Mathematical Logic. Set Theory deals with fundamental questions about the nature of math- ematics. • Set theory is a branch of mathematics. • Set theory includes a study of the infinite. • Set theory and logic a symbiotic relationship. Why we study Set Theory? • To build foundations for the other branches of mathematics. • To understand the fundamental sets of mathematics: N, Z, Q, R, C. 0.1 Ingredients of the language for Set Theory • Logical connectives: ∧ (and), ∨ (or), ¬ (not), → (implies), ↔ (iff); • Quantifiers: ∀ (forall) Universal Quantifier, ∃ (there exists) Existential Quantifiers; • Relational Symbols: ∈ and = . What do you mean by a Set? A set is a well defined class of objects. By a well define means that there exists a rule with the help of which it is possible to tell whether a given object belongs or does not belongs to the give collection. • Sets defined by their extent: Examples: {3, 5, 7}, {φ, φ}. • Sets described by their intent: Examples: {x ∈ N : x is prime}, {x ∈ R : x 2 > 5}. Set of Sets. If the element of a set are sets themselves, then such a set is said to be ’class of sets’ or a ’family of sets’. 1
Transcript
Mathematical Logic.
Set Theory deals with fundamental questions about the nature of math-ematics.
• Set theory is a branch of mathematics.
• Set theory includes a study of the infinite.
• Set theory and logic a symbiotic relationship.
Why we study Set Theory?
• To build foundations for the other branches of mathematics.
• To understand the fundamental sets of mathematics:
A set is a well defined class of objects. By a well define means that thereexists a rule with the help of which it is possible to tell whether a givenobject belongs or does not belongs to the give collection.
• Sets defined by their extent:Examples:
{3, 5, 7}, {φ, φ}.
• Sets described by their intent:Examples:
{x ∈ N : x is prime}, {x ∈ R : x2 > 5}.
Set of Sets. If the element of a set are sets themselves, then such a set issaid to be ’class of sets’ or a ’family of sets’.
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0.2 Axiom of Set Theory
• Axiom of Extension Tow sets are equal iff they have same members:
A = B if x ∈ A ⇔ x ∈ B.
• Axiom of Empty Set There is a set, φ with on element.
• Powerset Axiom If A is a set, there is a set P(A), whose membersare the subsets of A. Symbolically, P(A) = {T : T ⊆ A}. Obviously φand A are both element of P(a).
Singleton. A set consisting of a single element is called a singleton set.Subset of a set. If A,B are sets such that every element of A is also aelement of B, Then A is said to be subset of B. In other words, A is a subsetof B if x ∈ A ⇒ x ∈ B.Proper Subset.
Disjoint Set. If two sets A and B have no common element, then A andB are disjoint, or mutually exclusive.
Examples. Let A = {x : x2 + 1 = 0, x ∈ R}.Examples. Let A = {x : xis a straight line passing through three points on a circle.}.
Theorem 0.1. If a finite set A has n elements, then the power set of A has
2n element.
Proof. There are nCr subsets consisting r elements out of n elementsof set A, where r = 0, 1, . . . , n. Since A has only n elements then no subsetof A can have more then n elements. Hence P(A) = 1 +n C1 + . . . +n Cn =(1 + 1)n = 2n.
Theorem 0.2. If A is any set, then φ ⊆ A, i.e., then empty set is a subset
of every set.
Proof. Let φ & A. Then ∃ at least one element x such that x ∈ φ andx /∈ A. But ∀x, x /∈ φ, since φ is the null set. Therefore ∃ no element x suchthat x ∈ φ and x /∈ A. This contradict our initial assumption. Hence wemust have φ ⊆ A.
Theorem 0.3. If A and B are sets, then A = B iff A ⊆ B and B ⊆ A.
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Proof. If A = B, Then by the Axiom of Extension every element of Ais an element of B and every element of B is an element of A. Hence A ⊆ Band B ⊆ A.Conversely, if A ⊆ B and B ⊆ A, then every element of A is an elementof B and every element of B is an element of A. Hence by the Axiom ofExtension, A = B.
Theorem 0.4. If A ⊆ B and B ⊆ C, then A ⊆ C.
Proof. Let x be an arbitrary element of A. Since A ⊆ B, if we have x ∈A ⇒ x ∈ B. Also It is given that B ⊆ C. This implies that x ∈ B ⇒ x ∈ C.Hence x ∈ A ⇒ x ∈ c. Hence by the definition, A ⊆ C.
Union of sets. Let A and B be two sets. Then the union of A and Bis the set of all elements which are either in A or in B. Union of A and B isdenoted by A ∪ B.Mathematically,
A ∪ B = {x : x ∈ A or x ∈ B}.
Intersection of sets. Let A and B be two sets. Then the intersection of Aand B is the set of all elements which are common in A and B. Intersectionof A and B is denoted by A ∩ B.Mathematically,
A ∩ B = {x : x ∈ A and x ∈ B}.
Difference of sets. Let A and B be two sets. Then the difference of Aand B is the set of elements which belongs to A but not belongs to B. Thedifference between A and B is denoted by A − B, A \ B or A ∼ B.Mathematically,
A \ B = {x : x ∈ A and x ∈ Bc}.
Symmetric difference of two sets. Let A and B be two sets. Then thesymmetric difference of A and B is define as the union of A \ B and B \ Aand is denoted by A∆B.
Index Set. A non-empty set, I,(say) having the property that to eachof its element, i ∈ I, there is corresponds a set Ai. Such a set is known asindex set.Let A be the collection of sets such that to each member i of I there corre-sponds a member of Ai ∈ A. Then A is called an index family of sets withindex set I. Mathematically,
A = {Ai : i ∈ I}.
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Ordered Pair. Let A and B be two sets. Let a ∈ A and b ∈ B. Then (a, b)called an ordered pair.
Cartesian Product of sets. Let A and B be two sets, then the setof all distinct ordered pairs whose first coordinate in an element of A andsecond coordinate is an element of B is called the cartesian product of Aand B, and is denoted by A × B.Mathematically,
A × B = {(a, b) : a ∈ A, b ∈ B}.
Relation. Let A and B be two sets. Then a relation from A to B is asubset of A × B.Mathematically, R is a relation from A to B iff R ⊆ A×B. Example. LetZ be a set of all integers. The statement ”x is less than y”, where x, y ∈ Idetermines a relation in I. If we denote this relation by R, then
R = {(x, y) : x, y ∈ I, x < y}.
Equivalence Relation. Let R be a relation in a set A. Then R is anequivalence relation in A iff
• R is reflexive, i.e., ∀a ∈ A, aRa.
• R is symmetric, i.e., aRb ⇒ bRa.
• R is transitive, i.e., aRb and bRc ⇒ aRc.
0.3 Definition and Properties.
Binary Operations. A binary operation ∗ on a set is a rule that assignsto each ordered pair of elements of the set some element of the set.
More precisely, a binary operation on a set S is a binary function from Sand S to S, in other words a function f from the Cartesian product S×S toS. Sometimes, the term is used for any binary function. That f takes valuesin the same set S that provides its arguments is the property of closure.Binary operations are the keystone of algebraic structures studied in ab-stract algebra: they form part of groups, monoids, semigroups, rings, andmore. Many binary operations of interest in both algebra and formal logicare commutative or associative. Many also have identity elements and in-verse elements. Typical examples of binary operations are the addition (+)and multiplication (∗) of numbers and matrices as well as composition of
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functions on a single set.
Set A is said to be close under operation ∗, if (a ∗ b) ∈ G for all a, b ∈ G.Binary operations are often written using infix notation such as a∗b, a+b, ora · b rather than by functional notation of the form f(a, b). Sometimes theyare even written just by juxtaposition(combination): ab. Powers are usuallyalso written without operator, but with the second argument as superscript.
Commutative. A binary operation ∗ is called commutative if and onlyif a ∗ b = b ∗ a, for all a, b ∈ S.
Examples of operations that are not commutative are subtraction (−),division (/), and exponentiation.
The well-known binary operations of addition (+) and multiplication (∗)on the real numbers are commutative because a + b = b+ a and a ∗ b = b ∗ afor all real numbers.For example 8 + 3 = 11 = 3 + 8 and 5 ∗ 9 = 45 = 9 ∗ 5. An example ofa binary operation which is not commutative is exponentiation on the realnumbers, because in general ab ± ba, for instance 23 = 2 ∗ 2 ∗ 2 = 8 while32 = 3 ∗ 3 = 9.
Mathematically, A map or binary operation f : A × A → B from a setA to a set B is said to be commutative if,
f(x, y) = f(y, x) ∀x, y ∈ A.
Every element x belonging to the domain A of a binary operation ∗ com-mutes with itself. If a particular pair of elements a and b satisfy a ∗ b = b ∗aunder a certain binary operation then it is said that the two elements com-mute under that operation.A binary operation which is not commutative is called noncommutative.subtraction, division, exponentiation, matrix multiplication and functioncomposition fog are the example of noncommutative binary operations.
Example. Matrix multiplication is non-commutative
A =
[
0 12 3
]
and B =
[
1 23 4
]
;
Then,
AB =
[
3 411 16
]
6=
[
4 78 15
]
= BA.
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Associative. A binary operation ∗ is called associative iff (a∗b)∗c = a∗(b∗c)for all a, b, c ∈ S.
Important: Remember that changing the order of operations does notinvolve or permit changing the actual operations themselves by moving theoperands around within the expression.Why is it we need associativity to solve 3x=11 in the reals? Well, the inverseof 3 is 1/3 and so 1/3(3x) =11/3 and now we use associativity to rewritethis as (3/3) x = 11/3 etc.Sets that have an associative binary operation are known as semigroups. Inmany practical applications of studying binary operations on sets it is notunusual to discover they are associative but it is something that cannot beassumed. Indeed, one commonly known operation, the cross product onthree dimensional real vectors is not associative.
Identity Element. Let ∗ : A × A → A be a binary operation on A. Aelement e ∈ A is called an identity element for the operation ∗ if
e ∗ a = a = a ∗ e ∀a ∈ A.
Inverse Element. Let ∗ : A × A → A be a binary operation on A. Aelement a of a set A is said to be inversible for binary operation ∗ withidentity element e if there exists b ∈ A. such that
a ∗ b = e = b ∗ a.
b = a−1 is said to be inverse of a.
e ∗ a = a = a ∗ e ∀a ∈ A.
0.4 Group.
Groupoid. A groupoid(magma) is a basic kind of algebraic structure.Specifically, a magma consists of a set A equipped with a single binaryoperation A × A → A. A binary operation is closed by definition, but noother axioms are imposed on the operation.Groupoids are often used to capture information about geometrical objectssuch as manifolds. Groupoids were first developed by Heinrich Brandt in1926.
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Example. The algebraic structure (N,+) is a groupoid.Example. The algebraic structure (N,−) is not a groupoid.Semigroup. A semigroup is an algebraic structure consisting of a set Sclosed under an associative binary operation. In other words, a semigroupis an associative magma.Mathematically, Let (A, ∗) be an algebraic system where ∗ is a binary oper-ation on A. Then (A, ∗) ia called semigroup if the following conditions aresatisfied:
• ∗ is closed operation,
• for every choice of elements a, b, c ∈ G;
(a ∗ b) ∗ c = a ∗ (b ∗ c).
Example. Let Z+ be the set of positive integers, and + (addition) be theset operation on it. Then (Z+,+) is a semigroup.
Solution.
• Closure: If x1 and x2 are integers then x1 + x2 is an positive integer.
• Associativity: If x1, x2, and x3 are integers, then (x1 + x2) + x3 =x1 + (x2 + x3).
Monoid Let (A, ∗) be an algebraic system, where ∗ is a binary operationon A. Then (A, ∗) ia called monoid if the following conditions are satisfied:
• ∗ is closed operation,
• for every choice of elements a, b, c ∈ G;
(a ∗ b) ∗ c = a ∗ (b ∗ c).
• there exists an identity element e ∈ S; such that
a ∗ e = a = e ∗ a∀a ∈ S.
Example. Let Z be the set of integers, and + (addition) be the set opera-tion on it. Then (Z,+) is a monoid.
Solution.
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• Closure: If x1 and x2 are integers then x1 + x2 is an integer.
• Associativity: If x1, x2, and x3 are integers, then (x1 + x2) + x3 =x1 + (x2 + x3).
• Identity element: 0 is an integer and for any integer x, 0 + x =a + 0 = x.
Group. A semigroup with an identity in which every element has an inverseis termed as group. Mathematically, Let (G, ∗) be an algebraic system, where∗ is a binary operation on G. Then (G, ∗) ia called group if the followingconditions are satisfied:
• ∗ is closed operation,
• for every choice of elements a, b, c ∈ G;
(a ∗ b) ∗ c = a ∗ (b ∗ c).
• there exists an identity element e ∈ S; such that
a ∗ e = a = e ∗ a∀a ∈ S.
• every element of a ∈ S has an inverse in S, i.e. there is an elementb ∈ S such that
a ∗ b = e = b ∗ a.
This element b is usually denoted by a−1.
Example. A familiar group is the group of integers under addition. Let Zbe the set of integers, and let ”+” indicate the operation of addition. Then(Z,+) is a group.
Solution.
• Closure: If x1 and x2 are integers then x1 + x2 is an integer.
• Associativity: If x1, x2, and x3 are integers, then (x1 + x2) + x3 =x1 + (x2 + x3).
• Identity element: 0 is an integer and for any integer x, 0 + x =a + 0 = x.
• Inverse elements: If x is an integer, then the integer −a satisfies theinverse rules: x + (−x) = (−x) + x = 0.
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Example.
Many of the structures investigated in mathematics turn out to be groups.These include familiar number systems, such as the integers, the rationalnumbers, the real numbers, and the complex numbers under addition, aswell as the non-zero rationales, reals, and complex numbers, under multi-plication. Other important examples are the group of non-singular matricesunder multiplication and the group of invertible functions under composi-tion. Group theory allows for the properties of such structures to be inves-tigated in a general setting.
0.5 Properties of Groups.
Uniqueness of identity.
Lemma 0.5. A group G has exactly one identity element ‘e’ satisfying ex =x = xe for all x ∈ G.
Proof. Suppose that f is an element of G with property that fx = xfor all element x of G. The in particular f = fe = e. similarly one can showthat ‘e’ is the only element of G satisfying xe = x for all elements x of G.
Uniqueness of inverse.
Lemma 0.6. An element x of a group G has exactly one inverse x−1.
Proof. we know from the axioms that the group G contains at least oneelement x−1 which satisfies xx−1 = e and x−1x = e. If z is any element ofG which satisfy xz = e then z = ez = (x−1x)z = x−1(xz) = x−1e = x−1.Similarly, if w is any element of G which satisfies wx = e the w = x−1. Hencewe conclude that the inverse x−1 of x is uniquely determined, as required.
Lemma 0.7. Let x and y be elements of a group G. Then (xy)−1 = y−1x−1.
Similarly, (y−1x−1)(xy) = e, and thus (y−1x−1) is the inverse of xy, as re-quired.Note that (x−1)−1 = x for all elements x of a group G, since x has the
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properties that characterize the inverse of the inverse x−1 of x.
Abelian group. A group G is abelian if its binary operation is com-mutative.Example. (Z,+) is abelian group.Solution. (Z,+) is a group. As we shown earlier. Also it is abelian sincex1 + x2 = x2 + x1 for all x1, x2 ∈ Z.
Order of a group. The order of a group G, usually denoted by | G |or occasionally by o(G), is the number of elements in the set G. If the orderis not finite, then the group is an infinite group, denoted | G |= ∞.
Example. A set G = {0o, 120o, 240o}, with the binary operation ∗ whichis define as the rotation of a line at 120o in the plane, is a group (G, ∗), oforder 3.
Example. For each positive integer n the set of all nonsingular n × nmatrices is a group, where the group operation is matrix multiplication.These groups are not Abelian when n ≥ 2.
Integral power of element of a group. Let a ∈ G, and ‘∗′ multi-plicative binary operation. Then be closure property a, a ∗ a, a ∗ a ∗ a, . . . ,are all element of G.If n is a positive integers, we define an = a∗a∗ . . .∗a to n factors. Obviouslyan ∈ G.If e is the identity element of G, then we define
a0 = e.
If n is a positive integers, then −n is a negative integer.
a−n = (an)−1, (an)−1 ∈ G.
Thus we have define an, for all n ∈ Z.
a−n = (an)−1 = (a ∗ × ∗ a, n times)−1
= (a−1)(a−1) . . . (a−1) = (a−1)n.
Integral multiplication of element of a group. Let a ∈ G, and ‘+′
additive binary operation. Then be closure property a, a + a, a + a + a, . . . ,are all element of G.
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In place of a0 we write 0a. Thus we define 0a = e where e is the identity ofG.If n is a positive integer, then in place of a−n we write (−n)a. Thus wedefine (−n)a = −(na), where −(na) denotes the inverse of na ∈ G.
order of an element. The order of an element a in a group G is theleast positive integer n such that an = e, where an is composition dependingon the group operator on a by itself n times. If no such n exists, then theorder of a is said to be infinity.
Example. A set G = {0o, 120o, 240o}, with the binary operation ∗ whichis define as the rotation of a line in the plane, is a group, (G, ∗). Then orderof 0o is three, 120o is two and 240o is one.
0.6 Subgroup.
Let (G, ∗) be a group with binary operator ∗, and let H be a non emptysubset of G. Then we say that H is a subgroup of G if the following aresatisfied:
•a ∗ b ∈ H, ∀a, b ∈ H.
• the identity element of G is an element of H.
• the inverse of any element of H is itself an element of H.
Then H is itself a group.Example. Let G = {1,−1, i,−i} be a set. Then (G, ∗) is a group, where ∗is multiplication operation. Then H = {1,−1} is a subgroup of G.
Example. Let (Z,+) is a group, where + is addition operation. Then(A,+) is a subgroup of G, where A = {2n : n ∈ Z}.
Every group of G has a subgroups G improper subgroup itself and {e}trivial subgroup, where e is the identity element of G. A subgroup H of G issaid to be proper if H 6= G.Following lemma help us to show that a subset of a group is a subgroup.
Lemma 0.8. Let (G, ∗) be a group and then a subset H of a set G is a
subgroup of G iff
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• H is closed under the binary operation of G;
• the identity e of G is in H;
• for all x ∈ H it is true that x−1 ∈ H also.
Proof. Let H be a subgroup of G. Then by the definition of subgroupH is closed under the binary operation, must contains the identity elementin H and inverse element of every element of H must be in H.Conversely, Suppose H is a subset of a group G, such that all the conditionsgiven in lemma are true. For being a subgroup of G, associativity must holdalong with these given conditions. For this let x1, x2, x3 ∈ H, since H ⊆ G,we have x1, x2, x3 ∈ G, and we know that
x1 ∗ (x2 ∗ x3) = (x1 ∗ x2) ∗ x3 ∀ x1, x2, x3 ∈ G.
This implies
x1 ∗ (x2 ∗ x3) = (x1 ∗ x2) ∗ x3 ∀ x1, x2, x3 ∈ H.
This completes the proof of theorem.
Lemma 0.9. Let (G, ∗) be a group and then a non empty subset H of a set
G is a subgroup of G iff
a ∗ b−1 ∈ H, ∀a, b ∈ H.
Proof. The condition is necessary. Suppose H is a subgroup of G.Let a ∈ H, b ∈ H. Now each element of H must possess inverse because His itself a group.
i.e., b ∈ H ⇒ b−1 ∈ H.
The condition is sufficient. It is given
a ∗ b−1 ∈ H, ∀a, b ∈ H.
We have to prove that H is subgroup of G.
Existence of Identity. We have
a ∈ H,a ∈ H ⇒ a ∗ a−1 ∈ H ⇒ e ∈ H.
Thus identity e is an element of H.Existence of Inverse. We have
e ∈ H,a ∈ H ⇒ e ∗ a−1 ∈ H ⇒ a−1 ∈ H.
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Thus each element of H possesses inverse in H.
Closure Property. Let b ∈ H ⇒ b−1 ∈ H. Therefore applying thegiven condition, we have
a ∈ H, b−1 ∈ H ⇒ a ∗ (b−1)−1 ∈ H ⇒ a ∗ b ∈ H.
Associativity. The element of H are also element of G. Therefore associa-tivity must hold for H.
Lemma 0.10. Let (G, ∗) be a group and let H and K be subgroups of a
group G. Then H ∩ K is also a subgroup of G.
Proof. The identity element of G belongs to H ∩ K, since it belongsto both the subgroups H and K. If x and y are elements of H ∩ K, impliesthat x and y elements of both subgroups H and K. Then xy is an elementof both subgroups H and K, and therefore xy is an element of H ∩K. Alsothe inverse x−1 of an element x of H ∩K belongs to both subgroups H andK and thus belongs to H ∩ K.
More generally, the intersection of any collection of a subgroups of agiven group is itself a subgroup of that group.
Subgroup generated by x. Let (G, ∗) be a group and x be an elementof G. The subgroup generated by ‘x’ is the subgroup consisting of all ele-ments of G that are of the form xn for some integer n.
0.7 Cyclic Group.
A Group (G, ∗) is said to be cyclic group, with generator ′x′, if every elementof G is of the form xn for some integer n, and denote by < x > .
Example. The group Z of integers under addition is a cyclic groupgenerated by 1.
Example. The group of all rotations of the plane about the originthrough an integer multiple of 2π/n radians is a cyclic group of order n for
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all integers n. This group is generated by an anticlockwise rotation throughan angle of 2π/n radians.
Cyclic Subgroup. If (G, ∗) be a group and x ∈ G, then
H = {xn | n ∈ Z}
is a subgroup of G. This group is called cyclic subgroup of G generated by‘x’.
Theorem 0.11. Every cyclic group is abelian.
Proof. Let G be a cyclic group and let x be a generator of G so that
G =< x >= {xn | n ∈ Z}.
If x1 and x2 are any two elements of G, there exist integers r and s suchthat x1 = xr and x2 = xs. Then
x1 ∗ x2 = xrxs = xr+s = xs+r = x2 ∗ x1.
This completes the proof.
Lemma 0.12. Let (G, ∗) be a group and H be a subset of G. If H is a finite
set, Then (H, ∗) is a subgroup of (G, ∗), if ∗ is a closed operation on H.
Proof. Let x be an element of H. If ∗ is a close operation on H, theelements x, x2, x3, . . . are all in H. Because H is a finite set, we have xi = xj
for some i and j, i < j, i.e., xi = xi ∗ xj−i. It follows that xj−i is theidentity of (G, ∗), and it is included in the subset H. If j − i > 1, accordingto xj−i = x ∗ xj−i−1, we can conclude that xj−i−1 is inverse of x, and it isincluded in the subset of H. If j− i = 1, we have xi = xi ∗x. Thus x must bethe identity element and is its own inverse. Consequently, that ∗ is a closedoperation on H guarantees that (H, ∗) is a subgroup.
Theorem 0.13. Let (G, ∗) be a group and x ∈ G, then
H = {xn | n ∈ Z}
is a subgroup of G and is the smallest subgroup of G that contains x, that is
every subgroup containing ‘x′ contain H.
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Proof. To prove H be a subgroup of G, we must show that it is closeunder group operation of G, must have identity element and must containsinverse element of every element in H.Since xrxs = xr+s ∈ Z, ∀xr, xs ∈ Z. Thus H is closed under the groupoperation. Also x0 = e, so e ∈ H, and for xr ∈ H, x−r ∈ H and xrx−r = e.Hence H satisfied all the necessary and sufficient conditions for being a sub-group.Also every subgroup of G containing ‘x’ must contain H. This implies H bethe smallest subgroup of G containing ‘x′.
Lemma 0.14. Let (G, ∗) be a group and x be an element of G. Then the set
of all elements of G that are of the form xn for some integer n is a subgroup
of G.
Proof. Let H = {xn : n ∈ Z}. Then the identity element belongs to H,since it is equal to x0. The product of two elements of H is itself an elementof H, since xmxn = xm+n for al integers m and n. Also the inverse of anelement of H is itself an element of H since (xn)−1 = x−n for all integers n.Thus H is a subgroup of G.
0.8 Symmetric Group
Let En denote n−dimensional Euclidean space, so that E2 denotes theEuclidean plane, and E3 denotes three dimensional Euclidean space. Let Xbe a subset of En. Then a symmetry of X is a transformation T : En → En
of En which sends straight lines to straight lines, preserves all lengths andangles, and has the property that T (X) = X. The collection of all symme-tries of a geometric figure is a group, the symmetry group on X, the groupoperation being that of composition of transformations.Mathematically, Let X be any non empty set. A group arises from the setSX of all one-to-one mapping of X on to X, called the symmetry group on
X. this group is usually describe as
• SX is the set of all mappings of the non empty set X with itself.
• If σ, τ ∈ SX then we define στ to be the composition of the mappingσ and τ. Here we must verify that στ is a mappings of X with itself.Suppose x ∈ X; then as τ is onto, we can find x′′ ∈ X such thatx′′ = x. But σ is also onto, so we can find x′ ∈ X such that x′σ = x′′.
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Consequentlyx′(στ) = (x′σ)τ = x′′τ = x
and hence στ is onto. If x(στ) = y(στ), then (xσ)τ = y(στ) byhe definition of composition of mappings; this means τ is one-to-one,that xσ = yσ. This in turn implies, since σ is one-to-one, that x = y.Therefore στ is also one-to-one. Thus composition of mapping is abinary operation in SX .
• The identity mapping e : x → x is in SX and is an identity element ofSX .
• The groupoid (SX , ·) is a semigroup since composition of mappings isassociative.
• Let σ ∈ SX . Since σ is one-to-one and onto, then it implies σ has aninverse, τ in MX . Now, στ = e = τσ means σ is an inverse of τ. Weknow the only elements in MX which have inverse are those which areone-to-one and onto mapping. Therefore τ ∈ SX and τ is the requiredinverse of σ. It completes the proof.
Example. The symmetries of a rectangle that is not a square constitutea group of order 4. This group consists of the identity transformation, re-flection in the axis of symmetry joining the midpoints of the shorter side,reflection in the axis of symmetry joining the two longer sides, and rotationthough an angle of radian (180o). If I denote the identity transformation, Aand B denote the reflections in the two axes of symmetry, and C denotes therotation through π radians then A2 = B2 = C2 = I, AB = BA = C, AC =CA = B and BC = CB = A. This group is Abelian.
Permutation. An element of SX is a permutation of X.In particular case when X = {1, 2, . . . , n}, we write SX = Sn, and Sn iscalled the symmetric group of degree n. The number of element in Sn is| Sn |= n!, since there are n! different permutation of n elements.
Permutation Multiplication. It is natural binary operation, and de-fine on the permutations of a set. Let G be a set, and let σ and τ be thepermutations of G so that σ and τ are both one-to-one functions mappingG onto G.
16
Example. Let G = {1, 2, 3, 4, 5} and let σ and τ be two permutation ofA. Then
σ =
(
1 2 3 4 54 2 5 3 1
)
, τ =
(
1 2 3 4 53 5 4 2 1
)
.
Then,
στ =
(
1 2 3 4 54 2 5 3 1
)(
1 2 3 4 53 5 4 2 1
)
=
(
1 2 3 4 52 5 1 4 3
)
.
Lemma 0.15. The composite function στ will be a permutation if it is
one-to-one and onto G.
Proof. Let x1, x2, . . . , xn ∈ G. To show composite function στ on G isone-to-one. Let x1(στ) = x2(στ), then
(x1σ)τ = (x2σ)τ,
and since τ is one-to-one, we know that x1σ = x2σ. Since σ is also one-to-one, this implies that x1 = x2. Hence στ is one-to-one.To show thatστ is onto A, let x ∈ A. Since τ is onto A, there exists x′ ∈ Asuch that x′τ = x. Since σ is onto A, there exists x′′ ∈ A such that x′ = x′′σ.Then
x = x′τ = (x′′σ)τ = x′′(στ),
so στ is onto G.Equality of Two Permutations. Two permutations f and g of degree nare said to be equal if we have f(a) = g(a) ∀a ∈ SX .
Example. Let
f =
(
1 2 3 42 3 4 1
)
and g =
(
3 2 4 14 3 1 2
)
are two permutations of degree 4, then we have f = g. Hence we see thatboth f(a) = g(a) ∀a ∈ {1, 2, 3, 4}.
Theorem 0.16. Let G be a non empty set, and let SG be a collection of all
permutations of G. Then SG is a group under permutation multiplication.
Proof. To show that SG is a group. We have to show that SG satisfiesthe three axioms of group. Existence of identity and inverse is very simplyto prove. Now we have to show that function composition is associative, i.e.,
(στ)µ = σ(τµ),
17
for permutations σ, τ and µ of G. for this we have to show that each com-posite function maps each x ∈ G into the same image in G, i.e.,
x[(στ)µ] = x[σ(τµ)] ∀x ∈ G.
We havex[(στ)µ] = x[(στ)]µ = [(xσ)τ ]µ = x[σ(τµ).
Thus, (στ)µ and σ(τµ) map each x ∈ G into the same element [(aσ)τ ]µ andhence the same permutation.
Cycles and Products of Cycles. A permutation σ of a set G is acycle of length n if there exist x1, x2, . . . , xn ∈ G such that
x1 → x2, x2 → x3, . . . , xn → x1
and x → x for x ∈ G but x /∈ {x1, x2, . . . , xn}.Example.
f =
(
1 2 5 3 4 62 4 5 1 6 3
)
it can be represented by the cycle (1 2 4 3).Note. A cycle does not change by changing the place of its elements pro-vided their cyclic order is not changed. The cycles in a collection of cyclesare disjoint if there is no common between any two cycles. Multiplicationof disjoint cycles is commutative.Note that neither every permutation is not a cycle nor is the product of anytwo cycle is necessarily a cycle.In general, a cycle of length one is denoted by identity element, and a cycleof length two is called a transposition.Inverse of a permutation.Inverse of a cyclic permutation.The inverse of the cycle a = (x1, x2, . . . , xn) is the cycle b = (xn, xn−1, . . . , x1),since
xi(ab) = (xia)b = xi+1b = xi ∀i 6= n,
whilexn(ab) = (xna)b = x1b = xn.
Inver of a product of cyclic permutation. If f and g are any two cycles, thenwe have
(fg)−1 = g−1f−1.
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Also, if f, g, h are three cycles, then
(fgh)−1 = h−1g−1f−1.
If f and g are disjoint cycles then (gf)−1 = (fg)−1 = f−1g−1.Example.
Even or Odd Permutation. A permutation of a finite set is even andodd according to whether it can be expressed as the product of an evennumber of transposition or the product of an odd number of transpositions,respectively.Alternating Group. The subgroup of Sn consisting of the even permuta-tions of n letters is the alternating group An on n letters.
0.9 Cosets and Lagrange’s Theorem.
Let (G, ∗) be a group and H be a subgroup of G. A left coset of H in G isa subset of G that is of the form xH, where x ∈ G and
xH = {y ∈ G : y = xh for some h ∈ H}.
Similarly a right coset of H in G is a subset of G that is of the form Hx,where x ∈ G and
Hx = {y ∈ G : y = hx for some h ∈ H}.
Note that a subgroup H of a group G is itself a left coset of H in G.
Lemma 0.17. Let (G, ∗) be a group and H be a subgroup of G. Then the
left cosets of H in G have the following properties:
• x ∈ xH for all x ∈ G;
• if x and y are elements of G, and if y = xa for some a ∈ H, then
xH = yH;
• if x and y are elements of G, and if xH ∩ yH is non empty then
xH = yH.
19
Proof. Let x ∈ G. Then x = xe, where e is the identity element of G.Also, e ∈ H. It follows that x ∈ xH. This prove the first condition.Let x, y ∈ G, where y = xa for some a ∈ H. Then yh = (xa)h = x(ah) andxh = y(a−1h) for all h ∈ H. Moreover ah ∈ H and a−1h ∈ H for all h ∈ H,since H is a subgroup of G. It follows that yH ⊂ xH and xH ⊂ yH, andhence xH = yH. This completes the prove of second condition.Finally suppose that xH ∩ yH is a non empty for some elements x and yof G. Let z be an element of xH ∩ yH. Then z = xa for some a ∈ H, andz = yb for some b ∈ H. It follows from second condition that zH = xH andzH = yH. therefore xH = yH. This completes the proof.
Lemma 0.18. Let (G, ∗) be a group and H be a subset of G. Then each left
coset of H in G has the same number of elements as H.
Proof. Let H = {h1, h2, . . . , hn}, where h1, h2, . . . , hn are distinct, andlet x be an element of G. Then the left coset xH consists of the elementsxhj for j = 1, 2 . . . , n. Suppose that j and k are integers between 1 andn for which xhj = xhk. Then hj = x−1(xhj) = x−1(xhk) = hk, andthus j = k, since h1, h2, . . . , hn are distinct. It follows that the elementsxh1, xh2, . . . , xhn are distinct. We conclude that the subgroup H and theleft coset xH both have n elements, as required.
Lemma 0.19. Let xH and yH be two cosets of H. Either xH and yH are
disjoint or they are identical.
Proof. Let H be a subgroup of a group G. Let xH and yH are two leftcosets of H in G. Let xH and yH are not disjoint, and have f as a commonelement, i.e., there exists h1 and h2 in H such that f = xh1 = yh2. Then
xh1 = xh2
⇒ xh1h−1
1= yh2h
−1
1
⇒ xe = y(h2h−1
1)
⇒ x = y(h2h−1
1)
Since H be a subgroup, therefore h2h−1
1∈ H. Let h2h
−1
1= h3. Then x = yh3.
Now xH = yh3H = y(h3H) = yH. Since h3 ∈ H. This implies h3H = H.therefore left cosets are identical if they are not disjoint. In the similar waywe can show that any two right coset Hy and Hx are identical if they arenot disjoint.
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Lemma 0.20. Let (G, ∗) be a group and H be a subgroup of G. Let a, b ∈ G.Then
a ∈ Hb ⇔ Ha = Hb and a ∈ bH ⇔ aH = bH.
Proof. Let a ∈ Hb, this implies
ab−1 ∈ Hbb−1
⇒ ab−1 ∈ H
⇒ Hab−1 = H
⇒ Hab−1b = Hb
⇒ Ha = Hb.
Conversely, let Ha = Hb. If a ∈ Ha, therefore a ∈ Hb.
Theorem 0.21. Let H be any subgroup of G and let h ∈ H. Then Hh =H = hH.
Proof. Let h ∈ H, then to prove Hh = H. Suppose h′ is an arbitraryelement of H. Then h′h is an arbitrary element of Hh. Since H is a subgroup,we have
h′ ∈ H,h ∈ H ⇒ h′h ∈ H, By closure property.
Thus every element of Hh is also an element of H. Hence Hh ⊆ H.Again,
h′ = h′(h−1h) = (h′h−1)h ∈ Hh.
Since h ∈ H ⇒ h−1 ∈ H and h′ ∈ H,h−1 ∈ H ⇒ h′h−1 ∈ H.Thus every element h′ of H is also an element of Hh. Hence H ⊆ Hh.Hence Hh = H. In the very similar way we can prove H = hH, and henceHh = H = hH.
Theorem 0.22. Lagrange Theorem Let G be a finite group, and let H b
a subgroup of G. Then the order of H divides the order of G.
Proof. Let G be a group of finite order n. Let H be a subgroup of Gand let o(H) = m. Suppose h1, h2, . . . , hm are the m members of H.Let a ∈ G. Then Ha is the right coset of H in G and we have
Ha = {h1a, h2a, . . . , hma}
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Ha has m distinct members, since hia = hja ⇒ hi = hj . Therefore eachright coset of H in G has m distinct members. Also any two right cosetsof H in G are disjoint. Since G is a finite group, the number of distinctright cosets of H in G will be finite, say, equal to k. Then union of thesek distinct right cosets of H in G is equal to G. Thus, if Ha1,Ha2, . . . ,Hak
are k distinct right cosets of H in G. Then
G = Ha1 ∪ Ha2 ∪ . . . ∪ Hak
⇒ o(G) = km ⇒ n = km
⇒ k =n
m⇒ o(H) is a divisor of o(G).
Index. Let G be a group, and let H be a subgroup of G. If the number ofleft cosets of H in G is finite then the number of such cosets is referred asthe index of H in G, denoted by [G : H].
The proof of Lagrange’s Theorem shows that the index [G : H] of asubgroup H of a finite group G is given be [G : H] =| G | \ | H | .
Corollary 0.23. Any finite group of prime order is cyclic.
Proof. Let G be a group of prime order, and let x become element of Gthat is not the identity element. Then the order of x is greater than one anddivides the order of G. But then the order of x must be equal to the orderof G, since the latter i a prime number. Thus G is a cyclic group generatedby x.
0.10 Homomorphism.
A homomorphism f : G → K from a group (G, ∗) to a group (K, ∗) is afunction with property that f(x1 ∗ x2) = f(x1)f(x2) for all x1, x2 ∈ G.
Example. Let q be an integer. The function from the group Z of inte-gers to itself that sends each integer n to qn is a homomorphism.
Example. Identity mapping f : G → G, is a homomorphism.
Example. Let x be an element of a group G. The function that sendeach integer n to the element xn is a homomorphism from the group Z ofintegers to G, since xm+n = xmxn for all integers m and n.
22
Kernel. The kernel, ker f of the homomorphism f : G → K is the setof all elements of G that mapped by f onto the identity element of K.
ker f = {x ∈ G : f(x) = e′,where e′ be the identity element of K..}
Example. Let the group operation on the set {+1,−1} be the multipli-cation, and let f : Z → {+1,−1} be the homomorphism that sand eachinteger n to (−1)n. then the kernel of homomorphism f is the subgroup ofZ consisting of all even numbers.
0.11 Isomorphism.
An algebraic systems (G, ∗) is isomorphic to the algebraic systems (K, o),if one can obtain (G, ∗) from (K, o) by renaming the elements and/or theoperation in (K, o.)Mathematically, An isomorphism f : G → K between groups G and K is aone-to-one function f mapping from G on to K, such that for all x and y inG,
f(x ∗ y) = f(x)of(y).
The groups G and K are the isomorphic. The usual notation for isomor-phism is ≃ .Two groups G and K are isomorphic if there exists an isomorphism mappingG on to K.
Example. Let R be the group of real numbers with the operation ofaddition, and let R+ be the group of strictly positive real numbers with theoperation of multiplication. the function exp : R → R+ that sends each realnumber x to the positive real number ex is an isomorphism: it is both ahomomorphism of groups and a bijection. The inverse of this isomorphismis the function log : R+ → R that sends each strictly positive real numberto its natural logarithm.
Lemma 0.24. Let f : G → K is an isomorphism of G with K, and e is the
identity of G, then the f -image of an element a ∈ G is the inverse of the
f -image of a i.e., f(a−1) = [f(a)]−1.
Proof. Let e be the identity element of G and e′ be the identity elementof K. Then f(e) = e′. Now let a ∈ G, then a−1 ∈ G and aa−1 = e ∈ G. Wehave
e′ = f(e) = f(aa−1) = f(a)f(a−1).
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Therefore f(a−1) is the inverse of f(a) in K. Thus
f(a−1) = [f(a)]−1.
Lemma 0.25. Let f : G → K is an isomorphism of G with K. Then the
order of an element a ∈ G is equal to the order of its f -image f(a).
Proof. Let e be the identity element of G and e′ be the identity elementof K. Let order of an element a be finite and let it is equal to n, and letorder of f(a) is m. Then
an = e
⇒ f(an) = f(e)
⇒ f(a . . . a, n times) = f(e)
⇒ f(a) . . . f(a), n times = f(e)
⇒ [f(a)]n = f(e)
⇒ o(f(a)) ≤ n. (1)
Now,
[f(a)]m = f(e)
⇒ f(a) . . . f(a),m times = f(e)
⇒ f(a . . . a,m times) = f(e)
⇒ f(am) = f(e)
⇒ am = e
⇒ o(a) ≤ m. (2)
From eq.(1) and eq.(2) we have m = n.
Lemma 0.26. If f : G → K is an isomorphism of G with K, and e is the
identity of G, then ef is the identity in K. Also,
a−1f = (af)−1 ∀a ∈ G.
Proof. Let x′ ∈ K. Since f is onto, there is x ∈ G such that xf = x′.Then
x′ = xf = (ex)f = (ef)(xf) = (ef)x′.
Similarly,x′ = xf = (xe)f = (xf)(ef) = x′(ef).
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Thus for every x′ ∈ K, we have
(ef)x′ = x′x′(ef),
so ef is the identity of K. Also for a ∈ G, we have
ef = (a−1a)f = (a−1f)(af).
Similarly,ef = (aa−1)f = (af)(a−1f).
Thus a−1f = (af)−1.
0.12 Automorphism and Normal Subgroup.
Automorphism. An automorphism is an isomorphism mapping a grouponto itself.Inner Automorphism. For each x ∈ G, the mapping ix : G → G given byyix = x−1yx is an automorphism of G, the inner automorphism of G underconjugation by G.
Let A and B be subsets of a group G. then product AB of the sets Aand B is defined as
AB = {xy : x ∈ A, y ∈ B}.
We denote {x}A and A{x} by xA and Ax, for all elements x of G and sub-sets A of G. The Associative Law for multiplication of elements G ensuresthat (AB)C = A(BC) for all subsets A,B and C of G. We can therefore usethe notation ABC to denote the products (AB)C and A(BC); and we canuse analogous notation to denote the product of four or more subsets of G.If A,B and C are subsets of a group G, and if A ⊂ B then clearly AC ⊂ BCand CA ⊂ C.Note that if H is a subgroup of group G and if x is an element of G thenxH is the left coset of H in G that contains the element x. Similarly Hx isthe right coset of H in G that contains the element x.If H is a subgroup of G then HH = H. Indeed HH ⊂ H, since the productof two elements of a subgroup H is itself an element of H. Also H ⊂ HHsince h = eh for any element h of H, where e, the identity element of G,belongs to H.
Normal Subgroup. A subgroup N of a group G is said to be a normal
subgroup of G if xnx−1 ∈ N for all n ∈ N and x ∈ G.
25
The notation ‘N ⊳ G’ signifies ‘N is a normal subgroup of G’.
A non trivial group G is said to be simple if the only normal subgroupsof G are the whole of G and the trivial subgroup {e} whose only element isthe identity element e of G.
Lemma 0.27. Every subgroup of an abelian group is a normal subgroup.
Proof. Let N be a subgroup of an abelian group G. Then
xnx−1 = (xn)x−1 = (nx)x−1 = n(xx−1) = ne = n ∀n ∈ N, x ∈ G.
Thus x ∈ G,h ∈ N ⇒ xhx−1 ∈ N. Hence N is normal subgroup.
Example. Let S3 be the group of permutations of the set {1, 2, 3},and let H be the subgroup of S3 consisting of the identity permutation andthe transposition (12). Then H is not normal in G, since (23)−1(12)(23) =(23)(12)(23) = (13) and (13) does not belong to the subgroup H.
Proposition 0.28. A subgroup N of a group G is a normal subgroup of Giff xNx−1 = N for all elements x of G.
Proof. Suppose that N is a normal subgroup of G. Let x be an elementof G. Then xNx−1 ⊂ N. On replacing x by x−1 we get x−1Nx ⊂ N, andthus N = x(x−1Nx)x−1 ⊂ xNx−1. This each of the sets N and xNx−1 iscontained in the order, and therefore xNx−1 = N.Conversely if N is a subgroup of G with the property that xNx−1 = Nfor all x ∈ G, then it follows immediately from the definition of a normalsubgroup that N is a normal subgroup of G.
Proposition 0.29. Intersection of any two normal subgroups of a group is
again an normal subgroup.
Proof. Let H and K be two normal subgroups of a group G. Thisimplies H ∩K is also a subgroup of G. Let x ∈ G and n ∈ H ∩K. Then wehaven ∈ H, n ∈ K.Since H is an normal subgroup of G. Therefore
x ∈ G,n ∈ H ⇒ xnx−1 ∈ H.
Similarly, we can show xnx−1 ∈ K. Thus xnx−1 ∈ H ∩ K. Hence H ∩ K isaa normal subgroup.
26
Proposition 0.30. A subgroup H of G is normal iff xHx−1 = H, ∀x ∈ G.
Proof. Let xHx−1 = H, ∀x ∈ G. Then xHx−1 ⊆ H, ∀x ∈ G. ThereforeH is a normal subgroup of G.Conversely, Let H be an normal subgroup of G. Then
xHx−1 ⊆ H, ∀x ∈ G. (3)
Also x ∈ G ⇒ x−1 ∈ G. Therefore
x−1H(x−1)−1 ⊂ H, ∀x ∈ G
⇒ x−1Hx ⊂ H, ∀x ∈ G
⇒ x(x−1Hx)x−1 ⊂ xHx−1, ∀x ∈ G
⇒ H ⊂ xHx−1, ∀x ∈ G (4)
From equ.(3) and equ.(4), we have
xHx−1 = H, ∀x ∈ G.
Conjugate. Two subgroups H and K of a group G are conjugate if H =a−1Ka for some a ∈ G, i.e., if one is mapped onto the other by some innerautomorphism of G.
Lemma 0.31. Let G and K be groups, and let f : G → K be a homomor-
phism from G to K. Then the kernel ker f of f is a normal subgroup of
G.
Proof. Let x and y be elements of ker f. Then f(x)eK and f(y) = eK ,where eK denotes the identity element of K. But then f(xy) = f(x)f(y) =eKeK = eK . Thus xy belongs to kerf. Also f(x−1) = f(x)−1 = e−1
K eK . Thusx−1 belongs to ker f. We conclude that kar f is a subgroup of K. Moreoverker f is a normal subgroup of G, for if k ∈ G and x ∈ ker f then
f(kxk−1) = f(k)f(x)f(k)−1 = f(k)f(k−1) = eK .
Quotient Homomorphism. If N is a normal subgroup of some group Gthen N is the kernel of the quotient homomorphism f : G → G/N tat sendsx ∈ G to the coset xN. It follows therefore that a subset of a group G isnormal subgroup of G iff it the kernel of some homomorphism.
Proposition 0.32. Let G and K be subgroups, let f : G → K be a homo-
morphism from G to K, and let N be a normal subgroup of G. Suppose that
N ⊂ ker f. The the homomorphism f : G → K induces a homomorphism
f̂ : G/N → K sending xN ∈ G/N to f(x). Moreover f̂ : G/N → K is
injective iff N = ker f.
27
Proof. Let x and y be elements of G. Now xN = yN iff x−1y ∈ N. Alsof(x) = f(y) iff x−1y ∈ ker f. Thus if N ⊂ ker f then f(x) = f(y) wheneverxN = yN, and thus f : G → K induces a well define function since f̂ :G/N → K sending xN ∈ G/N to f(x). This function is a homomorphism,
since f̂((xN)(yN)) = f̂(xyN) = f(xy) = f(x)f(y) = ˆf(xN) ˆf(yN).Suppose now that N = ker f. Then f(x) = f(y) iff xN = yN. Thus thehomomorphism f̂ : G/N → K is injective. Conversely if f̂ : G/N → K isinjective then N must be the kernel of f.
Corollary 0.33. Let (G, o) and (K, ) be groups, and let f : G → K be a
homomorphism. Then f(G) ∼= C/ker f.
Lemma 0.34. Let f : G → K be a homomorphism. Then f(eG) = eK ,where eG and eK denotes the identity elements of groups G and K. Also
f(x−1) = f(x)−1 for all elements of G.
Proof. Let z = f(eG). Then z2 = f(eG)f(eG) = f(eGeG) = f(eG) = z.The result that f(eG) = eK now follows from the fact that an element z ofK satisfied z2 = z iff z is the identity element of K.Let x be an element of G. The element f(x−1) satisfies f(x)f(x−1) =f(xx−1) = f(eG) = eK , and similarly f(x−1)f(x)eK . The Uniqueness ofthe inverse of f(x) now ensures that f(x−1) = f(x)−1.
Lemma 0.35. Let (G, ∗) be a group, let H be a subgroup of G, and let
N be a normal subgroup of G. The the set HN is a subgroup of G, where
HN = {hn : h ∈ H, n ∈ N}.
Proof. The set HN clearly contains the identity element of G. Let xand y be elements of HN. We have to show that xy and x−1 belongs toHN. Now x = hu and y = kv for some elements h and k of H and for someelements u, v ∈ N. Then xy = (hk)(k−1ukv). But k−1uk ∈ N, since N isnormal. It follows that (k−1ukv) ∈ N, since N is a subgroup and (k−1ukv)is the product of the elements (k−1uk and v of N. Also hk ∈ H. It followsthat xy ∈ HN.We must also show that x−1 ∈ HN. Now x−1 = u−1h−1 = h−1(hu−1h−1).Also h−1 ∈ H, since H is a subgroup of G, and hu−1h−1 ∈ N, since N is anormal subgroup of G, as required.
Theorem 0.36. (First Isomorphism Theorem) Let (G, ∗) be a group,
let H be a subgroup of G, and let N be a normal subgroup of G. Then
HN
N∼=
H
N ∩ H.
28
Proof. Every element of HN/N is a coset of N, i.e., of the form hNfor some h ∈ H. Thus if f(h) = hN ∀h ∈ H then f : H → HN?N isa surjective homomorphism, and ker f = N ∩ H. But f(H) ∼= H/karf.Therefore HN/N ∼= H/(N ∩ H).
Theorem 0.37. (Second Isomorphism Theorem) let M and N be nor-
mal subgroups of a group G, where M ⊂ N. Then
G
N∼=
G/M
N/M.
Proof. There is a well-define homomorphism f : G/M → G/N thatsends xM to xN for all x ∈ G. Moreover the homomorphism f is surjec-tive, and kerf = N/M. But f(G/M) ∼= (G/M)/kerf. Therefore G/N isisomorphic to (G/M)/(N/M).
