Elementary Group Theory - CMIpavithra/pavithranHomepage/NOtes_files/gro… · ELEMENTARY GROUP THEORY 3.1. STRUCTURE OF A GROUP 8. Notice that in the row corresponding to a, we have

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Chapter 3

Elementary Group Theory

3.1 Structure of a Group

Group: A Group (G,�) is a set G, along with a binary operation � such that:

• Closure: It is closed under the binary operation: (a � b) ∈ G ∀a, b ∈ G

• Associative: The binary operation on the elements of G is associative: a � (b � c) = (a � b) � c∀a, b, c ∈ G

• Identity: ∃ an unique element e ∈ G such that: a � e = e � a = a ∀a ∈ G. e is called the Identityelement of G.

• Inverse: ∃ b ∈ G such that: ab = ba = e ∀a, b ∈ G where e is the Identity element of G. b is calledthe inverse1 of a denoted as a−1.

Abelian: A group is called Abelian if, in addition to the above, the below property also holds:

• Commutative: a � b = b � a ∀a, b ∈ G.

Order of a Group: The order of a group (G,�), denoted as |(G,�)| is the cardinality of the set G: |(G,�)| =#G. Order of an element of the group: The order of an element g ∈ (G,�) is n where gn = e.

3.1.1 Cayley Table

As the group is defined along with a binary operation �, we need to define this operation for every pair ofelements in G. To do this in a compact manner we have the following table.

1Note that the inverse of an element depends upon the individual element and is not unique for a group, unlike the identityelement.

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3.1. STRUCTURE OF A GROUP CHAPTER 3. ELEMENTARY GROUP THEORY

Table 3.1: Cayley Table for (G,�). The (i, j) entry of the table gives gi � gj . g1 is taken as the identityelement e.

X g1 g2 g3 g4 . . .

g1 g1 � g1 g1 � g2 g1 � g3 g1 � g4 . . .

g2 g2 � g1. . . . . . g2 � g4 . . .

g3 g3 � g1 . . . g3 � g3 . . . . . .

g4 g4 � g1 . . . . . . g4 � g4 . . .

......

......

.... . .

Constructing the Cayley Table

Note that:

1. a � b = a � c implies b = c ∀a, b, c ∈ G.proof: Pre-Multiplying a−1 on both sides: a−1 � (a � b) = a−1 � (a � c). Since ∃ a uniqe identity e,a−1a = e ∀a ∈ G. Hence e�b = e�c ⇒ b = c. Hence the result a binary operation on two elements willproduce distinct results, unless they are same. Since along a particular row (or a particular column),each pair of elements involved in the binary operation is distinct, so is the result of the binary operation.Hence each element along a particular row (or column) is unique, as there are #G positions along arow (or column).

2. Now in the previous case, putting a → a−1 and c → b−1, we see that a � a−1 = a � b−1 impliesa−1 = b−1. Hence each element in G has a unique inverse.

3. ∀a, b ∈ G for which [a, b] = 0, as a � b = b � a, (a � b) is symmetric w.r.t the diagonal. Since ∀a ∈ G,�a, a−1

�= 0 and

�a � a−1

�= e, all e’s are placed symmetric w.r.t the diagonal.

We now take a fixed example. Consider G = {e, a, b, c, d, f} with the binary operation ’·’. Let us constructthe cayley table for (G, ·).

1. As there are 5 elements (apart from e) and each of them must have an unique inverse, we see that (sincethere are an odd number of elements) at least one of them must be its own inverse. Just by choice wetake a, b, c to be thier own inverses: a−1 = a, b−1 = b, c−1 = c. For the other two elements d and f , weassume them to be inverses of each other: d−1 = f an f−1 = d. Hence a·a = b·b = c·c = d·f = f ·d = e.We now begin by placing e’s. Notice that they are symmetric about the diagonal.

2. Also note that g · e = g ∀g ∈ G. Hence the first row and first column are trivially filled.

3. Just by choice, we take a · b = c. From this, we get: a · b = c

a · b = c (3.1)a · c = a · (a · b)

= (a · a) · b= b ∴ a · c = b (3.2)

using (eq. 3.2): b · c = (a · c) · c= a · (c · c)= a ∴ b · c = a (3.3)

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CHAPTER 3. ELEMENTARY GROUP THEORY 3.1. STRUCTURE OF A GROUP

4. Notice that the first row has two vaccant positions. Using (statement. 1), these two positions must befilled by d and f . Hence consider the following two possibilities:

• a · d = d:

d = a · dd · f = (a · d) · f

e = a · (d · f)∴ e = a

The above statement is incorrect as it states that the identity element is not unique.

• Hence the only other option is a · d = f .

f = a · d (3.4)f · f = (a · d) · f

= a · (d · f)= a ∴ f · f = a (3.5)

Using (eq. 3.4): f · d = (a · d) · de = a · (d · d)

a · e = a · (a · (d · d))= (a · a) · (d · d)

a = d · d ∴ d · d = a (3.6)Using (eq. 3.6): a · (f · d) = d · d

(a · (f · d)) · f = (d · d) · f(a · f) · (d · f) = d · (d · f)

a · f = d ∴ a · f = d (3.7)

5. Similarly, we need to determine: (c · a), (c · b) and (b · a):

Using (eq. 3.1): c · a = (a · b) · a= a · (b · a)

a · (c · a) = (a · a) · (b · a)Using a · c = c · a : c = (b · a) ∴ b · a = c (3.8)

6. Hence the first row elements as well as thier symmetric couterparts are determined. Notice that thepositions marked by ’♣’ have to be filled either with d or f as the row containing them has the othersymbols. But neither can be used as the column containing each has both d and f . This states thatthe initial assumption a · b = c is wrong.


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Table 3.2: Cayley Table for G = {e, a, b, c, d, f}. Positions marked as ’−’ have not yet been filled.X e a b c d f

e e a b c d f

a a e c b f d

b b c e a ♣ ♣

c c − − e − −

d d − − − − e

f f − − − e a

7. Notice that a · b = a, a · b = b are invalid as we would then get e = a and e = b respectively. Witha · b = c also ruled out, we see that the only two options are a · b = d and a · b = f . Let us take a · b = d.With this we have:

a · b= d (3.9)Using (eq. 3.9): a · (a · b) = a · d

b = a · d ∴ a · d = b (3.10)Using (eq. 3.9): (a · b) · b = d · b

a = d · b ∴ d · b = a (3.11)Using (eq. 3.10): b · f = (a · d) · f

b · f = a ∴ b · f = a (3.12)Using (eq. 3.12): b · (b · f) = b · a

f = b · a ∴ b · a = f (3.13)Using (eq. 3.13): (b · a) · a = f · a

b = f · a ∴ f · a = b (3.14)

Filling the table now gives:


e e a b c d f

a a e d − b −

b b f e − − a

c c − − e − −

d d − a − − e

f f b − − e −


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8. Notice that in the row corresponding to a, we have two vaccancies for a · c and a · f . These must befilled using c or f , as the other elements are already contained in this row. Also as a · c �= c (for if itwas, then we would have e = a, which is incorrect), the only assignments for the vaccancies are:

a · c= f (3.15)a · f= c (3.16)

Using (eq. 3.15): (a · c) · c = f · ca = f · c ∴ f · c = a (3.17)

Using (eq. 3.16): (a · f) · d = c · da = c · d ∴ c · d = a (3.18)

Using (eq. 3.18): c · (c · d) = c · ad = c · a ∴ c · a = d (3.19)

Using (eq. 3.20): (c · a) · a = d · ac = d · a ∴ d · a = c (3.20)

Filling the table, we get:


e e a b c d f

a a e d f b c

b b f e − − a

c c d − e a −

d d c a − − e

f f b − a e −

9. Notice that in the row corresponding to b, there are two vaccancies for b · c and b · d which must befilled using c and d as other elements in this row contain the rest of the elements. The option of puttingb · c = c (and hence b · d = d) is ruled out since we will then get b = e. Hence the only option of fillingthe two positions is to put:

b · c = d (3.21)b · d = c (3.22)

Using (eq. 3.21): (b · c) · c = d · cb = d · c ∴ d · c = b (3.23)

Now, the only vaccant position in the row corresponding to d (for d · d) must be filled with f .

d · d = f (3.24)Using (eq. 3.22): (b · d) · f = c · f

b = c · f ∴ c · f = b (3.25)

Finally, in the row corresponding to f , the only vaccancy (for f · f) must be replaced by d.

f · f = d (3.26)


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Hence we have the final cayley table:

Table 3.5: Final Cayley Table for G = {e, a, b, c, d, f}.X e a b c d f

e e a b c d f

a a e d f b c

b b f e d c a

c c d f e a b

d d c a b f e

f f b c a e d

3.1.2 Subgroups

Subgroup: (H,�) is a subgroup of (G,�) if H ⊆ G and (H,�) satisfies the group properties.It is called an Abelian subgroup if it also satisfies the commutative law.We can now infer few general properties:

• Every subgroup of a group contains the identity element of the group. As (H,�) is also a group, it hasan identity element. Moreover since the identity element of a group is unique, (G,�) and (H,�) mustcontain the same identity element, which is eG. Therefore (H,�) contains eG.

• For every group, the set containing just the identity element (of a group) is a subgroup (of that group).This is becuase identity element is contained in the group (otherwise the group properties would otbe satisfied for this group). It has an identity by definition, it is its own inverse and is closed underthe group operation by definition. The subgroup containing just the identity element is called a trivialsubgroup.

Cosets

The coset of a subgroup (along with an element of the group) is the set containing the results of the binaryoperation between the given element and every element of the subgroup. Since the group in general is non-Abelian, we see that if (H,�) is a subgroup of (G,�) then for some g ∈ (G,�) and h ∈ (H,�), g � h neednot be equal to h � g. Therefore we need to specify which ’sided’ a particular binary operation is. Hencehave ’left’ and ’right’ cosets.Left and Right Cosets: The left and right cosets of (H,�) in (G,�) are defined by, for some g ∈ (G,�):

Left Coset : g � (H,�) = {g � h|h ∈ (H,�)} (3.27)Right Coset : (H,�) � g = {h � g|h ∈ (H,�)} (3.28)

We can now look at a few properties of cosets.

1. From the definitions in (eq. 3.27) and (eq. 3.28), the number of elements in the left and right cosets ofa subgroup in a group is equal to the order of the subgroup.

2. Assumption: g � (H,�) = (H,�) � g = (H,�) ∀g ∈ (H,�).Justification: Since (H,�) forms a group it is closed under the � operation. Hence from (eq. 3.27)we see that ∀g, h ∈ (H,�) g � h ∈ (H,�), and similarly from (eq. 3.28) we have ∀g, h ∈ (H,�)


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h � g ∈ (H,�). Hence we see that all the elements in g � (H,�) and (H,�) � g are in (H,�),∀g ∈ (H,�). Hence we have the justification.

3. Assumption: ∀g ∈ (G,�), the idenitity element of (G,�): eG is contained in g � (H,�) and (H,�)�g.Justification: It suffices to show that ∃g ∈ (G,�) such that eG = g � h and eG = h � g. This is trueas both (G,�) and (H,�) contain eG, so setting g = h = eG, we provide the justification.

4. Assumption: Every element of (G,�) is present in exactly one of the left cosets of (H,�) in (G,�).Justification: If an element is present in two different left cosets, say g1 � (H,�) and g2 � (H,�), of(H,�) in (G,�) then ∃g1, g2, h ∈ (G,�) with g1 �= g2 such that g1h = g2h. From (eq. 1) of (sec. 3.1.1)we see that if g1 �= g2, then g1h �= g2h. Also note that g1h = g2h implies that g1 = g2, which meansthat the two left cosets are identical. Hence we have the justification.

5. Assumption: Every pair of cosets (of (H,�) in (G,�)) are either disjont or identical.Justification: In the previous statement we have showed that no two left cosets can share the sameelement unless the left cosets are identical, implying that every pair of left cosets is disjoint, unless theyare alike. Hence we have the justification.

6. Assumption: ∃ no g ∈ (G,�) which is not present in any left coset of (H,�) in (G,�).Assumption: Consider the coset g � (H,�). It suffices to show that g ∈ g � g (H,�) ∀g ∈ (G,�). Forg ∈ (H,�) this statement is trivially true. Else, it means that ∃h ∈ (H,�) such that g = g � h. As(H,�) contains the identity element eG, we see that the previous statement is true, thereby justifyingthe assumption.

7. From the assumptions in (statement. 5) and (statement. 6), we see that the cosets of a subgroup ina group, are all disjoint and they cover all the elements of the group. In other words, they partitionthe group, with each partition being a coset. Notice that # elements in (G,�) = |(G,�)|. From(statement. 1) we have that # elements in a left coset = |(H,�)|. Hence we see that # left cosets

needed to partition (G,�) =|(G,�)||(H,�)| . This quantity is denoted by [(G,�) : (H,�)] and called the

index of (H,�). the theorem named after this equation is called Lagrange’s Theorem.Index of a Subgroup: The index of a subgroup of a group is the # left cosets of the subgroup requiredto partition the group.

[(G,�) : (H,�)] =|(G,�)||(H,�)| (3.29)

Normal Subgroup: A normal subgroup (N, �) of a group (G,�) is such that the right and left cosets of x in(N, �) are equal ∀ (G,�).

Normal Subgroup (N, �) of (G,�) : a � N = N � a ∀a ∈ (G,�) (3.30)

3.1.3 Quotient Groups

We see clearly that the coset does not form a group. So we consider the set of all cosets of a normalsubgroup (left or right, it does not matter). Let (N, �) be a normal subgroup of (G,�). Cosider the setS = {a � N |a ∈ (G,�)}. We claim this set forms a group. To check this we have:

• Identity, Inverse and Associativity are trivially satisfies as the one of the cosets contain the identityand as (G,�) is a group, the elements in it have thier also inverses.

• Closure: We need to show that: (a � N) � (b � N) ∈ S ∀(a � N), (b � N) ∈ S.For this it suffices to show that (a � N) � (b � N) = (c � N) for some c ∈ (G,�). Now notice that:

(aN) � (bN) = a � ((Nb) � N)normal subgroups coset = (a � b) � (N � N)

= (a � b) N


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Now since a, b ∈ (G,�) we have: (a � b) ∈ (G,�), thereby justifying the assumption.Hence we see that S forms a group under the � operation. This group formed by the elements of S iscalled the quotient group of (G,�) and is represented as: (G,�) / (N, �).Quotient Group: The quotient group of (G,�) is the set {g � N |g ∈ (G,�)} containing all the cosets(right or left) of its elements in of its normal subgroup. It is represented as (G,�).More generally, since the cosets of elements in a subgroup partition the group, the quotient group isthe group formed by the partition of (G,�) along with the operation � defined analogously.

3.1.4 Normalizers and centralizers

For any two elements A, B of a group: (G,�), we note that unless the group is Abelian, the result of thebinary operation � of the two elements depends upon the order in which they are considered: A�B �= B�A.The difference between these two inequal quantites is defined at the commutator between A and B.Commutator: The commutator for any two elements A and B of a group (G,�) is defined as: [A, B] =A � B −B � A.From the definition of the commutator we can verify that ∀g1, g2 ∈ (G,�): [g1, g1] = 0 and [g1, g2] = − [g2, g1].Based on this commutator we have the following sets associated with the group elements:Center of a group: The center of a group Z (G,�) is the set of all elements in (G,�) that commute with allthe elements in (G,�). Hence Z (G,�) = {g ∈ (G,�) |g � x = x � g∀x ∈ (G,�)}.Since in an Abelian group, all the elements commute with the each other, we see that the centre of an abeliangroup is the group itself, i,e; Z (G,�) = (G,�) ∀ Abelian groups (G,�).For any subgroup (of a group) we define the following two groups:Normalizer: The normalizer of a subgroup (H,�) of (G,�) is the set:

N (H,�) =�g| (g � hi) � g−1 ∈ (H,�) ∀g ∈ (G,�)

�(3.31)

Centralizer: The centralizer of a subgroup (H,�) of (G,�) is the set:

Z (H,�) =�g| (g � hi) � g−1 = hi ∀g ∈ (G,�)

�(3.32)

We can now look at some properties of the centralizers and the normalizers of a subgroup.

1. Immediately one can see that the elements in Z (H,�) form a subset of those in N (H,�).

2. The centalizer of a subgroup forms a subgroup (of a group) of the underlying group, i,e; Z (H,�) and(H,�) are subgroups of (G,�).Justification: Notice that the definition of the centralizer can also be given as the set of elements of thesubgroup that commute with each element in the group. Z (H,�) = {g|g � h = h � g,∀g ∈ (H,�)}.With this denition we can verify the group properties of Z (H,�).

• Identity: Trivially identity commutes with all the elements of the group and hence it is in Z (H,�).

• Inverse: If x ∈ Z (H,�) then x−1 ∈ Z (H,�).Justification: We see that x � y = y � x, ∀y ∈ Z (H,�). Notice that if It suffices to show:x−1 � y = y � x−1 ∀g ∈ (G,�). Since x ∈ (G,�) which is a group, ∃ x−1 ∈ (G,�) such thatx � x−1 = eG.

x � y = y � x�x−1 � (x � y) �

�x−1 =

�x−1 � (y � x) � x−1

��x−1 � x

��

�y � x−1

�=

�x−1 � y

��

�x � x−1

�

y � x−1 = x−1 � y

Hence justifying the assumption.

• Closure: a � x ∈ Z (H,�) ∀x, a ∈ Z (H,�)Justification: It suffices to show that (a � x) � y = y � (a � x). Notice that since x, y and a are


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CHAPTER 3. ELEMENTARY GROUP THEORY 3.2. GROUP OPERATIONS

elements of Z (H,�), they commute with all elements in (G,�).

(a � x) � y = a � (y � x)= y � (a � x)

Hence justifying the theorem.

Hence from the above statements it can be seen that Z (H,�) is group, moreover it is a subgroup of(G,�).

Note that the center of a group is not to be confused with the centralizer of a subgroup in a group. Theformer is the set of elements in (G,�) which commute with every element in (G,�), while the latter is theset of all elements in (G,�) that comuute with every element in the subgroup (H,�). Hence the latter isdefined with respect to a subgroup unlike the former. However, both the center (of a group) as well as thecentralizer (of any subgroup in that group) are subgroups of the underlying group.

3.2 Group Operations

3.2.1 Direct product of groups

Direct product: The direct product of two groups (H,�) and (K,�), represented by (H,�) × (K,�) is agroup containng (h, k), ∀h ∈ (H,�) , k ∈ (K,�).We can define the group operations on (H,�)× (K,�) are:

(h1, k1) � (h2, k2) = (h1 � h2, k1 � k2) (3.33)

The idenity element of this direct product group is the tuple containing the idenity elements of the individualgroups:(eH , eK). The inverse of an element is also the tuple containig the inverse of the correspoding elementsfrom the two groups (in the product).

3.2.2 Homomorphism

Homomorphism: For any two groups (G,�) and (H,�), a group homomorphism is a function f : (G,�) →(H,�) whose action on the elements of a, b ∈ G is given by:

f (a � b) = f (a) � f (b) (3.34)

Notice that it preserves the group structure, i,e; if the elements in G form a group, so do the elements inthe set H. We can now see some properties of f . Let eG and eH denote the identity elements of the groups(G,�) and (H,�) respectively.

Using (eq. 3.34): f (a � eG) = f (a) � f (eG)f (a) = f (a) � f (eG)

(f(a))−1 � f (a) = (f(a))−1 � (f (a) � f (eG))

eH =�(f(a))−1 � f (a)

�� f (eG)

eH = f (eG) (3.35)

Hence we see that the identity element of (G,�) is mapped to the identity element of (H,�). We now seethe mapping of the inverse of an element in (G,�). From (eq. 3.34), for u ∈ (G,�):

f (u) � f�u−1

�= f (eG)

Using (eq. 3.35): = eH�[f (u)]−1 � f (u)

�� f

�u−1

�= [f (u)]−1 � eH

∴ f�u−1

�= [f (u)]−1 (3.36)


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3.2. GROUP OPERATIONS CHAPTER 3. ELEMENTARY GROUP THEORY

Hence we see that f maps the inverse of every element to the inverse of its image in (H,�).

Types of Homomorphisms

Isomorphism: It is a homomorphism f : (G,�) → (H,�), where f is one-to-one. Hence f−1 too is ahomomorphism.Automorphism: It is a iomorphism from a group onto itself: f : (G,�) → (G,�).Endomorphism: It is a homomorphism from a group onto itself: f : (G,�) → (G,�). Note: f is notone-to-one.

Kernel

We now consider the elements in (G,�) that are mapped to the same element in (H,�) by the homomorphism.As the identity element eH is unique to (H,�), the properties of this element can be referrd to as the propertiesof the group. Hence, we consider all the elements in (G,�) that are mapped to eH

2. The set containing allsuch elements is called the kernel of f , denoted as ker(f).

ker(f) = {g ∈ (G,+q) |f(g) = eH} (3.37)

The kernel is useful in associating a homomorphism to a set. That is we can can check the containment of anelement in a set by checking the action of the correspoding homomorphism (associated with that set) on theelement in question. When we talk of (G,�) and (H,�) being linear codes, or vector spaces over the field Fq

we see that �,� become addition modulo q: +q. Notice that the identity element here is the null vector. Sothe kernel of the homomorphism (in this case it is a linear map represented by a matrix) is the set of vectorsin (G,+q) that are mapped to the null vector in (H,+q). We now pick a linear map such that the kernel ofthis linear map is the linear code. The advantage of doing this is that we can quickly identify a code elementby checking if it gives the null vector upon action of this linear map. Such a linear map (represented by amatrix) is called the Parity Check Matrix of the linear code. It is used to check for errors in the codewords.If there is an error in the the codeword, the vector undergoes a translation such that the new vector no morebelongs to the vector space, and hence does not lie in the kernel or the pairity check matrix. Therefore uponaction of this matrix it will not give the null vector, thereby indicating the presence of an error.

3.2.3 Conjugation

Two elements of a group a, b ∈ (G,�) are said to be conjugate to each other if ∃g ∈ (G,�) such that:g � a = b � g. Restating the previous statement, we have: b is conjugate to a if ∃g ∈ (G,�) such thatb = (g � a) � g−1. We can further formalize this by looking at the RHS of previous equation as a functionof a: b = f(a), where f(a) = (g � a) � g−1. This function, or automomorphism (as it takes an element in(G,�) to itself) is called inner automorphism or conjugation.Conjugation or Inner-Automorphism: It is an automorphism defined by f : (G,�) → (G,�) such that forsome a, g ∈ (G,�), f(a) = g �

�a � g−1

�.

We now consider the set of all elements (in a group) that are conjugate to a given element (from the samegroup). That is the set {b|g � a = b � g, g ∈ (G,�)}. Note that since a is the free variable in the definitionof this set, it must be indexed by a. We can also write this set as: Sa =

�g �

�a � g−1

�|g ∈ (G,�)

�. Such

a set is called the conjugacy class of a.Conjugacy Class: Conjugacy class of an element (in a group) is the set containing all the elements from thatgroup which are conjugate to the given element.

Cl(a) =�g �

�a � g−1

�|g ∈ (G,�)

�(3.38)

We can now look at some properties of this set:2Note that, from (eq. 3.35), eG is one of them.


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CHAPTER 3. ELEMENTARY GROUP THEORY 3.3. GROUP ACTIONS

1. Assumption: If g is an element of an Abelian group, ∀g ∈ (G,�), #Cl(a) = 1, that is Cl(a) is asingleton set.Justification: To see this, we try to write the definitions of the abelian group, and the conjugacy classof an element, in the same form.

Cl(a) = {b|g � a = b � g, g ∈ (G,�)}G = {b|g � b = b � g, g ∈ (G,�)}

Notice that the conjugacy class of an element from an Abelian group has t satisfy Observing the twoconditions in the above definitions, we see that the only ’solution’, or satisfying assignment for a (it is theonly free variable) is a = b. Hence we see that: Cl(a) = {b|g � a = b � g, a = b, g ∈ (G,�)} ⇒ {a}.Hence we justify our assumption.

3.3 Group Actions

3.3.1 Generating set of a group

As the elements of a group satisfy certian properties, given a set X it must be possible to construct (ormechanically generate from this set) a subset following these properties. In otherwords, we can take a set ofelements and then generate a group such that group properties are satisfied by construction. Such a set iscalled a generating set of a group and the elements of this set are called generators.Generating Set of a Group: A generating set of a group (G,�) is a set X such that every element of (G,�)can be expressed as a combination (using �) of finite subset of X. We donte it by: (G,�) = �X�.Since a group is also a set of elemets, it can also be used to generate another group. More generally we canuse more than one group to generate a single group. That is elements in groups (H,�) , (G,�) can be usedto generate the elements of (G,�) In this case we say that (G,�) is a direct sum of subgroups (H,�) and(K,�). But for this to be possible notice that (H,�) has to be a normal subgroup of (G,�).Direct sum of Groups: A group (G,�) is a direct sum of groups (H,�) and (K,�), represented as (G,�) =(H,�) ⊕ (K,�) if the generating set of (G,�) ∈ H ∪ K, and (H,�) and (K,�) are normal subgroups of(H,�).

3.3.2 Symmetric group

We now try to explore the properties of a given set X using groups. By exploring the properties, we meanto look at all possible relations between the elements of the set. For this we need to consider the set of allfunctions {h : X → X}, since each function relates one element of X with another. For the sake of simplicity,we avoid relations between a given element and many other elements. Hence we only consider all possibleone-to-one functions (or bijections). This set containing all possible bijections from X to X also forms agroup under the composition (of two functions) operation.

Symmetric Group: A symmetric subgroup on a set X is a group formed by the set G cotaining all possiblebijections f : X → X, under the binary operation ◦ which denoted the composition of two functions. Thisgroup (G, ◦) satisfies the group properties.

We can now verify the group properties of this symmetric group:

• Closure: ∀f, g ∈ (G, ◦) (f ◦ g) ∈ (G, ◦).For this it suffices to show that (f ◦ g) is also a bijection. Notice that ∀x ∈ X, (f ◦ g) (x) = f (g(x)).Since g is a bijection g : X → X we see that g(x) is injective and surjective [g (X) = (X)]. Hence allvalues of g(x) are distinct (∀ distinct x ∈ X). As f is again a bijection from X to X, it will take eachof these distinct g(x) (corresponding to distinct x) to some f(x) ∈ X. Therefore all values of f(x) aredistinct (∀ distinct x ∈ X). Hence f ◦ g is injective. As f(X) = X, we see that f (g(X)) = X. Hencef ◦ g is both surjective and injective, thereby bijective, hence justifying the assumption.


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3.3. GROUP ACTIONS CHAPTER 3. ELEMENTARY GROUP THEORY

• Associative: ∀α,β, γ ∈ (G, ◦), α ◦ (β ◦ γ) = (α ◦ β) ◦ γ.The above statement is true as we have: [α ◦ (β ◦ γ)]x = [(α ◦ β) ◦ γ]x = α (β (γ(x))) ∀x ∈ X. Thisis in general true for any three functions. Hence we justify the assumption.

• Identity: ∃eG ∈ (G, ◦) such that ∀f ∈ (G, ◦), eG ◦ f = f ◦ eG = f .It suffices to show that ∃ a bijection eG such that ∀f ∈ (G, ◦) and ∀x ∈ X, we have: f (eG(x)) =eG (f(x)) = f(x). We can now see that eG is nothing but the identity map: eG ≡ IX , which clearly isa bijection and hence ∈ (G, ◦). Hence we justify the assumption.

• Inverse: ∀f ∈ (G, ◦) ∃g ∈ (G, ◦) such that f ◦ g = g ◦ f = eG.It suffices to show that ∀f ∈ (G, ◦) ∃g ∈ (G, ◦) such that ∀x ∈ X we have: f (g(x)) = g (f(x)) = x.We can now see that g is nothing but the inverse map: f−1, which clearly is a bijection since f is abijection. Hence we see that there is an inverse in (G, ◦) for every element in it, thereby justifying theassumption.

3.3.3 Action of Group on a set

In the above section (sec. 3.3.2) we considered a group of bijections on a set X. Now we consider a generalgroup (Q,�) and a homomorphism: h from (Q,�) to the symmmetric group of X: (G, ◦). Every element of(Q,�) is mapped to a bijection on X. As a result we can describe the action of the bijection (on some x ∈ X)as the action of the element of (Q,�) (which has been mapped to this bijection by the homomorphism h).Therefore the homomorphism is defined as: h : (Q,�) → (G, ◦) such that:

∀q1, q2 ∈ (Q,�) , h (q1 � q2) = h (q1) ◦ h (q2) (3.39)

Note that h(q) is a bijection ∀q ∈ (Q,�), that is in the above expression: h (q1) : X → X. The homomorphismmaps the identity of (Q,�) to identity element of (G, ◦) the inverse of every element in (Q,�) is mapped tothe inverse of the corresponding element’s map.

h (eQ) = IX

∀q ∈ (Q,�) , h�q−1

�= (h(q))−1

Now when we want to describe the operation [h (q1)] (x) for x ∈ X, we denote it as action of q1 on x: q1·x. Sim-ilarly {[h(q)] (x)|q ∈ (Q,�)} can be denoted as: {q · x|q ∈ (Q,�)} and hence Q·X = {q · x|x ∈ X, q ∈ (Q,�)}.This operation is called action of a group on a set, or group action. The group action can be described asa function that takes an element of (Q,�) and an element of X, giving another element of X. Hence thegroup action on a set is described as another homomorphism from (Q,�)×X to X.Group Action: The Group Action of (Q,�) on a set X is defined as a homomorphism: f : (Q,�)×X → Xsuch that: for x ∈ X, q ∈ (Q,�) we have: f (q, x) = q · x which represents q · x ≡ [h(q)] (x) where h is asdefined in (eq. 3.39).

3.3.4 Orbits and Stabilizers

We now need to look at the geometric picture by considering X to be a set of points (in R, C, or anything.)and the action of elements q ∈ (Q,�) (which is the action of h(q) as defined in (eq. 3.39). When q · x = x�

where x� ∈ X we say that x has been transported to the point x� and the path taken by x is the set {x, x�}.Consider the set (Q,�) · x = {q · x|q ∈ (Q,�)}. This says all the points which can be obtained by acting theelements of (Q,�) on x. In other words it gives the path traced by the point x ∈ X upon action of elementsin (Q,�). This path represented by the set of points is called the Orbit of x, denoted by O(Q,�)(x).

Orbit : Orbit of an element x ∈ X : O(Q,�)(x) = (Q,�) · x ≡ {q · x|q ∈ (Q,�)} (3.40)

Since x is a finite set, it may happen that the path traced by x contains x itself. Suppose the path taken byx has points {x, x1, x2, . . . xm, x . . . xn, x . . . xk, x . . . } we have:


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CHAPTER 3. ELEMENTARY GROUP THEORY 3.3. GROUP ACTIONS

xq1−→ x1

q2�···�qm−1−−−−−−−−→ xmqm−−→ x

qm+1�···�qn−1−−−−−−−−−−→ xnqn−→ x

qn+1�···�qk−1−−−−−−−−−→ xk → . . . which can now be rewrittenas: x

q1�q2�···�qm−1�qm−−−−−−−−−−−−−−→ xqm+1�···�qn−1�qn−−−−−−−−−−−−−→ x

qn+1�···�qk−1...−−−−−−−−−−−→ . . . . We now see that the set of opera-tors: Sx (Q,�) = {q1 � q2 � · · · � qm−1 � qm, qm+1 � · · · � qn−1 � qn, q2 � · · · � qk−1 . . . } leave the point xinvariant. Now since q1, q2, . . . , qm, . . . , qn, . . . ∈ (Q,�) which s closed under �, we see that the operatorsin Sx (Q,�) are also in (Q,�). Moveover, the identity operator eQ ∈ Sx (Q,�) (since it corresponds tothe identity mapping) and since each of these operators correspond to a bijection (on X), an inverse can bedefined easily which also is a bjection that leaves x invariant. Hence Sx (Q,�) contains the inverse of elementelement in it. Note that the set Sx (Q,�) is closed under �, has an identity element and evey element in thisset has its inverse in the same set. Therefore the set Sx (Q,�) along with the operation � forms a group,and trivially a subgroup of (Q,�). This subgroup is called the Stabilizer Subgroup of (Q,�).Stabilizer Subgroup: The stabilizer subgroup Sx (Q,�) (of a group) consists of elements that leave an elementx ∈ X invaraint.

Sx (Q,�) = {q ∈ (Q,�) |q · x = x} (3.41)

3.3.5 Orbit Stabilizer theorem

We now have a theorem similar to lagrange’s theorem in (eq. 3.29), relating the sizes of O(x), Sx (Q,�) and(Q,�).

Theorem: For any group, the number of elements in the orbit, with respect to any element in the set and thestabilizer subgroup (of this group) with respect to the same element of the set are related by:

|O(x)|× |Sx (Q,�)| = |(Q,�)| (3.42)

Proof: Let us denote Sx (Q,�) ≡ S(x). Consider the slight modification of S(x) (eq. 3.41) as:

Hy(x) = {q ∈ (Q,�) |q · x = y} (3.43)∴ Hx(x) = S(x) (3.44)

We now claim that the sets Hy(x) for different y are disjoint.Assumption: ∀y1, y2 ∈ X, y1 �= y2 the sets Hy1(x) and Hy2 are disjoint.Justification: It suffices to show that if ∃q ∈ Hy1 ∩ Hy2 , then y1 = y2. This is true since in the former casewe will have: q · x = y1 and q · x = y2, which clearly implies y1 = y2, thereby justifying the assumption.We now have: (Q,�) =

�

y∈O(x)

Hy(x) and hence:

|(Q,�)| =�

y∈O(x)

|Hy(x)| (3.45)

We now claim that each set Hy(x) for every value of y ∈ X has the same cardinality which can be equatedto that of #Hx(x) which is, from (eq. 3.44): |S(x)|.Assumption: For every y ∈ X, #Hy(x) = |S|.Justification: It suffices to show that ∃ a bijection f : S → Hy(x), ∀y ∈ X. We now try to construct thisbijection. For a fixed t ∈ Hy(x) and ∀h ∈ S define the bijection as: f(h) = t�h. Notice that (t � h) ∈ Hy(x)since (t � h) · x = t · (h · x) which from (eq. 3.41) = t · x ⇒ y using (eq. 3.43). To show that f is a bijection,we need to show that f is injective and surjective.Injective: For f (h1) = f (h2), then t · h1 = t · h2 which implies h1 = h2. So if h1 �= h2 then f (h1) �= f (h2).Surjective: We need to show that every element in Hy(x) is covered in the image of f , for some x ∈ X. Itsuffices to show that every element in Hy(x) can be represented in the form of the image of f . Equivalently


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3.3. GROUP ACTIONS CHAPTER 3. ELEMENTARY GROUP THEORY

we can show that ∀u ∈ Hy(x), ∃t ∈ Hy(x) such that t−1 · u ∈ S.

From (eq. 3.43) : u · x = y & t · x = y (3.46)

∴ x = t−1 · yUsing (eq. 3.46) : = t−1 · (u · x)

∴ x =�t−1 · u

�· x

Now from (eq. 3.41) we have:�t−1 · u

�∈ S(x). Hence we have the justification.

Now we see that f is a bijection and ∀y ∈ X,#Hy(x) = |S|. In (eq. 3.45) we can replace the sum by aproduct as all the entities being summer over have the same value. We then have the statement as in thetheorem (eq. 3.42). Hence we have proved the theorem.


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