Groups

Rupesh Tiwari

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Groups

• 1. Introduction
• 2.Normal subgroups, quotien groups.
• 3. Homomorphism.

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1.Introduction

• 1.1. Binary Operations
• 1.2.Definition of Groups
• 1.3.Examples of Groups
• 1.4.Subgroups

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1.Introduction

• 1.1. Binary Operations
• 1.2.Definition of Groups
• 1.3.Examples of Groups
• 1.4.Subgroups

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1.Introduction

1.1.Binary Operations

A binary operation on a set is a rule for combining two elements of the set. More precisely, if S iz a nonemty set, a binary operation on S iz a mapping f : S × S → S. Thus f associates with each ordered pair (x,y) of element of S an element f(x,y) of S. It is better notation to write x y for f(x,y), refering to as the binary operation.

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1.Introduction

1.2.Definition of Groups

A group (G, ・) is a set G together with a binary operation ・ satisfying the following axioms.

1. The operation ・ is associative; that is,

(a ・ b) ・ c = a ・ (b ・ c) for all a, b, c ∈ G.

(ii) There is an identity element e ∈ G such that

e ・ a = a ・ e = a for all a ∈ G.

(iii) Each element a ∈ G has an inverse element a⁻¹ ∈ G such that a^-1・ a = a ・ a⁻¹ = e.

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1.Introduction

If the operation is commutative, that is,

if a ・ b = b ・ a for all a, b ∈ G,

the group is called commutative or abelian, in honor of the

mathematician Niels Abel.

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1.Introduction

1.3.Examples of Groups

• Example 1.3.1. Let G be the set of complex numbers {1,−1, i,−i} and let ・ be the standard multiplication of complex numbers. Then (G, ・) is an abelian group. The product of any two of these elements is an element of G; thus G is closed under the operation. Multiplication is associative and commutative in G because multiplication of complex numbers is always associative and commutative. The identity element is 1, and the inverse of each element a is the element 1/a. Hence

1⁻¹ = 1, (−1)⁻¹ = −1, i⁻¹ = −i, and (−i)⁻¹ = i.

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1.Introduction

• Example 1.3.2. The set of all rational numbers, Q, forms an abelian group (Q,+) under addition.The identity is 0, and the inverse of each element is its negative. Similarly,

(Z,+), (R,+), and (C,+) are all abelian groups under addition.

• Example1. 3.3. If Q^∗, R^∗, and C^∗ denote the set of nonzero rational, real, and complex numbers, respectively, (Q^∗,・),

(R^∗,・), and (C^∗, ・) are all abelian groups under

multiplication.

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1.Introduction

• Example 1.3.4. A translation of the plane R² in the direction of the vector (a, b) is a function f :R² → R² defined by f (x, y) = (x + a, y + b). The composition of this translation with a translation g in the direction of (c, d) is the function

f g:R² → R², where

f g(x, y) = f (g(x, y))= f (x + c, y + d)= (x + c + a, y + d + b).

This is a translation in the direction of (c + a, d + b). It can easily be verified that the set of all translations in R² forms an abelian group, under composition. The identity is the identity transformation 1_R² :R² → R², and the inverse of the translation in the direction (a, b) is the translation in the opposite direction (−a,−b).

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1.Introduction

• Example1.3.5. If S(X) is the set of bijections from any set X to itself, then (S(X), °) is a group under composition. This group is called the symmetric group or permutation group of X.

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1.Introduction

• Proposition 1.3.1. If a, b, and c are elements of a group G, then

(i) (a⁻¹)⁻¹ = a.

(ii) (ab)⁻¹ = b⁻¹a⁻¹.

(iii) ab = ac or ba = ca implies that b = c. (cancellation law)

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1.Introduction

• 1.4.Subgroups

It often happens that some subset of a group will

also form a group under the same operation.Such

a group is called a subgroup. If (G, ・) is a

group and H is a nonempty subset of G, then

(H, ・) is called a subgroup of (G, ・) if the

following conditions hold:

(i) a ・ b ∈ H for all a, b ∈ H. (closure)

(ii) a⁻¹ ∈ H for all a ∈ H. (existence of inverses)

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1.Introduction

• Conditions (i) and (ii) are equivalent to the single condition:

(iii) a ・ b⁻¹ ∈ H for all a, b ∈ H.

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Proposition 1.4.2. If H is a nonempty finite subset of a group G and ab ∈ H for all a, b ∈ H, then H is a subgroup of G.

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Example 1.4.1 In the group ({1,−1, i,−i}, ・), the subset {1,−1} forms a subgroup because this subset is closed under multiplication

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1.Introduction

• Example 1.4.2 .The group Z is a subgroup of Q,Q is a subgroup of R, and R is a subgroup of C. (Remember that addition is the operation in all these groups.)
• However, the set N = {0, 1, 2, . . .} of nonnegative integers is a subset of Z but not a subgroup, because the inverse of 1, namely, −1, is not in N. This example shows that Proposition 1.4.2 is false if we drop the condition that H be finite.
• The relation of being a subgroup is transitive. In fact, for any group G, the inclusion relation between the subgroups of G is a partial order relation.

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1.Introduction

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• Definition. Let G be a group and let a ∈ G. If a^k = 1 for some k ≥ 1, then the smallest such exponent k ≥ 1 is called the order of a; if no such power exists, then one says that a has infinite order.

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• Proposition 1.4.3 . Let G be a group and assume that a ∈G has finite order k. If aⁿ = 1, then k | n. In fact, {n ∈Z : aⁿ = 1} is the set of all the multiples of k.

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1.Introduction

• Definition. If G is a group and a ∈ G, write

<a > = {aⁿ : n∈ Z} = {all powers of a } .

It is easy to see that <a > is a subgroup of G .

< a > is called the cyclic subgroup of G generated by a. A group G is called cyclic if there is some a ∈ G with G = < a >; in this case a is called a generator of G.

• Proposition 1.4.4. If G= <a > is a cyclic group of order n, then a^k is a generator of G if and only if gcd(k; n)= 1.
• Corollary 1.4.5. The number of generators of a cyclic group of order n is φ(n).

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1.Introduction

• Proposition 1.4.6. Let G be a finite group and let a ∈ G. Then the order of a is the number of elements in <a >.

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• Definition. If G is a finite group, then the number of elements in G, denoted by |G|, is called the order of G.

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2.Normal subgroups,quotient� groups

• 2.1.Cosets
• 2.2.Theorem of Lagrange
• 2.3.Normal Subgrops
• 2.4.Quotient Groups

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2.Normal subgroups,quotient� groups

• 2.1.Cosets
• Let (G, ·) be a group with subgroup H. For a, b ∈ G, we say that a is congruent to b modulo H, and write a ≡ b mod H if and only if ab⁻¹ ∈ H.
• Proposition 2.1. 1.The relation a ≡ b mod H is an equivalence relation on G. The equivalence class containing a can be written in the form Ha = {ha|h ∈ H}, and it is called a right coset of H in G. The element a is called a representative of

the coset Ha.

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2.Normal subgroups,quotient� groups

• Example 2.1.1. Find the right cosets of A₃ in S₃.

Solution. One coset is the subgroup itself A₃ = {(1), (123), (132)}. Take any element not in the subgroup, say (12). Then another coset is A₃(12) = {(12), (123) (12), (132) (12)} = {(12), (13), (23)}.Since the right cosets form a partition of S₃ and the two cosets above contain all the elements of S₃, it follows that these are the only two cosets.

In fact, A₃ = A₃(123) = A₃(132) and A₃(12) = A₃(13) = A₃(23).

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2.Normal subgroups,quotient� groups

• Example 2.1.2. Find the right cosets of H = {e, g⁴, g⁸} in C₁₂ = {e, g, g², . . . , g¹¹}.
• Solution. H itself is one coset. Another is Hg = {g, g⁵, g⁹}. These two cosets have not exhausted all the elements of C₁₂, so pick an element, say g², which is not in H or Hg. A third coset is Hg² = {g², g⁶, g¹⁰} and a fourth is Hg³ ={g³, g⁷, g¹¹}.

Since C₁₂ = H ∪ Hg ∪ Hg² ∪ Hg³, these are all the cosets

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2.Normal subgroups,quotient� groups

• 2.2.Theorem of Lagrange
• As the examples above suggest, every coset contains the same number of elements. We use this result to prove the famous theorem of Joseph Lagrange (1736–1813).
• Lemma 2.2.1. There is a bijection between any two right cosets of H in G.

Proof. Let Ha be a right coset of H in G. We produce a bijection between Ha and H, from which it follows that there is a bijection between any two right cosets.

Define ψ:H → Ha by ψ(h) = ha. Then ψ is clearly surjective. Now suppose that ψ(h₁) = ψ(h₂), so that h₁a = h₂a. Multiplying each side by a⁻¹ on the right, we obtain h₁ = h₂. Hence ψ is a bijection.

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2.Normal subgroups,quotient� groups

• Theorem 2.2.2. Lagrange’s Theorem. If G is a finite group and H is a subgroup of G, then |H| divides |G|.

Proof. The right cosets of H in G form a partition of G, so G can be written as a disjoint union

G = Ha₁ ∪ Ha₂ ∪ ·· ·∪ Ha_k for a finite set of elements a₁, a₂, . . . , a_k ∈ G.

By Lemma 2.2.1, the number of elements in each coset is |H|. Hence, counting all the elements in the disjoint union above, we see that |G| = k|H|. Therefore, |H| divides |G|.

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2.Normal subgroups,quotient� groups

• If H is a subgroup of G, the number of distinct right cosets of H in G is called the index of H in G and is written |G : H|. The following is a direct consequence of the proof of Lagrange’s theorem.
• Corollary 2.2.3. If G is a finite group with subgroup H, then

|G : H| = |G|/|H|.

• Corollary 2.2.4. If a is an element of a finite group G, then the order of a divides the order of G.

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2.Normal subgroups,quotient� groups

• 2.3.Normal Subgrops
• Let G be a group with subgroup H. The right cosets of H in G are equivalence classes under the relation a ≡ b mod H, defined by ab⁻¹ ∈ H. We can also define the relation L on G so that aLb if and only if b⁻¹a ∈ H. This relation, L, is an equivalence relation, and the equivalence class containing a is the left coset aH = {ah|h ∈ H}. As the following example shows, the left coset of an element does not necessarily equal the right coset.

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2.Normal subgroups,quotient� groups

• Example 2.3.1. Find the left and right cosets of H = A₃ and K = {(1), (12)} in S₃.
• Solution. We calculated the right cosets of H = A₃ in Example 2.1.1.

Right Cosets

H = {(1), (123), (132)}; H(12) = {(12), (13), (23)}

Left Cosets

H = {(1), (123), (132}; (12)H = {(12), (23), (13)}

In this case, the left and right cosets of H are the same.

• However, the left and right cosets of K are not all the same.

Right Cosets

K = {(1), (12)} ; K(13) = {(13), (132)} ; K(23) = {(23), (123)}

Left Cosets

K = {(1), (12)};(23)K = {(23), (132)}; (13)K = {(13), (123)}

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2.Normal subgroups,quotient� groups

Definition: A subgroup H of a group G is called a normal subgroup of G if g⁻¹hg ∈ H for all g ∈ G and h ∈ H.

Proposition 2.3.1. Hg = gH, for all g ∈ G, if and only if H is a normal subgroup of G.

Proof. Suppose that Hg = gH. Then, for any element h ∈ H, hg ∈ Hg = gH. Hence hg = gh₁ for some h₁ ∈ H and g⁻¹hg = g⁻¹gh₁ = h₁ ∈ H. Therefore,H is a normal subgroup.

Conversely, if H is normal, let hg ∈ Hg and g⁻¹hg = h₁ ∈ H. Then hg = gh₁ ∈ gH and Hg ⊆ gH. Also, ghg⁻¹ = (g⁻¹)⁻¹hg⁻¹ = h₂ ∈ H, since H is normal, so gh = h₂g ∈ Hg. Hence, gH ⊆ Hg, and so Hg = gH.

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2.Normal subgroups,quotient� groups

• If N is a normal subgroup of a group G, the left cosets of N in G are the same as the right cosets of N in G, so there will be no ambiguity in just talking about the cosets of N in G.

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• Theorem 2.3.2. If N is a normal subgroup of (G, ·), the set of cosets G/N = {Ng|g ∈ G} forms a group (G/N, ·), where the operation is defined by (Ng₁) · (Ng₂) = N(g₁ · g₂). This group is called the quotient group or factor group of G by N.

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2.Normal subgroups,quotient� groups

• Proof. The operation of multiplying two cosets, Ng₁ and Ng₂, is defined in terms of particular elements, g₁ and g₂, of the cosets. For this operation to make sense, we have to verify that, if we choose different elements, h₁ and h₂, in the same cosets, the product coset N(h₁ · h₂) is the same as N(g₁ · g₂). In other words, we have to show that multiplication of cosets is well defined. Since h₁ is in the same coset as g₁, we have h₁ ≡ g₁ mod N. Similarly, h₂ ≡ g₂ mod N. We show that Nh₁h₂ = Ng₁g₂. We have h₁g ₁⁻¹ = n₁ ∈ N and h₂g ₂⁻¹ = n₂ ∈ N, so h₁h₂(g₁g₂)⁻¹ = h₁h₂g ₂⁻¹g ₁⁻¹ = n₁g₁n₂g₂g₂ ⁻¹ g ₁⁻¹ = n₁g₁n₂g ₁⁻¹. Now N is a normal subgroup, so g₁n₂g ₁⁻¹ ∈ N and n₁g₁n₂g ₁⁻¹ ∈ N. Hence h₁h₂ ≡ g₁g₂ mod N and Nh₁h₂ = Ng₁g₂. Therefore, the operation is well defined.

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2.Normal subgroups,quotient� groups

• The operation is associative because (Ng₁ · Ng₂) · Ng₃ = N(g₁g₂) · Ng₃ = N(g₁g₂)g₃ and also Ng₁ · (Ng₂ · Ng₃) = Ng₁ · N(g₂g₃) = Ng₁(g₂g₃) = N(g₁g₂)g₃.
• Since Ng · Ne = Nge = Ng and Ne · Ng = Ng, the identity is Ne = N.
• The inverse of Ng is Ng⁻¹ because Ng · Ng⁻¹ = N(g · g⁻¹) = Ne = N and also Ng⁻¹ · Ng = N.
• Hence (G/N, ·) is a group.

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2.Normal subgroups,quotient� groups

• Example 2.3.1. (Z_n, +) is the quotient group of (Z,+) by the subgroup nZ = {nz|z ∈ Z}.
• Solution. Since (Z,+) is abelian, every subgroup is normal. The set nZ can be verified to be a subgroup, and the relationship a ≡ b mod nZ is equivalent to a − b ∈ nZ and to n|a − b. Hence a ≡ b mod nZ is the same relation as a ≡ b mod n. Therefore, Z_n is the quotient group Z/nZ, where the operation on congruence classes is defined by [a] + [b] = [a + b].

(Z_n,+) is a cyclic group with 1 as a generator .When there is no confusion, we write the elements of Z_n as 0, 1, 2, 3, . . . ,

n − 1 instead of [0], [1], [2], [3], . . . , [n − 1].

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3.Homorphisms.

• 3.1.Definition of Homomorphisms
• 3.2.Examples of Homomorphisms
• 3.3.Theorem on Homomorphisms

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3.Homorphisms

• 3.1.Definition of Homomorphisms
• If (G, ・) and (H, * ) are two groups, the function f :G → H is called a group homomorphism if

f (a ・ b) = f (a) * f (b) for all a, b ∈ G.

• We often use the notation f : (G, ・) → (H, *) for such a homorphism. Many authors use morphism instead of homomorphism.
• A group isomorphism is a bijective group homomorphism. If there is an isomorphism between the groups (G, ・) and

(H, *), we say that (G, ・) and (H, *) are isomorphic and write (G, ・) ≅ (H, * ).

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3.Homorphisms

• 3.2.Examples of Homomorphisms

- The function f : Z → Z_n , defined by f (x) = [x] iz the group homomorphism.

- Let be R the group of all real numbers with operation addition, and let R⁺ be the group of all positive real numbers with operation multiplication. The function f : R → R⁺ , defined by f (x) = e^x , is a homomorphism, for if x, y ∈ R, then

f(x + y) = e^x+y = e^x e^y = f (x) f (y). Now f is an isomorphism, for its inverse function g :R⁺ → R is lnx. There-fore, the additive group R is isomorphic to the multiplicative group R⁺ . Note that the inverse function g is also an isomorphism: g(x y) = ln(x y) = lnx + lny = g(x) + g(y).

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3.Homorphisms

• 3.3.Theorem on Homomorphisms
• Proposition 3.3.1. Let f :G → H be a group morphism, and let e_G and e_H be the identities of G and H, respectively. Then

(i) f (e_G) = e_H .

(ii) f (a⁻¹) = f (a)−1 for all a ∈ G.

• Proof. (i) Since f is a morphism, f (e_G)f (e_G) = f (e_G e_G) = f (e_G) = f (e_G)e_H . Hence (i) follows by cancellation in H

(ii) f (a) f (a⁻¹) = f (a a⁻¹) = f (e_G) = e_H by (i). Hence f (a⁻¹) is the unique inverse of f (a); that is f (a⁻¹) = f (a)⁻¹

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3.Homorphisms

• If f :G → H is a group morphism, the kernel of f , denoted by Kerf, is defined to be the set of elements of G that are mapped by f to the identity of H. That is, Kerf ={g ∈ G|f (g) = e_H }
• Proposition 3.3.2. Let f :G → H be a group morphism. Then:

(i) Kerf is a normal subgroup of G.

(ii) f is injective if and only if Kerf = {e_G}.

• Proposition 3.3.3. For any group morphism f :G → H, the image of f , Imf ={f (g)|g ∈ G}, is a subgroup of H (although not necessarily normal).

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3.Homorphisms

• Theorem 3.3.4. Morphism Theorem for Groups. Let K be the kernel of the group morphism f :G → H. Then G/K is isomorphic to the image of f, and the isomorphism

ψ: G/K → Imf is defined by ψ(Kg) = f (g).

• This result is also known as the first isomorphism theorem.
• Proof. The function ψ is defined on a coset by using one particular element in the coset, so we have to check that ψ is well defined; that is, it does not matter which element we use. If Kg ^, = Kg, then g’ ≡ g mod K so g ^,g⁻¹ = k ∈ K = Kerf . Hence g ^, = kg and so

f (g ^,) = f (kg) = f (k)f(g) = e_Hf (g) = f (g).

Thus ψ is well defined on cosets.

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3.Homorphisms

• The function ψ is a morphism because

ψ(Kg₁Kg₂) = ψ(Kg₁g₂) = f (g₁g₂) = f (g₁)f (g₂) = ψ(Kg₁)ψ(Kg₂).

• If ψ(Kg) = e_H, then f (g) = e_H and g ∈ K. Hence the only element in the kernel of ψ is the identity coset K, and ψ is injective. Finally, Imψ = Imf ,by the definition of ψ. Therefore, ψ is the required isomorphism between G/K and Imf

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3.Homorphisms

• Example 3.3.1. Show that the quotient group R/Z is isomorphic to the circle group W = {e^iθ ∈ C | θ ∈ R }.

Solution. The set W consists of points on the circle of complex numbers of unit modulus, and forms a group under multiplication. Define the function

f :R → W by f (x) = e^2πix. This is a morphism from (R,+) to

(W, ·) because

f (x + y) = e^2πi(x+y) = e^2πix · e^2πiy = f (x) · f (y).

The morphism f is clearly surjective, and its kernel is {x ∈ R|e^2πix = 1} = Z.

Therefore, the morphism theorem implies that R/Z ≅ W.

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