Mathematics For Statisticians-A Brief note about Group Theory

An Introductory Note On Group Theory

The term "abstract" is a highly subjective one,which refers to the fact that the "art of not representing things pictorially". But in relation to current research activity in abstract algebra,it could be described as "not too abstract".
Origin of Abstract Algebra

Now we shall discuss some terms:

Definition: Let A be a non-empty set. A binary operation‘0’ on A is a mapping  f : A x A-> A.
 In other words , a binary operation ‘0’ on A is a rule of correspondence that assigns to each
ordered pair (a1,a2) A x A, some element of A, which we shall denote by a1 0 a2. Note that a10a2 need not be distinct from a1 or a2.
Example:- Subtraction is a binary operation on Z but not on N; division is a binary operation on
the set Q* of all non-zero rational number but not on Z.

Definition: Let 0 be a binary composition on the set A.
(A,0) is called a mathematical system.
"0" is commutative iff x0y = y0x holds ,for all x,y∈ A.
"0" is associative iff x0(y0z) = (x0y)0z holds for all x,y,z∈ A.

Definition: A nonempty set S with an associative binary operation ‘.’defined on S forms a semigroup. A semigroup M that contains an identity element is called a Monoid . A monoid in which every element is invertible is called a Group . Thus (G,.) is a group iff following conditions are satisfied:
(a) . is associative,
(b) (G,.) has an identity element, generally denoted by e and
(c) every element x∈ G has an inverse element of x in G such that x.x^{-1}=e.
If, in addition, (G,.) is commutative , (G,.) is an abelian group.

Elementary properties of Group
Theorem: Let (G,.) be a group. The following properties hold:
If G has a left identity e and a right identity f, then e=f.    In particular, identity element in G is unique.
If an element x of G has a left inverse y and a right inverse z, then y=z. In particular, x^(-1),inverse element to x, is unique. e-1 = e, (x^-1)^(-1) = x, x∈ G.
(Cancellation Laws) for a,b,c∈ G.
 a.c = b.c ⇒ a = b (right cancellation property)
 c.a = c.b ⇒ a = b (left cancellation property).

Permutation Groups
Definition: Let A. A permutation on A is a bijective mapping of A onto itself. If A={1,2,…,n},
then the group Sn formed by the set of all permutations on A under composition of functions as composition is called Symmetric Group on n symbols . Order of Sn, that is , the number of elements of Sn , is n!.

Theorem: If n is a positive integer ,n≥3, then  Sn is a non-commutative group.

Let (G,0) be a group and  H ⊆ G. H is closed under 0 iff ∀ h1,h2∈H, h1 0 h2 ∈ H. If H be closed
under 0 , then the restriction of 0 to H x H is a mapping from H x H into H. Thus the binary
operation 0 on G induces binary operation, also denoted by 0 , on H.

Definition: Let (G,0) be a group and  H ⊆ G. (H , 0) is a subgroup of (G , 0) iff H is closed under 0 and
 (H , 0) is a group.
Every group G has at least two subgroups , namely, {e} and G. These are called trivial subgroups.
Other subgroups, if there be any, are called nontrivial subgroups of G.

Ex. (2Z,+) is a subgroup of (Z,+) , where 2Z is the set of all odd integers. Since the set 2Z+1 of all odd integers is not closed under addition of integers, (2Z+1 , +) is not a subgroup.
Though N is closed under addition, (N , +) does not form a group; hence (N,+) is not a subgroup of (Z , +).

Theorem: Let G be a group and  H ⊆ G. Then H is a subgroup of G iff a.b^(-1) ∈ H, ∀ a,b ∈ H.
Proof: Let H⊆G and let a.b^(-1) ∈H, ∀ a,b ∈ H.
Since H, let a∈H.
By assumption, e = a.a^(-1) ∈ H.
Let b∈H. Then b^(-1)= e.b^(-1)∈ H.
Also, for a,b∈H, a,b^(-1)∈ H. and hence a.b=a.(b^-1)^-1∈H.
Associativity, being a common property, holds in (H,.) since it holds in (G,.).
Thus H is a subgroup of G.
Converse part is trivial.

Theorem: Let G be a group and H⊆G, H finite. Then H is a subgroup of G iff a.b ∈ H , ∀ a,b ∈H.
Proof:(sufficiency) Let h∈H.Then A={h,h2,h3,…}⊆H.
Since H is finite, there exist integers r and s, 0≤r<s and hr=hs. Hence by cancellation property in G, e=hs-r ∈H.
Now e=hs-r=h.hs-r-1 implies h-1=hs-r-1∈H
[ Note that  s-r-1 is non-negative integer] .
Let a,b∈H. Then a,b-1∈H. Hence by hypothesis, a.b-1∈H.
Thus by the previous theorem, H is a subgroup of G.

Ex: Prove that every subgroup of (Z,+) is of the form nZ={na:a∈Z}, where n is a nonnegative integer.
Sol:  Let H be a subgroup of Z. If H={0}, then H=0Z.
Let {0} ⊊ H. Let 0  a ∈H . Since H is a subgroup, -a∈H.
Thus H contains positive integers.
By well-ordering principle of N, the set A = {a∈H : a∈N} ⊆ N has a smallest element, say, n. We shall prove that H={nr : r∈Z}.
Since n∈H and H is a subgroup, {nr : r∈Z} ⊆ H.
Again, let b∈H. By division algorithm for integers, there exist p,r∈Z such that b = pn + r, 0≤r<n.
So r = b - pn ∈H (since H is a subgroup).
Since r<n and n is the smallest positive integer such that n∈H,  so r=0.
Thus b = pn ∈ nZ.
Thus H⊆nZ.
Combining the two, H=Nz.

Definition: A group G is a cyclic group iff there exists a∈G such that G=(a)={an:n∈Z}. Such an element a is called a generator of G.
Ex:(Z,+) is a cyclic group since Z=(1). (Zn , +n) is cyclic with [1] as one of its generators.
And (R,+) is not cyclic: a rational can not generate irrational and vice versa. The multiplicative group
of all four fourth roots of unity is cyclic with i as one of its generators.

Theorem: Every cyclic group is commutative ; converse may not hold.
Proof: Let G=(a) and let b,c∈G. Then b=an,c=am, there exist integers m,n. Thus, proving that G is abelian.
For the converse, consider the Klein’s 4-group: G={e,a,b,c} with the binary operation 0 defined by the property a2=b2=c2=e2=e. Then G is abelian but not cyclic.

Theorem:.  Every group of prime order is cyclic.
Proof: Let 0(G)=p, p prime. Thus p≥2. Let ea∈G. Then 0((a)) divides 0(G)=p ,0((a))=0(a)>1. Thus
0((a))=p=0(G), (a)⊆G. Hence G=(a).