### IIT JAM MATHS

DOUBLE INTEGRAL
BETA AND GAMMA FUNCTION
VOLUME
SURFACE AREA
LIMITS
CONTINUITY
Continuity
DIFFERENTIABILITY
APPLICATION OF DERIVATIVES
MEAN VALUE THEOREM
Part B: FUNCTION OF TWO VARIABLES
INTRODUCTION
DIFFERENTIAL EQUATIONS OF FIRST ORDER AND FIRST DEGREE
ORTHOGONAL TRAJECTORY
DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS
CAUCHY EULER EQUATIONS
DIFFERENTIAL EQUATION OF SECOND ORDER
Section 5: LINEAR ALGEBRA
VECTOR SPACE AND LINEAR EQUATIONS
ORTHOGONALITY
EIGENVALUES AND EIGENVECTORS
LINEAR TRANSFORMATION
Section 6: GROUP THEORY
Unit Group
Subgroup
CYCLIC GROUP
PERMUTATION GROUP
ISOMORPHISM OF GROUPS
EXTRNAL DIRECT PRODUCT
HOMOMORPHISM OF GROUP
SET THEORY
Set Theory
REAL NUMBERS
TOPOLOGY ON REAL LINE
Open Sets
REAL SEQUENCES
INFINITE SERIES
POWER SERIES

# Dihedral Groups The Group of symmetric operations of a regular n-sided polygon is called dihedral group. It is denoted by the symbol $D_{n}$.

The elements of $D_{n}$ consists of n rotations and n reflections.

Let us consider the set of symmetric operations of equilateral triangle with centroid O.

It has six symmetric operations : Three operations of rotational symmetry and three operations of reflection symmetry.

The three operations of rotational symmetry :

(a) $R_{0}$ – Rotation through 0° (identity element)

(b) $R_{120}$ – Rotation through 120°

(c) $R_{240}$ – Rotation through 240°

The three operations of reflection symmetry are

(a) Reflection about $L_{1} - R_{L_{1}}$

(b) Reflection about $L_{2} - R_{L_{2}}$

(c) Reflection about $L_{3} - R_{L_{3}}$

Let $G = ({R_{0}, R_{120}, R_{240}, R_{L_{1}}, R_{L_{2}}, R_{L_{3}}})$ be the set of all symmetric operations of a triangle. On G, define binary operation as the composition of two symmetric operations. Then, the Cayley table w.r.t. composition of symmetric operations is

The table is completely filled in without introducing any new symmetric operations. This means that if A and B are in G then AB also belongs to G. So, G has closure property. $R_{0}$ is an identity element. Since there exist $R_{0}$ in every row and every columns, so every element has its inverse. The six operations are maps, so, the composition of symmetric operation are associative. So, the associativity holds in G. Since, Cayley table is not symmetric about diagonal. So, G is not abelian. So, G is a non-abelian group of order 6. This is also called dihedral group and G is denoted by $D_{3}$

Group of Symmetries of a Square

Let us consider the symmetries of a square

Consider the following symmetric operations : $R_{0}$ : Rotation through 0° about the axis perpendicular to the plane passing through the centre of the square (no change in position) $R_{90}$ : Rotation in the anticlockwise direction through 90°, about the axis through the centre and perpendicular to the plane of the square $R_{180}$ : Rotation in the anticlockwise direction through 180°, about the axis through the centre and perpendicular to the plane of the square. $R_{270}$ : Rotation in the anticlockwise direction through 270°, about the axis through the centre and perpendicular to the plane of the square.

H : Flip (or reflection) about the horizontal axis in the plane of the square.

V : Flip (or reflection) about the vertical axis in the plane of the square.

D : Flip (or reflection) of 180° anticlockwise, about the main diagonal.

D’ : Flip (or reflection) of 180° anticlockwise, about the other diagonal.

Two symmetric operations are equivalent if their net effect is the same. For example, a rotation through 90° anticlockwise followed by a rotation through 180° anticlockwise is equivalent to a rotation through 270° anticlockwise. It is also equivalent to a rotation through 180° anticlockwise followed by a rotation through 90° anticlockwise. Moreover, the effect of $R_{180}$ followed by V is equivalent to H as is shown below :

It can be verified that any motion of the square which makes it fit back into the original space is equivalent to one of the above eight motions.

Example :

Let $D_{4} = (R_{0}, R_{90}, R_{180}, R_{270}, H, V, D, D')$ be the set of all symmetric operations of a square. On $D_{4}$ define an operation as the composition of symmetric operations, i.e. ab means symmetric operations b followed by symmetric operations a.

The Cayley table is

Step 1 (Closure) : Observe that for all $a, b \in D_{4}, ab \in D_{4}$. Thus $D_{4}$ is closed with respect to composition of symmetric operations, so “composition of symmetric operations” is a binary operation on $D_{4}$.

Step 2 (Associativity) : Clearly “composition of symmetric operations” is associative, because every symmetric operation is a mapping from the set of vartices to itself, and composition of mapping is associative.

Step 3 (Identity element) : $R_{0} \in D_{4}$ is such that $a R_{0} = R_{0}a = a,$ for all $a \in D_{4}$, so that $R_{0}$ is an identity element of $D_{4}$

Step 4 (Inverse) : From the table, we see that $R_{0}, R_{180}, H, D, V$ and D’ are their own inverses; whereas $R_{90}$ and $R_{270}$ are inverses of each other. Hence, every element has an inverse.

Step 1 to 4 prove that $D_{4}$ is a group with respect to composition of symmetric operations.

From the table, $DH = R_{270}$ and $HD = R_{90} = D',$ so, that $DH \neq HD$. Hence $D_{4}$ is not Abelian.

Show that product of two rotations is a rotation, product of two reflection is a rotation, whereas product of a rotation and a reflection is a reflection. $D_{4} = (R_{0}, R_{90}, R_{180}, R_{270}, H, V, D, D')$

The binary operation in $D_{4}$ is composition of motion. The order of dihedral group $D_{n}$ is 2n. The order of all reflections is 2 because if we compose the same reflection twice, we get identity.

In $D_{4}$, $R_{0} = 1$ $R_{180} = 2$ $R_{90} = |R_{270}| = 4$

Representation of $\mathbf{D_{n}}$

Consider a regular n-sided polygon centred at the origin in the xy-plane and lebel the vertices consecutively from 1 to n in a clockwise manner. Let r be the rotation clockwise about the origin through $\dfrac{2 \pi}{n}$ radians. Let s be the reflection about the line of symmetry about vertex 1 and the origin. $D_{n} = (1, r, r^{2}, \cdots, r^{n}, s, sr, sr^{2}, sr^{3}, \cdots sr^{n-1})$

(a) $1, r, r^{2}, \cdots r^{n - 1}$ are all distinct and $r^{n - 1}$ so |r| = n.

(b) |s| = 2

(c) $s \neq r^{i}$ for any i

(d) $sr^{i} \neq sr^{i}$ for all $0 \leq i \leq n - 1$

(e) $rs = sr^{-1}$

(f) $r^{n} = 1$ so, $r^{-1} = r^{n - 1}$

So, $rs = sr^{n - 1}$

Also, $r^{2}s = rsr^{-1}$ $= sr^{-2}$ $r^{3}s = rsr^{-2}$ $= sr^{-3}$ $\vdots$ $r^{k} s = sr^{n} = sr^{n} \cdot r^{k} \ \ \ \ (\mathrm{as} \ r^{n} = e)$ $= sr^{n - k}$

So, $r^{k}s = sr^{n - k}$

Also, $r^{k}s = sr^{-k}$