Table of contents 
Preliminaries 
Preliminaries 
Preliminaries 
Group 
Abelian Group 
NonAbelian Group 
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The basics of set theory : sets, ∩, U, ∈, etc. should be familiar to the reader. Our notation for subsets of a given set A will be
B = {a ∈ A I . . . (conditions on a) . . . }.
The order or cardinality of a set A will be denoted by IAl. If A is a finite set, the order of A is simply the number of elements of A.
It is important to understand how to test whether a particular x ∈ A lies in a subset B of A. The Cartesian product of two sets A and B is the collection A x B = {(a, b)  a ∈ A, b ∈ B}, of ordered pairs of elements from A and B.
We shall use the following notation for some common sets of numbers:
We shall use the notation f : A → B or A B to denote a function f from A to B and the value of f at a is denoted f(a) (i.e., we shall apply all our functions on the left). We use the words function and map interchangeably. The set A is called the domain of f and B is called the codomain of f. The notation f : a → b or a → b if f is understood indicates that f(a) = b, i.e., the function is being specified on elements.
The set f(A) = {b ∈ B I b = f(a), for some a ∈ A} is a subset of B, called the range or image of f (or the image of A under f). For each subset C of B the set f^{1}(C) = {a ∈ A I f(a) ∈ C} consisting of the elements of A mapping into C under f is called the preimage or inverse image of C under f. For each b ∈ B, the preimage of {b} under f is called the fiber of f over b. Note that f^{1} is not in general a function and that the fibers of f generally contain many elements since there may be many elements of A mapping to the element b.
If f : A → B and g : B → C, then the composite map g of : A → C is defined by (g o f)(a) = g(f(a)).
Let f : A → B.
Proposition : Let f : A → B.
A permutation (rearrangement) of a set A is simply a bijection from A to itself.
If A ⊆ B and f : B → C, we denote the restriction of f to A by fl_{A}. When the domain we are considering is understood we shall occasionally denote fl_{A} again simply as f even though these are formally different functions (their domains are different).
If A ⊆ B and g : A → C and there is a function f : B C such that fl_{A} = g, we shall say f is an extension of g to B (such a map f need not exist nor be unique).
Let A be a nonempty set.
Proposition : Let A be a nonempty set.
The following properties of the integers Z (many familiar from elementary arithmetic)
where r_{n} is the last nonzero remainder. Such an r_{n} exists since b > r_{0} > r_{1} > ... > Ir_{n}l is a decreasing sequence of strictly positive integers if the remainders are nonzero and such a sequence cannot continue indefinitely. Then r_{n} is the g.c.d. (a, b) of a and b.
Example : Suppose a = 57970 and b = 10353. Then applying the Euclidean Algorithm we obtain:
57970 = (5)10353 + 6205
10353 = (1)6205 + 4148
6205 = (1)4148 + 2057
4148 = (2)2057 + 34
2057 = (60)34 + 17
34 = (2)17
which shows that (57970, 10353) = 17.
Another application of the division algorithm that will be important to us is modular arithmetic. Modular arithmetic is an abstraction of a method of counting that you often use.
If a and b are integers and n is a positive integer, we often write a = b mod n whenever n divides a  b.
(ab) mod n = ((a mod n)(b mod n)) mod n.
Similarly, (a + b) mod n = ((17 mod 10) + (23 mod 10)) and 10
Examples (7 + 3) mod 10 = 10 mod 10 = 0,
(17 • 23) and 10 = ((17 mod 10)(23 mod 10)) mod 10
(7 • 3) mod 10 = 21 mod 10 = 1.
Definition
Examples :
A group is an ordered pair (G, *) where G is a set and * is a binary operation on G satisfying the following axioms:
Examples
Example : The set of integers Z (so denoted because the German word for numbers is Zahlen), the set of rational numbers 𝕢 (for quotient), and the set of real numbers 𝕣 are all groups under ordinary addition. In each case, the identity is 0 and the inverse of a is a.
Example : The set of integers under ordinary multiplication is not a group. Since the number 1 is the identity, property 3 fails. For example, there is no integer b such that 5b = 1.
Example : The subset {1,1, i, i} of the complex numbers is a group under complex multiplication. Note that 1 is its own inverse, whereas the inverse of i is i, and vice versa.
Example : The set 𝕢^{+} of positive rationals is a group under ordinary multiplication. The inverse of any a is 1/a 5 a^{1}.
Example : The set S of positive irrational numbers together with 1 under multiplication satisfies the three properties given in the definition of a group but is not a group. Indeed, V√2 • V√2 = 2, so S is not closed under multiplication.
Example : A rectangular array of the form is called 2 x 2 matrix. The set o f all 2 x 2 matrices with real entries is a group under component wise addition. That is,
The identity is and the inverse of
Example : The set 𝕫_{n} = {0, 1, ..., n  1} for n > 1 is a group under addition modulo n. For any j > 0 in 𝕫_{n}, the inverse of j is n  j.
This group is usually referred to as the group of integers modulo n.
As we have seen, the real numbers, the 2 x 2 matrices with real entries, and the integers modulo n are all groups under the appropriate addition.
Table : Summary of Group Examples (F can be any of Q, R, C, or Z_{p}; L is a reflection)
By the associative law, a_{1} * (a * a_{2}) = (a_{1} * a) * a) * a_{2}; hence a_{1} = a_{2}. This shows that in every group, each element has exactly one inverse.
The first formula tells us that the inverse of a product is the product of the inverses in reverse order. The next formula tells us that a is the inverse of the inverse of a. The proof of (i) is as follows:
(ab)(b^{1}a^{1}) = a[(bb^{1})a^{1}] by the associative law
= a[ea^{1}] because bb^{1} = e
= aa^{1}
= e
Since the product of ab and b ’ a 1 is equal to e, it follows by Theorem 2 that they are each other’s inverses. Thus, (ab)^{1} = b^{1}a^{1}. The proof of (ii) is analogous but simpler: aa^{1} = e, so by Theorem 2 a is the inverse of a^{1}, that is, a= (a^{1})^{1}.
The associative law states that the two products a(bc) and (ab)c are equal; for this reason, no confusion can result if we denote either of these products by writing abc (without any parentheses), and call abc the product of these three elements in this order.
We may next define the product of any four elements a, b, c, and d in G by abed = a(bcd)
By successive uses of the associative law we find that a(bc)d = ab(cd) = (ab)(cd) = (ab)cd
Hence the product abed (without parentheses, but without changing the order of its factors) is defined without ambiguity.
In general, any two products, each involving the same factors in the same order, are equal. The net effect of the associative law is that parentheses are redundant.
Having made this observation, we may feel free to use products of several factors, such as a_{1}a_{2} ... a_{n}, without parentheses, whenever it is convenient. Incidentally, by using the identity (ab)^{1} = b^{1}a^{1} repeatedly, we find that
If G is a finite group, the number of elements in G is called the order of G. It is customary to denote the order of G by the symbol G.
The group G is said to be abelian if a • b = b • a for all a, b ∈ G.
An abelian group is a nonempty set A with a binary operation + defined on A such that the following conditions hold:
EXAMPLE OF SEMIGROUP BUT NOT MONOID : Nonzero real numbers under the operation * defined by a*b = ab.
EXAMPLE OF MONOID BUT NOT GROUP : {0,1} under multiplication.
In mathematics, a nonabelian group, also sometimes called a noncommutative group, is a group (G, *) in which there are at least two elements a and b of G such that a * b ≠ b * a. The term nonabelian is used to distinguish from the idea of an abelian group, where all of the elements of the group commute.
Example : Show that the set {1,  1, i,  1} where i = is a finite abelian group for multiplication of complex numbers.
G = {1, 1, i i}.
We prepare the composition table for (G, x) as follows :
On observing the table, it is clear that :
 All the entire in the table are elements of G. So multiplication of complex numbers has induced a binary composition in G.
 The number 1 is the identity element for the multiplication composition.
 The inverses of 1,  1 , i,  i are 1,  1 ,  i and i respectively.
Since the associativity and commutativity of the multiplication of numbers is obvious, so G is a finite (of order 4) abelian group for multiplication.
Example : Show that the set {z ∈ C : z = 1} is a multiplicative group.
Let G = {z ∈ C : z = 1} and z_{1}, z_{2} ∈ G; then
z_{1}, z_{2} ∈ G ⇒ z_{1} = 1, z_{2} = 1
⇒ z_{1}z_{2} = 1.1 = 1
⇒ z_{1},z_{2} = 1 [∵ z_{1}z_{2} = z_{1}z_{2}]
⇒ z_{1}, z_{2} ∈ G
∴ G is closed for multiplication.
[G_{1}] Since multiplication operation is associative in numbers, therefore this operation will also be associative in G.
[G_{2}] ∵ 1 = 1 ⇒ 1 ∈ G, which is the identity for multiplication.
[G_{3}] For any element z of G
z ∈ G ⇒ z = 1
⇒
which is the multiplicative inverse of z.
Thus there exist inverse of every element in G.
Hence G is a multiplicative group.
Example : Prove that the set of n, nth roots of unity is a multiplicative finite abelian group.
Here 1^{1/n} = (1 + i0)^{1/n} = (cos 0 + i sin 0)^{1/n}
= (cos 2rπ + i sin 2rπ) ^{1/na}, where r is any integer
[by DeMoivre’s theorem]= e^{i(2rπ)/n }r = 0, 1,2, .... (n  1)
If we take e^{i(2rπ)/n} = θ^{r}, then the set of n complex roots
G = {1, θ, θ^{2}, θ^{3}, .... θ^{n1}}
If a, b ∈ G, then a^{n} = 1 and b^{n} = 1
Now (ab)^{n} = a^{n} b^{n} = 1.1 = 1
∴ ab is also the nth root of unity, therefore ab ∈ G
i.e., a, b ∈ G ⇒ ab ∈ G
Hence G is closed for multiplication.
[G_{1}] Associativity : Complex roots are associative for multiplication.
[G_{2}] Identity element 1 for multiplication is in G.
[G_{3}] If any element θ^{r}, 0 ≤ r ≤ (n  1) is in G, then there exist an element θ^{nr} in G such that θ^{nr}•θ^{r} = θ^{n} = i (identity) [∵ 0 is the nth root of 1]
i.e., there exist inverse θ^{nr} of θ^{r}
[G_{4}] Commutativity : Complex numbers are commutative for multiplication. Hence (G, x) is a finite abelian group of order n.
Example : Show that Z_{5} = {0, 1, 2, 3, 4} is an abelian group for the operation *+_{5}’ defined as follows :
The composition table of (Z_{5}, +_{5}) is as follows :
From the table it is observed that
 All the elements are members of Z_{5} therefore ‘+_{5}’ is a binary composition in Z_{5}.
 0 is the identity element for the composition.
 The inverse of 0, 1, 2, 3, 4 are 0, 4, 3, 2, and 1 respectively.
Again the base of '+_{5}’ is the addition composition of numbers which is associative and commutative. Therefore '+_{3}’ is also associative and commutative.
Hence (Z_{5},+_{5}) is a commutative group.
Example : Show that the set {f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}} is a finite nonabelian group for the operation “composite of functions” where f_{i}, i = 1, 2,..., 6 are transformations on the infinite complex plane defined by :
Let G = {f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}} and o denote composite of functions. Now calculating all the possible products under this operation, we have the following:
From these we obtain the following composition table of (G, o) :
The group axioms can be easily verified from the above composition table. There fore (G, o) is a finite group.
More over since the table is not symmetrical about the leading diagonal so it is a non commutative finite group.
Example : Show that the set Q^{+} of the positive relational numbers forms an abelian group for the operation * defined as :
a * b = ab/2
∴ * is a binary composition in Q^{+}
Verification of group axioms in (Q^{+}, *) :
[G_{1}] Associativity : Let a, b ∈ Q^{+}, then
and
∴ (a *b )*c = a *(b *c)
[G_{2}] Existence of identity :
2 ∈ Q^{+} is the identity element of the operation *
because for a ∈ Q^{+}, 2*a = a*2 = a.2/2 = a
[G_{3}] Existence of Inverse :
For every a ∈ Q^{+}, (4/a) ∈ Q^{+ }(∵ a ≠ 0) which is the inverse of a
Since
So every element of Q^{+} is invertible.
[G_{4}] Commutativity : For any a, b ∈ Q^{+}
[∵ multiplication is commutative]
= b*a
∴ the composition * is commutative.
Hence (Q^{+}, *) is an abelian group.
Example : If G = {(a, b)  a, b ∈ R, a ≠ 0} and o is the operation defined in G as (a, b) o (c, d) = (ac, be + d) ; then show that (G, o) is a nonabelian group.
Verification of group axioms in (G, o) :
[G_{1}] : Let (a, b); (c, d); (e, f) be any elements of G, then [(a, b) o (c, d)] o (e, f) = (ac, be + d) o (e, f)
= ((ac)e, (be + d)e + f)
= (ace, bee + de + f)
and (a, b) o [(c, d) o (e, f)] = (a, b) o (ce, de + f)
= (a(ce), b(ce) + de + f)
= (ace, bee + de + f)
∴ [(a, b) o (c, d)] o (e, f) = (a, b) o [(c, d) o (e, f)]
Therefore o is associative.
[G_{2}] : Here (1, 0) ∈ G is the identity element of the operation because for every (a, b) ∈ G (1, 0) o (a, b) = (1a, 0a + b) = (a, b)
and (a, b) o (1, 0) = (a1, b1 + 0) = (a, b)
[G_{3}] : If (a, b) ∈ G, then a ≠ 0
∴
Again
and
∴ is the inverse of (a, b)
Thus the inverse of every element of G also exist in G.
Hence (G, o) is a group.
[G_{4}] : The composition of the group is not commutative because if a, b, c, d are different real numbers, then (a, b) o (c, d) = (ac, bc + d)
≡ (ca, da + b) = (c, d) o (a, d)
⇒ (a, b) o (c, d) ≠ (c, d) o (a, b)
As such (G, o) is a non abelian group.
Example : Show that G = is a commutative group under matrix multiplication.
Let be any two elements of G, where a, b ∈ R_{0}; then AB = ∵ a ∈ R_{0}, b ∈ R_{0} ⇒ ab ∈ R_{0}
∴ AB = G which shows that G is closed for matrix multiplication.
Verification of group axioms in (G, x) :
[G_{1}] : Since Matrix multiplication is associative.
Therefore this is associative in G also.
[G_{2}] : E = is the identity element because
[G_{3}] : For every (∵ a ∈ R_{0} ⇒ 1/a ∈ R_{0})
which is inverse of A because
(Identity)
[G_{4}] : Since AB = and ab = ba (by commutativity of multiplication of real numbers),
Therefore, AB = BA
As such matrix multiplication is commutative in G.
Hence G is a commutative group for matrix multiplication.
Example : Show that the set of all the matrices of the form is an abelian group for matrix multiplication.
Let
and A_{α, }A_{β}, ∈ G, where α, β ∈ R, then,
= A_{α+β} ∈ G [∵ α + β ∈ R]
∴
Consequently G is closed for matrix multiplication.
Verification of group axioms in (G, x) :
[G_{1}] : Matrix multiplication is associative.
Therefore, this is associative in G also.
[G_{2}]: For α = 0
which is the identity for matrix multiplication.
[G_{3}]: For every A_{α} ∈ G, A_{(α)}, ∈ G which is the inverse of A_{α}, because A_{α} x A_{(α)} = A_{αα} = A_{0 }and A_{(α)} x A_{α} = A_{(α + α)} = A_{0}
Therefore, the inverse of every element of G exists in G.
[G_{4}]: For any two elements A_{α} and A_{β} of G
A_{α} x A_{β} = A_{α+β} [proved above]
= A_{β+α} [∵ α, β ∈ R ⇒ α + β = β + α]
Therefore, matrix multiplication is commutative in G.
As such G is a commutative group for matrix multiplication.
Example : The set G = {1, 2, 3, 4...(p  1)}, x_{p}. p being prime is an abelian group of order (p  1) with respect to multiplication modulo p.
Let a, b ∈ G, then 1 ≤ a ≤ (p  1), 1 ≤ b ≤ (p  1)
By definition, a x_{P} b = r, where r is the least non negative remainder when ordinary product is divided by p.
Now since p is prime, therefore ab is not divisible by p
Consequently, r ≠ 0, 1 ≤ r ≤ (p  1)
therefore a x_{p} b ∈ G ∀ a, b ∈ G
⇒ G is closed for the multiplication modulo p.
[G_{1}] : Associativity : Let a, b, c ∈ G then
a x_{p} (b x_{p} c) = a x_{p} (bc) [∵ b x_{p} c = be (mod p)]
= the least nonnegative remainder obtained on dividing a(bc) by p
= the least nonnegative remainder obtained on dividing (ab)c by p
= (ab) x_{p} c = (a x_{p} b) x_{p} c [∵ ab = a x_{p} b(mod p)]
Therefore multiplication modulo p is associative.
[G_{2}] : 1 ∈ G is the identity element because for every a ∈ G
1 x_{p} a = a x_{p} 1 = a
[G_{3}] : Let s ∈ G then 1 ≤ s ≤ (p  1).
Consider the following (p  1) products :
1 x_{p} s, 2 x_{p} s, 3 x_{p} s, ... (p  1) x_{p} s
All these are elements of G and no two of these are equal since if 1 ≤ i ≤ (p  1), 1 ≤ j ≤ (p  1) and i > j, then i x_{p} s = j x_{p} s
⇒ (i s) and (j s) leave the same nonnegative remainders when divided by p
⇒ (i s  j s) is divisible by p
⇒ (i  j)s is divisible by p
But this is not possible because
1 ≤ (i  j) ≤ (p  1), 1 ≤ s ≤ (p  1) and p is prime.
∴ i x_{p} s ≠ j x_{p} s
and (p  1) distinct elements of G and so one of the products must be equal to 1.
Let r x_{p} s = 1 ⇒ is the inverse of s
Therefore every element of G is invertible.
and (p  1) distinct elements of G and so one of the products must be equal to 1.
Let r x_{p} s = 1 ⇒ r is the inverse of s
Therefore, every element of G is invertible.
[G_{4}] : a x_{p} b = the least nonnegative remainder obtained on dividing ab by p
= the least nonnegative remainder obtained on dividing (ba) by p
= b x_{p} a (∵ ba = ab)
Therefore, multiplication modulo p is commutative.
Hence G is an abelian group of order (p  1).
Remark. If p is not prime, then G will not be a group because G will not be closed for multiplication.
Example : The set of residue classes modulo m is an abelian group with respect to the addition of residue classes.
Let I = { . . . , 3 , 2 , 1 , 0 , 1, 2, 3, ...} be the set of integers. If a ∈ I, then [a] is a residue class modulo m of I if [a] = {x : x ∈ I and x  a is divisible by m}.
Let I_{m} be the set of all residue class of I mod m
i.e., I_{m} = {[a] : a ∈ I}. We have [a] = [b] ⇔ a = b (mod m). The set l_{m} has m distinct elements [0], [1], [2]......[m  1]. Thus we have
l_{m} = {[0], [1], [2], .... [m  1]}.
Let a and b are any two integers, then we define the addition of residue classes [a] and [b] as follows :
[a] + [b] = [a + b].
If [a] = [c] and [b] = [d], then it can be easily seen that [a] + [b] = [c] + [d].
Therefore, our addition of residue classes is welldefined.
Now we shall show that l_{m} is a group with respect to addition of residue classes.
If [a], [b] ∈ l_{m}, then by definition [a] + [b] = [a + b]. Since a + b is an integer, therefore [a + b] ∈ l_{m}. Thus l_{m} is closed with respect to addition of residue classes.
Associativity : Let [a], [b], [c] be any three elements of l_{m}. Then [a] + ([b] + [c]).
= [a] + [b + c] [by def. of addition of residue classes]
= [a + (b + c)] [by def. of addition of residue classes]
= [(a + b) + c] [∵ addition of integers is a associativity]
= [a + b] + [c] = ([a] + ([b] + [c].
Commutativity : Let [a], [b] ∈ l_{m}. Then [a] + [b] = [a + b] = [b + a] = [b] + [a].
Existence of Identity : We have [0] ∈ I_{m}. If [a] ∈ l_{m}, then we have [0] + [a] = [0 + a] = [a] = [a] + [0]. Therefore, the residue class [0] is the identity element.
Existence of Inverse : We have [0] ∈ l_{m}. If [a] ∈ l_{m,} then we have [0] + [a] = [0 + a] = [0 + a] = [a] + [0]. Therefore, the residue class [0] is the identity element.
Existence of Inverse : Let [a] ∈ l_{m} be arbitrary. Since a ∈ I  a ∈ I, therefore [a] is also an element of l_{m}. We have [a] + [a] = [a + a] = [0] = [a] + [a]. Thus, [a] is the inverse of [a].
Thus l_{m} is an abelian group with respect to addition of residue classes.
Since the number of distinct elements in l_{m} is m, therefore the order of this group is m,
Note : If [r] = l_{m} and 0 ≤ r < m, then the inverse of [r] is [m  r]. We have [r] + [m  r] = [r + (m  r)] = [m] = [0].
Note that m I m ⇒ [m] = [0].
1. Uniqueness of the Identity
In a group G, there is only one identity element.
Proof. Suppose both e and e’ are identities of G. Then,
1. ae = a for all a in G, and
2. e’a = a for all a in G.
The choice of a = e’ in (1) and a = e in (2) yields e’ e = e’ and e’ e = e. Thus, e and e’ are both equal to e’ e and so are equal to each other.
Theorem : (Uniqueness of inverse)
The inverse of an element in a group is unique.
Proof. Let a be any element of the group (G, *) which has two inverses b and c in the group.
a^{1} = b ⇒ ba = e = ab
and a^{1} = c ⇒ ca = e = ac
Now ba = e ⇒ (ba) c = ec
⇒ b(ac) = c [by G_{1} and G_{2}]
⇒ be = c [by (2)]
⇒ b = c [by G_{3}]
Therefore, the inverse of every element of a group is unique.
Remark. The inverse of the identity of a group is itself.
Theorem : If G is a group then for a, b ∈ G:
(a) (a^{1})^{1} = a (b) (ab)^{1} = b^{}^{1} a^{1}
i.e. the inverse of the produced of two elements is the product of their inverses in the reverse order.
Proof, (a) Since a^{1} is the inverse of a, therefore aa^{1} = e = a^{1} a
⇒ a^{1} a = e = aa^{1}
⇒ inverse of a^{1} = a, i.e. (a^{1})^{1} = a.
Remark. For the additive operation  (a) = a.
(b) Since a, b, a^{1}, b^{1}, ab, b^{1} a^{1} all are element of G, therefore
(ab)(b^{1}a^{1}) = a(bb^{1})a^{1} [by G_{1}]
= aea^{1 }[by G_{3}]
= aa^{1} [by G_{2}]
= e
∴ (ab)(b^{1}a^{1}) = e ...(1)
Again (b^{1}a^{1})(ab) = b^{1}(a^{1} a)b
= b^{1}eb [by G_{1}]
= b^{1}b [by G_{3}]
= e [by G_{2}]
∴ (b^{1}a^{1})(ab) = e ...(2)
From (1) and (2),
(ab)(b^{1}a^{1}) = e = (b^{1}a^{1})(ab)
⇒ (ab)^{1} = b^{1} a^{1}
Generalised reversal law :
By principle of induction, the above theorem can be generalised as :
(a_{1}a_{2}a_{3}...a_{n})^{1} =
Remark : If the composition is addition (+) then this can be written as :
(a + b) = (b) + (a)
Remark : If G is a commutative group, then for a, b ∈ G
(ab)^{1} = a^{1} b^{1}
Theorem : Cancellation law :
If a, b, c are elements of a group G, then :
(a) ab = ac ⇒ b = c
(b) ba = ca ⇒ b = c
Proof. ∵ a ∈ G ⇒ a^{1} ∈ G [by G_{3}]
∴ ab = ac ⇒ a^{1} (ab) = a^{1} (ac)
⇒ (a^{1} a)b = (a^{1} a)c [by G_{1}]
⇒ eb = ec [by G_{3}]
⇒ b = c [by G_{2}]
Similarly, it can be proved that ba = ca ⇒ b = c
Theorem : If a, b are elements of a group G, then the equations ax = b and ya = b have unique solutions in G.
Proof. ∵ a ∈ G ⇒ a^{1} ∈ G [by G_{3}]
∴ a ∈ G, b ∈ G ⇒ a^{1}b ∈ G
Now a(a^{1} b) = (aa^{1})b [by G_{1}]
= eb [by G_{3}]
= b
Therefore, x = a^{1} b is a solution of the equation ax = b in G.
Uniqueness : Let the equation ax = b have two solutions x = x_{1} and x = x_{2} in G, then ax_{1} = b and ax_{2} = b
⇒ ax_{1} = ax_{2}
⇒ x_{1} = x_{2 }[by left cancellation law]
Therefore, the solution of ax = b is unique in G.
Similarly, its can also be proved that the solution of the equation ya = b is unique in G.
Theorem : Alternate definition of a group :
If for all elements a, b of a semigroup G, equations ax = b and ya = b have unique solutions in G, then G is a group.
Proof. G being semi group, is a non empty set. Therefore let a ∈ G, then the equations ax = a and ya = a will have unique solutions in G. Let these solutions be denoted by e_{1} and e_{2} respectively, then
ae_{1} = a and e_{2}a = a ...(i)
Again if b be any other element of g, then by the given property
x, y ∈ G so that ax = b and ya = b ...(ii)
Now ya = b ⇒ (ya)e_{1} = be_{1}
⇒ y(ae_{1}) = be_{1} [by associativity in G]
⇒ ya = be_{1} [by (i)]
⇒ b = be_{1} [by (ii)]
∴ e_{1} is the right identity in G.
Again ax = b ⇒ e_{2}(ax) = e_{2}b
⇒ (e_{2}a)x = e_{2}b [by associativity in G]
⇒ ax = e_{2}b [by (i)]
⇒ b = e_{2}b [by (ii)]
∴ e_{2} is the left identity in G.
Since e_{1} is right identity in G and e_{2} is in G ⇒ e_{2}e_{1} = e_{2}
Also e_{2} is left identity in G and e_{1} is in G ⇒ e_{2}e_{1} = e_{1} ⇒ e_{1} = e_{1}
Hence there exists an identity e_{1} = e_{2} = e (say) in G.
Again using the given property for the elements a, e e G we find that the equations ax = e and ya = e have unique solutions in G.
Let these solutions be x_{a} and y_{a} respectively. So
ax_{a} = e and y_{a}a = e ...(iii)
⇒ x_{a} and y_{a} are right and left inverses of a in G.
Now ax_{a} = e ⇒ y_{a}(ax_{a}) = y_{a}e
⇒ (y_{a}a)x_{a} = y_{a} [by associativity in G]
⇒ ex_{a} = y_{a} [by (iii)]
⇒ x_{a} = y_{a}
∴ there exist the inverse of a in G.
Since the identity exist and every element of G is invertible in the semigroup G, therefore G is a group.
Remark. If G is a semi group such that for a, b in G, only the equation ax = b (or ya = b) has a unique solution in G, then G may not be a group.
This can be observed by the following example :
Example : Let G be any nonempty set having atleast two elements.
Define a binary operation (*) in G as follows :
a*b = b ∀ a, b ∈ G
∵ For a, b, c ∈ G (a*b)*c = b*c = c
and a*(b*c) = a*c = c
∴ (a*b)*c = a*(b*c)
Therefore the composition * of G is associative.
Hence (G, *) is a semi group.
Again we see that a * b = b ⇒ x = b i s a solution of ax = b in G.
But trivially (G, *) is not a group as there is a no right identity in G.
Theorem : If G is finite semigroup such that for any a, b, c ∈ G
(a) ab = ac ⇒ b = c
and (b) ba = ca ⇒ b = c
then G is a group.
Proof. Since G is a non void finite set, so let
G = {a_{1}, a_{2}, a_{3}......a_{n}}
If a ∈ G, then the n products aa_{1}, aa_{2}, ..., aa_{n} ...(1)
are all elements of G (∵ G is a semi group)
Again all these are distinct elements because
if aa_{i} = aa_{j}, then by the given property
aa_{i}, = aa_{j} ⇒ a_{i} = a_{j}
So all elements of (1) are n distinct elements of G, placed possibly in a different order i.e. G = {aa_{1}, aa_{2}......aa_{n}}
Now if b ∈ G, then b is one of the above n products.
Therefore, let aa_{r} = b, a_{r} ∈ G
Hence we see that for any a, b ∈ G, the equation ax = b has a unique solution in G.
Similarly by considering the n products a_{1}a,a_{2}a....a_{n}a it can be shown that for every pair a, b ∈ G, the equation ya = h has a unique solution in G. Hence by theorem G is a group.
Remark : If G is an infinite semi group satisfying cancellation law but it is not a group.
Example : The positive integers under multiplication form a cancellative semigroup, but is not a group.
Theorem : Definition of a group based on left axioms.
A finite semigroup G is a group iff :
(i) there exists e ∈ G such that ea = a, ∀ a ∈ G (Left Identity)
(ii) there exists b ∈ G such that ba = e, ∀ a ∈ G (Left Inverse)
Proof. If G is a group, then by the definition of a group the above properties (i) and (ii) must obviously hold true.
Conversely, if a semi group G satisfies both the properties (i) and (ii), then we have to prove that G is a group. For this we have to show that e is the identity in G and b is the inverse of a i.e.
ea = a = ae, ∀ a ∈ G
and ba = e = ab
Let a ∈ G, then by the given property (ii), there exists an element b such that
ba = e ...(1)
Again by the same property there exists c such that
cb = e ...(2)
Now for a, b ∈ G ab = e(ab) [by property (i)]
= (cb)(ab) [by (2)]
= c(ba)b [by associativity in G]
= (ce)b [by property (ii)]
= c(eb)
= cb [by property (i)]
= e [by (2)]
∴ ba = e = ab which prove that b is the inverse of a.
Further, we see that ae = a(ba) [∵ e = ab]
= (ab)a [by associativity in G]
= ea [∵ ab = e]
= a [by property (i)]
∵ ea = a = ea which proves that e is the identity in G.
Therefore, G is a group.
Theorem : Definition of a group based on right axioms :
A semigroup G is group iff :
(i) ∃ e ∈ G such that ae = a, ∀ a ∈ G
(ii) ∃ b ∈ G such that ab = e, ∀ a ∈ G
This can be proved similar to the previous theorem.
Caution : It should be noted that a semi group G may be such that G contains a left identity e and each element a in G has a right inverse related to e. In such a case G may fail to be a group. This will be clear by the following example :
Example : Let G be a set containing atleast two elements.
Define a composition (•) in G as follows :
a • b = b, ∀ a, b ∈ G
Obviously G is a semi group. Let e be any fixed element of G. Then by definition, for every a ∈ G, ea = a. So e is a identity of a.
Also for each a ∈ G, ae = e ⇒ e is a right inverse of a wrt e.
But G is not a group as G does not have any right identity.
Finite and Infinite group
A group (G, *) is said to be finite if its underlying set G is a finite set and a group which is not finite is called an infinite group.
Order of a Group
The number of elements of a group (finite or infinite) is called its order. We will use IGI to denote the order of G.
Thus, the group Z of integers under addition has infinite order, whereas the group U(10) ={1, 3, 7, 9} under multiplication modulo 10 has order 4.
Integral powers of an element in a group
Let (G, *) be a group and a ∈ G, then a*a, a*a*a, a*a*a, are all elements of G and shall be written as a^{2}, a^{3}, a^{4} respectively.
Thus (a) for any positive integer n,
a^{n} = a*a*a*...*a (n times)
(b) If n is a negative integer : Let n = n where m is a positive integer, then a" can be defined as follows :
a^{n} = a^{m} = (a^{1})^{m} = a^{1} * a^{1} *...* a^{1} (m times)
Therefore a^{2} = a^{1} * a^{1}, a^{3} = a^{1} * a^{1} * a^{1} etc.
More over we define a^{0} = e, ∀ a ∈ G where e is the identity in G.
Law of indices : If a ∈ G and m, n are integers, then by mathematical induction, the following laws can be easily established :
(a) a^{m + n} = am an (b) (am)n = an m
Note. When the composition of a group is addition (+), then a^{n} is written as n and in that case if m and n are positive integers, then na = a + a + ... + a (n times)
(m)a = (a) +...+ (a) (m times)
(m + n)a = ma + na
and n(ma) = (nm)a
Order of an Element
The order of an element g in a group G is the smallest positive integer n such that g^{n} = e. (In additive notation, this would be ng = 0). If no such integer exists, we say g has infinite order. The order of an element g is denoted by Igl.
So, to find the order of a group element g, we need only compute the sequence of products g, g^{2}, g^{3},..., until you reach the identity for the first time. The exponent of this product (or coefficient if the operation is addition) is the order of g. If the identity never appears in the sequence, then g has infinite order.
Remark. From the above definition, it is clear that for any element a of a group if a^{n} = e then 0(a) ≤ n.
Example : In the group (Z,+), the order of 0 is 1 and the order of every non zero integer a is infinite, because for any a ∈ Z(a ≠ 0) there exists no positive integer n such that na = 0.
Example : In the group (Q_{0}, *), O(1) = 1, O(1) = 2 and order of all the remaining elements is infinite.
Example : In the multiplicative group {1, ω, ω^{2}} (ω^{3} = 1)
0(1) = 1, O((ω) = 3, 0(ω^{2}) = 3.
Example : In the multiplicative group { 1 ,  1 , i,  i } since 1^{1} = 1, (1)^{2} = 1, i^{4} = 1, and (i)^{4} = 1 so O(1) = 1, O(1) = 2, O(i) = 4 and O(i) = 4.
Example : In the group [{0, 1, 2, 3}, +_{4}]
Here 0 is the identity element
0 +_{4} 0 = 0 ⇒ O(0) = 1
1 +_{4} 0 = 1, 1 +_{4} 1 = 2, 1 +_{4} 2 = 3, 1 +_{4} 3 = 0 → O(1) = 4
2 +_{4} 0 = 2, 2 +_{4} 1 = 3, 2 +_{4} 2 = 0 ⇒ O(2) = 3
3 +_{4} 0 = 3, 3 +_{4} 1 = 0 ⇒ O(3) = 2
Example : In Klein’s 4 group every element except identity is of order 2.
Properties of the order of an element of a group
Theorem : The order of the identity of every group is 1 and it is the only element of order 1.
Proof. Let e be the identity of any group G.
∵ e^{1} = e, ∴ O(e) = 1
Again if a ∈ G and O(a) = 1, then
O(a) = 1 ⇒ a^{1} = e ⇒ a = e
Therefore O(a) = 1 ⇒ a = e
Theorem : The order of every element of a finite group is finite and less than or equal to the order of the group i.e.,
O(a) ≤ O(G), ∀ a ∈ G
Proof. Let G be a finite group of order n and a ∈ G.
We see that a^{0}, a^{1}, a^{2}, .... a^{n }are all elements of G But G contains n elements and the number of these elements is (n + 1), so all can not be distinct. Therefore atleast two of these elements are equal.
Let a^{i} = a^{i}, 0 ≤ j ≤ i ≤ n
⇒ a^{i}a^{j} = a^{j} a^{j}
⇒ a^{ij }= a^{0} = e
⇒ a^{r} = e, where 0 < r = i  j ≤ n
⇒ O(a) ≤ r ≤ n = O(G)
Hence O(a) is also finite and less than or equal to O(G).
Remark : The above theorem can also be stated as follows :
If G is a finite group of order n, then for any a ∈ G, there exists a positive integer r ≤ n such that a^{r} = e.
Theorem : If order of an element a of a group (G, *) is n, then a^{m} = e, iff m is a multiple of n.
Proof. First of all, suppose that a^{m }= e^{ }
Now since m and n are integers, so by division algorithm in integer, there exist integers q and r such that
m = nq + r, 0 ≤ r < n
∴ a^{m} = e ⇒ a^{nq+r} = e ⇒ a^{nq} a^{r }= e
⇒ (a^{n})^{q} a^{r} = e [∵ (a^{m})^{n} = a^{mn}]
⇒ e^{q}a^{r} = e [∵ O(a) = n ⇒ a^{n} = e]
⇒ a^{r} = e ⇒ r = 0 [∵ 0 ≤ r < n]
⇒ m = nq i.e., m is a multiple of n.
Conversely : Let m be a multiple of n i.e. m = nq(q then m = nq ⇒ a^{m} = a^{nq} = (a^{n})^{q} = e^{q} = e
Therefore, a^{m} = e ⇔ m is a multiple of O(a).
Cor. : The order of any integral power of an element a of a group (G,*) can not exceed the order of the element.
Proof. Let a be any element of a group G and O(a) = n.
Then for any k ∈
(a^{k})^{n} = a^{nk} = (a^{n})^{k} = e^{k} = e
⇒ O(a^{k}) ≤ n ⇒ O(a^{k}) ≤ O(a)
Cor. : The order of an element a of a group (G, *) is equal to that of its inverse a^{1} i.e., O(a) = O(a^{1}).
Proof. Let O(a) = n and O(a^{1}) = m
Now a^{1} is an integral power of a ⇒ 0(a^{1}) ≤ O(a)
⇒ m ≤ n ...(1)
Again a = (a^{1}) ⇒ a is the integral power of a^{1}
⇒ O(a) ≤ O(a^{1})
⇒ n ≤ m ...(2)
(1) and (2) ⇒ m = n, i.e. O(a^{1}) = O(a)
Cor. : If any element of a group G is of order n, then order of a^{k} is n/m where m is g.c.d. of n and k.
You have two things to show. Namely that:
The first part is easy:
(a^{k})^{n/gcd(n,k)} = (a^{n})^{k/gcd(n,k) }= e^{k/gcd(n,k)} = e
For the second part, let be such that (a^{k})^{m} = a^{km} = e. Since the order of a is n, it follows that n I km. Therefore we also have
Now gcd(n/gcd(n, k), k/gcd(n, k)) = 1 (try to prove this), so it follows that n/gcd(n, k) I m
Hence in particular n/gcd(n, k) ≤ m.
Theorem : For any element a of a group G :
O(a) = O(x^{1} ax), ∀ x ∈ G
Proof. Let a ∈ G, x ∈ G, then
(x^{1} ax)^{2} = (x^{1} ax)(x^{1} ax)
= x^{1} (xx^{1})ax [by associativity]
= x^{1 }aeax = x^{1} (aea)x
= x^{1} a^{2}x
Again let (x^{1}ax)^{n1} = x^{1} a^{n1} x , where(n  1)
⇒ (x^{1} ax)^{n1} (x^{1} ax) = (x^{1} a^{n1}x) (x^{1} ax)
⇒ (x^{1} ax)^{n} = x^{1}a^{n1} (xx^{1})ax = x^{1}a^{n1} (eax)
= x^{1}a^{n1}ax = x^{1} a^{n}x
Therefore, by induction method,
(x^{1} ax)^{n} = x^{1} a^{n} x,
Now let O(a) = n and O(x^{1} ax) = m, then (x^{1} ax)^{n} = x^{1}a^{n}x = x^{1} ex = e
⇒ 0(x^{1} ax) ≤ n ⇒ m ≤ n ...(1)
Again 0(x^{1} ax) = m ⇒ (x^{1}ax)^{m} = e ⇒ x^{1} a^{m} x = e
⇒ x(x^{1} a^{m}x)x^{1} = xex^{1} = e
⇒ (xx^{1}) a^{m}(xx^{1}) = e
⇒ ea^{m} e = e ⇒ a^{m} = e
⇒ O(a) ≤ m ⇒ n ≤ m ...(2)
(1) and (2)
⇒ n = m ⇒ O(a) = O(x^{1} ax)
If O(a) is infinite, then 0(x^{1} ax) will also be infinite.
Cor. : If a and b are elements of a group G; then O(ab) = O(ba)
Let n and m be the order of ab and ba, respectively. That is, (ab)^{n} = e, (ba)^{m} = e,
where e is the identity element of G.
We compute
= a(ba)^{n1} b.
From this, we obtain(ba)^{n1} = a^{1} b^{1} = (ba)^{1}, and thus we have (ba)^{n} = e.
Therefore the order m of ba divides n.
Similarly, we see that n divides m, and hence m = n.
Thus the orders of ab and ba are the same.
Proof. ∵ b^{1} (ba)b = (b^{1}b)(ab) [by associativity]
= e(ab) = ab
Therefore by the above theorem, O(ba) = O(b^{1}(ba) b) = O(ab)
Theorem : If the order of an element a of a group G is n, then the order of a^{p} is also n provided p and n are relatively prime.
Proof. Let O(a^{p}) = m
∵ O(a) = n ⇒ a^{n} = e ⇒ a^{np} = e^{P} = e
⇒ (a^{p})^{n} = e ⇒ O(a^{p}) ≤ n
⇒ m ≤ n ...(1)
Again since p and n are relatively prime, therefore GCD of p and n = 1
⇒ there exist two integers x and y such that px + ny = 1
⇒ a^{1} = a^{px+ny} = a^{px} • a^{ny}
= a^{px} .(a^{n})^{y} = a^{px}.e^{y}
= a^{px} .e = a^{px}
a^{m} = (a^{px})^{m} = a^{mpx} = (a^{mp})^{x} = e^{x} = e
⇒ O(a) ≤ m
⇒ n ≤ m ...(2)
(1) and (2) ⇒ n ≤ m ⇒ O(a) = O(a^{p})
Example : If for every element a of a group G, a^{2} = e; then prove that G is abelian.
Let a, b ∈ G, then by the given property
a ∈ G ⇒ a^{2} = e ⇒ aa = e
⇒ a^{1} = a ...(1)
Similarly b ∈ G ⇒ b^{1} = b ...(2)
Now a ∈ G, b ∈ G ⇒ ab ∈ G
⇒ (ab)^{1} = ab [by the given property]
⇒ b^{1}a^{1} = ab [by reversal law]
⇒ ba = ab [by (1) and (2)]
∴ G is an abelian group.
Example : If a, b are any two elements of a group G; then show that G is an abelian group if (ab)^{2} = a^{2} b^{2}.
Let G be an abelian group, then G is abelian ⇒ ab = ba, ∀ a, b ∈ G
⇒ (ab)(ab) = (ba)(ab)
⇒ (ab)^{2} = b(aa)b [by associativity]
⇒ (ab)^{2} = (ba^{2})b
⇒ (ab)^{2} = (a^{2}b)b [∵ G is abelian]
⇒ (ab)^{2} = a^{2}(bb)
⇒ (ab)^{2} = a^{2} b^{2}Conversely : For a, b ∈ G, (ab)^{2} = a^{2} b^{2}, then (ab)^{2} = a^{2} b^{2} ⇒ (ab) (ab) = (aa) (bb)
⇒ a(ba)b = a(ab)b [by associativity]
⇒ G is an abelian group.
Therefore G is an abelian group ⇔ (ab)^{2} = a^{2} b^{2}.
Example : If a is an element of a group G, then show that : a^{2} = a ⇔ a = e
Let a ∈ G and a^{2} = a then a^{2} = a ⇒ aa = ae
⇒ a = e [by left cancellation law]
Conversely : If a = e, then
a = e ⇒ aa = ea
⇒ a^{2} = a
Therefore a^{2} = a ⇔ a = e
Remark. For any element a of a group G, if a^{2} = a, then a is called an idempotent.
The identity element of the given group is 1, therefore O(a) = 1. For other elements, we see that
2^{2} = 4, 2^{3} = 3, 2^{4} = 1 ⇒ O(2) = 4.
3^{2} = 4, 3^{3} = 2, 3^{4} = 1 ⇒ O(3) = 4.
and 4^{2} = 2 ⇒ O(4) = 2.
Example : If a, b are two elements of a group G such that a^{m} b^{n} = ba, ∀ m, n ; then prove that : a^{m}b^{n2}, a^{m2}b^{n}, ab^{1 }are element of the same order in G.
a^{m}b^{m2} = a^{m}b^{n}b^{2} = (a^{m}b^{n})b^{2} = (ba)b^{2}
= (ba)(b^{1}b^{1}) [∵ a^{m}b^{n} = ba]
= b(ab^{1})b^{1} = (b^{1})^{1} (ab^{1}) (b^{1})
∴ O(a^{m}b^{n2}) = O(ab^{1}) [∴ O(a) = O(x^{1} ax), ∀ x ∈ G] ...(1)
Again a^{m2}b^{n} = a^{2} (a^{m}b^{n}) = a^{1}a^{1} ba [∵ a^{m}b^{n} = ba]
a^{m2}b^{n} = a^{1}(a^{1}b)a
∴ O(a^{m2} b^{n}) = O(a^{}^{1}(a^{1}b)a)
= O(a^{1}b) [∵ O(a) = 0(x^{1} ax), ∀ x ∈ G]
= O(a^{1} b)^{1} [∵ O(x) = 0(x^{1})]
= O(b^{1}a) = O(eb^{1} a)
= O(a^{1} ab^{1} a)
= O( a^{1} (ab^{1})a) = O(ab^{1}) ...(2)
From (1) and (2),
O(a^{m}b^{n2}) = O(a^{m2}b^{n}) = O(ab)^{1}
Example : If a is the only element of order 2 in a group G, then show that ax = xa, ∀ x ∈ G.
We know that for any element a in a group G.
O(a) = O(x^{1} ax), ∀ x ∈ G
∴ O(a) = 2 ⇒ O(x^{1} ax) = 2
But a is the only element of order 2 in G, therefore
O(a) = O(x^{1} ax) ⇒ a = x^{1} ax
⇒ xa = x(x^{1} ax)
⇒ xa = ax, ∀ x ∈ G
Example : If G is a group such that (ab)^{m} = a^{m}b^{n} for three consecutive integers m, m + 1, m + 2 for all a, b ∈ G; show that G is abelian.
a, b ∈ G, then as given
(ab)^{m} = a^{m}b^{m} ...(i)
(ab)^{m+1} = a^{m+1} b^{m+1} ...(ii)
(ab)^{m+2} = a^{m+2} b^{m+2} ...(iii)
Now (ab)^{m+2} = (ab)^{m+1} ab
⇒ a^{m+2} b^{m+2} = a^{m+1} b^{m+1 }ab [by (ii) and (iii)]
⇒ a^{m+1} ab^{m+1} b = b^{m+1 }b^{m+1} ab
⇒ ab^{m+1} = b^{m+1} a [by cancellation law]
⇒ a^{m}(ab^{m+1}) = a^{m}(b^{m+1} a)
⇒ (a^{m}a)b^{m+1} = a^{m}(b^{m}b)a
⇒ a^{m+1}b^{m+1} = a^{m}b^{m}(ba)
⇒ (ab)^{m+1} = (ab)^{m} (ba)
⇒ (ab)^{m} (ab) = (ba)^{m} (ba)
⇒ ab = ba
∴ G is an abelian group.
Example : If a, b are elements of an abelian group G, then prove that :
(ab)^{n} = a^{n}b^{n},
Case (i) When n = 0, then (ab)° = e = ee [by definition]
= a° b°
Case (ii) when n > 0; If n > 1, then (ab)^{1} = ab = a^{1} b^{1}
∴ The result is true for n = 1.
Let the result be true for n = k, then (ab)^{k} = a^{k} b^{k}
⇒ (ab)(ab)^{k} = (ab)(a^{k} b^{k})
⇒ (ab)^{k+1} = a(ba^{k})b^{k }[by associativity]
⇒ = a(a^{k}b)b^{k }[by commutativity in G]
⇒ = (aa^{k})(bb^{k}) = a^{k+1} b^{k+1}
Thus if the result is true for n = k, it is also true for n = k + 1.
Hence by principle of induction it is true for all integers.
Case (iii) When n < 0, then let n =  m , where m ∈ Z+, then
(ab)^{n} = (ab)^{}^{m} = [(ab)^{m}]^{1}
= (a^{m}b^{m})^{1} [by case (ii)]
= (b^{m} a^{m})^{1 }[by commutativity in G]
= (a^{m})^{1} (b^{m})^{1} = a^{m} b^{m}
= a^{n} b^{n}
Combining all the three cases, we conclude that G is commutative ⇒ (ab)^{n} = a^{n}b^{n}, ∀
Example : Prove that every group of order 4 is an abelian group.
Let G = {e, a, b, c} be a group of order 4 with e as its identity. Its composition table is as under :
Since every element of the group appears only once in each row of the composition table, therefore in the second row
ab = e or ab = b or ab = c
But ab = b ⇒ a = e which is not possible (∵ a ≠ e)
∴ ab ≠ b, Hence ab = e or, ab = c
Similarly in the third row ba ≠ a
∴ ba = e or ba = c
Since every element of the group appears only once in each column of the composition table, therefore
and
Similarly it can be seen that
ac = ca and be = cb
Hence G is an abelian group.
Example : Consider under addition modulo 10. Since 12 = 2, 22 = 4, 32 = 6, 42 = 8, 52 = 0, we know that 2 = 5. Similar computations show that 0 = 1, 7 = 10, 5 = 2, 6 = 5.
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