathit{ldm}$ and so $k|\mathit{lm}$.
But $k$ and $l$ are relatively prime.
Hence $k|m$ so that $m\ge k$.
Thus $k$ is the least positive integer such that ${\left(a^s\right)}^k=e$.
Hence the order of $a^s=k=n/d.$ □

Corollary 1.6.9. The order of any power of $a$ cannot exceed the order of $a.$

Corollary 1.6.10. Let $G$ be a finite cyclic group of order $n$ generated by an element $a$. Then $a^s$ generates a cyclic group of order $n/d$ where $d$ is the g.c.d of $n$ and $s.$

Corollary 1.6.11. Let $G$ be a finite cyclic group of order $n$ generated by an element $a.a^s$ is a generator of $G$ if and only if $s$ and $n$ are relatively prime. Hence the number of generators of a cyclic group of order $n$ is $\varphi \left(n\right)$ where $\varphi \left(n\right)$ is the number of positive integers less than $n$ and relatively prime to $n.$
For example, consider the group $(Z_{12},\oplus )$. $\varphi \left(12\right)=4$. Hence the group has exactly 4 generators and they are 1, 5, 7 and 11.

Solved Problems

Problem 1.6.12. If $G$ is a finite group with even number of elements then $G$ contains at least one element of order 2.

Solution. Clearly $a$ is an element of order $2$

$$\iff a^2=e\iff a^{-1}=a.$$

Hence it is enough if we prove that there exists an element different from $e$ in $G$ whose inverse is itself.
Let $S=\{a:a\in G,a\ne a^{-1}\}$. Then $a\in S\Rightarrow a^{-1}\in S$ and $a\ne a^{-1}$.
Hence $S$ contains an even number of elements.
Also $e\notin S$. Hence $S\cup \left\{e\right\}$ contains an odd number of elements.
Since the order of the group is even, there exists at least one element $a\notin S\cup \left\{e\right\}$ such that $a=a^{-1}$

Problem 1.6.13. The order of a permutation $p$ is the l.c.m of the lengths if its disjoint cycles.

Solution. Let $p=c_1{c}_2\cdots c_r$ where the $c_i$’s are mutually disjoint cycles of lengths $l_i$.
Now, let $p^m=e$. Since product of disjoint cycles is commutative,

$$e=p^m={(c_1{c}_2\dots <!--\; FUNCTION\; APPLICATION\; -->c_r)}^m=c_1^m{c}_2^m\cdots c_r^m.$$

Now, since the elements moved by one cycle are left fixed by all the other cycles,

$$c_1^m=c_2^m=\cdots =c_r^m=e.$$

Now, $c_1^m=e\Rightarrow l_1|m$ since the order of $c_1=l_1$.
Similarly $l_2,l_3,\dots <!--\; FUNCTION\; APPLICATION\; -->,l_r$ divide $m$.
Thus $m$ is a common multiple of $l_1,l_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,l_r.$
Thus the order of $p$ is the least such $m$ which is obviously the l.c.m of $l_1,l_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,l_r.$

Problem 1.6.14. If $a$ is a generator of the cyclic group $G$ and if there exist two unequal integers $m$ and $n$ such that $a^m=a^n$, prove that $G$ is a finite group.

Solution. Since $m$ and $n$ are unequal we may assume that $m>n$. Hence $m-n$ is a positive integer. Also

$$a^m=a^n\Rightarrow a^{m-n}=e.$$

Therefore the order of $a$ is finite and so $G=\langle a⟩$ is a finite group.