BVRICE
Group Theory
GROUP THEORY
The chapters in this Unit are
1)
2)
3)
4)
5)
6)
7)
Groups
Sub Groups
Cosets
Normal Subgroups
Homomorphism’s
Permutations (permutation groups)
Cyclic Groups
1.GROUPS
KEY POINTS:
1)Binary operation : Let S be any non-empty set, then S X S contains all ordered pairs(a,b) where a,b ∈ S.
Any mapping f:S X S → S is called a binary operation on S.
Examples: + and . are binary operations on ‘N’
2) A Binary operation * on a set S is said to be associative if (a*b)*c = a*(b*c) ∀ a,b,c ∈ S
3) A Binary operation on a set S is said to be commutative if a*b = b*a ∀ a,b ∈ S
4) A pair (S,*) is called as a binary system if S is any non-empty set and * is binary operation on S.
We also call a binary system as an algebraic structure.
5) Suppose (S,*) is a binary system and let e ∈ S, then e is called as
(i) left identity if e*a = a for all a∈ 𝑆
(ii) right identity if a*e = a for all a∈ 𝑆
If e is both left and right identity element, then e is called identity in S w.r.to *.
6) Suppose (S,*) is a binary system and let a ∈ S; let e be the identity element in S. Then an element b in S
is said to be ‘the inverse of a’ if a*b = b*a = e.
Note: if a*b=e is only satisfied, we call b as right inverse of a, and
If b*a =e is only satisfied, we call b as left inverse of a.
7) A binary system (S,*) is said to be a semi group if * is associative on S.
8) A binary system (S,*) is called a monoid if * is associative on S and S has identity element w.r.to *.
Examples: (W,+) is a monoid; (N,+) is a semi group but not a monoid because it has no identity w.r.to +
Group: A binary system (G,*) is called a group if
(i) * is associative on G i.e. (a*b)*c = a*(b*c) ∀ a,b,c ∈ G
(ii) there is an element e in G such that e*a = a*e = a for all a ∈G
(iii) for each a in G there is an element b in g such that a*b = b*a = e.
Abelian group: A group (G, *) is called as an Abelian group if * is commutative on G i.e. a*b = b*a ∀ a,b ∈G.
Examples: (Z,+), (Q,+), (R,+) are Abelian groups.
Order of a Group: The number of elements in a group (G,*) (either G is finite or infinite) is called order of
that group it is denoted by 0(G) or |G|.
Order of an element in a Group: Suppose (G,*) is a Group with identity element ‘e’. Let a∈G, if there
exists a least positive integer n such that an = e then we say that order of a = n denoted by 0(a)=n. If there
is no such integer then we say that a has order infinite or zero order.
Note: order of identity element in any group is “1”
Cancellation laws: suppose (S,*) is a binary system. For a,b,c ∈S
If (i) a*b = a*c then b=c and (ii) b*a = c*a then b=c (i)and (ii) are called as left, right cancellation laws.
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Group Theory
Theorem 1: In a group G, for a,b,x,y ∈G, the equations ax = b and ya = b have unique solutions.
Proof: Suppose (G,.) is a Group with identity element ‘e’ and let a,b ∈G
By inverse law in G, we have a-1 ∈G also by closure law; we have a-1b ∈G.
Now consider ax = b
Also if x= a-1b then a(a-1b) = (a a-1)b (associative law)
Now consider ax = b ⟹ a-1(ax) = a-1b
⟹(a-1a)x = a-1b
=eb
⟹ex =a-1b
=b
⟹x = a-1b
∴ x = a-1b is a solution of the equation ax =b
Uniqueness: if possible suppose let x1, x2 be two solutions of the equation ax =b
Then ax1 =b and ax2 =b ⟹ ax1 =ax2 ⟹x1=x2 (left cancellation law)
∴ The solution is Unique
Similarly we can prove that y = ba-1 is the solution for the equation ya = b
Hence the theorem.
Note: The condition in the above theorem is a necessary and sufficient condition for a group.
Theorem 2:A finite semi group (G,.) satisfying cancellation laws is a group (or)
A finite set G with a binary operation ’.‘ is a group if . is associative and cancellation laws hold in G.
Proof: Suppose G={ a1,a2, ..an} is a semi group with n distinct elements. Suppose G has cancellation laws.
We have to prove that (G,.) is a group.
Let a ∈G; now, consider the products {aa 1, aa2, aa3, …. aan}. By closure law all these elements belong to G.
Also all these elements are distinct.
{ ∵ suppose aai = aaj for i ≠j then ai = aj by using left cancellation law
Which is a contradiction to all the elements in G are distinct
}
Since G is finite all the above product elements must be equal to elements of G in some order.
Now let b∈G, therefore ∃ some i such that aai = b.
Hence the equation ax = b has solution in G
Similarly we can prove that the equation ya = b has solution in G by taking products {a1a,a2a, …ana}
∴ by theorem 1 (G,.) is a group.
Hence the theorem
Note: Inverse of an element in a Group is Unique.
Suppose (G,*) is group with identity element ‘e’
Let a∈G. now if possible suppose b and c are inverses for a.
∴ a*b = b*a = e and a*c = c*a = e
Now consider b= b*e
= b * (a *c)
(e is identity)
( e value)
= (b *a) * c
(associativity)
= e *c
(b * a = e)
=c
( e is identity)
∴ Inverse of an element in a group is unique.
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Group Theory
Theorem 3: In a Group (G,.) o(a) = o(a-1) ∀ a ∈G
Proof: Suppose (G.) is a group with identity element ‘e’ and let a ∈G. suppose o(a) = p and o(a-1) = q
We have to prove that p = q
∴ By definition ap = e and (a-1)q = e and p,q are least positive integers with this property.
Now consider ap = e
Now consider (a-1)q = e
⟹ (aq)-1 = e
⟹ ( ap ) -1= e-1
⟹aq = e-1
⟹ (a-1)p = e
(inverse e is e)
(by definition of order of a -1)
⟹q≤p
⟹aq = e
(inverse of e is e)
⟹p ≤q
(by the definition of o(a) )
Hence p = q
Hence the theorem.
Theorem 4: let (G,.) is a group and a ∈G with o(a) = n. then am= e iff n|m.
Proof: consider given data, i.e. o(a) = n ⟹an = e and n is the least positive integer with this property
We have to prove that am= e iff n|m.
Converse part: suppose n|m
∴ By definition ∃ a positive integer q such that m = nq
Now consider am = anq = (an)q = eq = e
Main part: suppose am= e
We have n and m belong to Z; by division algorithm in integers ∃ integers q and r such that
m = nq + r with 0 ≤ r <n
here r = 0 is must.
{ ∵ suppose r ≠ 0, then we have am= e
⟹a
nq +r
=e
⟹(a)nq .ar = e
⟹ (e)q.ar =e
⟹ar =e and o < r < n
which is a contradiction to order of a is n }
hence m = nq i.e. by definition n | m.
hence theorem.
Theorem 5: The order of every element of a finite group is finite and is less than or equal to order of the group.
Proof: suppose (G,.) is a finite group and let a ∈G.
First we prove that order of a exists and is finite then o(a) ≤0(G)
We have by closure law in G, a, a2, a3, a4, … all elements ∈G.
Since G is a finite set the above elements must repeat.
Hence ∃ r and s such that r ≠ 𝑠 and r>s with ar = as
⟹a r-s = a0
⟹am = e ( let r-s = m and a0 = e)
⟹let n be the least positive integer with an =e
∴ order of a exists and it is finite
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Now we prove that o(a) ≤0(G)
If possible suppose o(a) = n and 0(G) = m with n >m
Now consider the powers of a i.e. a, a2, a3, a4, …an; We have by closure law all these elements belong to G,
And they are distinct.
{ ∵ suppose they are not distinct hence they must repeat.
Hence ∃ r and s such that r ≠ 𝑠 and r>s with ar = as and 1≤ s < r ≤ n
⟹a r-s = a0 = e
Here 0 < r-s < n with ar-s = e
which is a contradiction to o(a) = n }
But this is impossible when n>m ( Because we cannot insert n distinct elements in m places when n>m)
Hence our supposition is wrong.
∴o(a) ≤0(G)
Hence the theorem.
SUBGROUPS
Complex: Any non-empty subset of a group is called a complex.
Complex multiplication: if M and N are two complexes of a group G then their product is given by
MN = {x y/ x ∈ M, y∈N}
Inverse of a complex: If M is a complex of a group G. then its inverse is given by M -1 = { x-1/x∈M}.
Note: (i) Multiplication of complexes is associative in any group.
(ii) if M and N are complexes then (MN)-1 = N-1M-1.
Subgroup: suppose (G,.) is a group. Then any complex H of G is called a subgroup of G if H itself is a
group with the operations defined on G. if H is a subgroup of g then we denote it as H < G (when H is a
proper subset of G) or H ≤G.
Ex: (Z,+) is a subgroup of (Q,+)
Note:(i) For any group G; {e} and G itself are subgroups. These are called trivial or improper subgroups of G
(ii)The identity element in a group and the identity elements in its any subgroup are same.
Theorem1: necessary and sufficient condition theorem:
Statement: Suppose H is a non-empty complex of a group G. the necessary condition for to be subgroup
of G is ∀ a,b, ∈ H, a.b-1 ∈H where b-1 is the inverse of b in G.
Proof: Given that (G,.) is a group and H is any non empty complex of G.
We have to prove that H is a subgroup of G iff ∀ a,b, ∈ H, a.b-1 ∈H
Main part: (necessary part): suppose that H is a subgroup of G.
Let a,b, ∈ H, since H is subgroup of G so H itself is a group. Since H is a group, by inverse law b -1 ∈H.
Hence by closure law a.b-1∈H.
Converse part: (sufficient part):
Conversely, suppose ∀ a,b, ∈ H, a.b-1 ∈H. →(1)
Since H is non-empty ,let a ∈H
In the equation (1) take a in place of b then a.a-1 ∈H i.e. identity element e ∈H.
Taking a=e and b=a in equation (1) we get e.a-1 ∈H i.e. a-1 ∈H.
Again taking b∈H then b-1 ∈H
Now a∈H and b-1 ∈H , using (1) a.(b-1)-1 ∈H
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Since all the elements H are in G, associative law holds clearly in H.
Hence by definition H is a group w.r.to ‘ .’
∴ H is a subgroup of G.
Hence the theorem
Theorem2: Intersection of two subgroups is a subgroup.
Proof: suppose (G,.) is a group and M and N are two subgroup of G
Since each subgroup contains identity element ‘e’ so M∩N contains ‘e hence M∩N ≠ ∅ .
Now let a.b ∈ M∩N then a.b ∈ M and a.b ∈ N
Since M and N subgroups of G by necessary and sufficient condition theorem ab -1∈M and ab-1∈N
Hence ab-1∈ M∩N. hence M∩N is a subgroup of G (using necessary and condition theorem)
Hence the theorem.
Note:(1) From the above theorem we can say that the intersection a finite collection of subgroups of a
group is again a subgroup.
(2) The union of two subgroups of a group need not be a subgroup. For this consider the example
Let the group be (Z6, +6) = {0,1,2,3,4,5} and S={0,2,4} and T={0,3} then S and T are subgroups of Z6.
But S U T is not a subgroup of Z 6. Because 2, 3 ∈ S U T but 2+3 = 5 ∉ S U T i.e. closure law fails
Theorem3: Union of two subgroups is subgroup iff one is contained in the other.
Proof: suppose (G,.) is a group and M and N are two subgroups of G
We have to prove that M U N is a subgroup of G iff either M ⊂N or N⊂ M.
Converse part: suppose either M ⊂N or N⊂ M.
Now if M ⊂N then MUN = N also if N⊂ M then M U N = M. in any of the case M U N is a subgroup of G
because M and N subgroups of G.
Main part: suppose M U N is a subgroup of G.
Now we prove that either M ⊂N or N⊂ M.
In a contrary way suppose that M⊄N and N⊄M
Since M⊄N, there exists x ∈M such that x ∉N, Similarly since N⊄M there exists y ∈N such that y ∉ M
But both x and y ∈M U N
Since M U N is a sub group, by closure law xy ∈MUN.
⟹either x.y ∈M or xy ∈N or xy ∈ both M and N
Suppose xy∈M, we have x∈M since M is group x-1 ∈M
Since xy∈M and x-1 ∈M, by closure law x-1 xy∈M
i.e. y∈M. which is a contradiction
M
xy∈M and
x-1∈M
Suppose xy∈N, we have y∈ 𝑁 since N is group y-1 ∈N
Since xy∈N and y-1 ∈N, by closure law xy y-1∈M
N
i.e. x∈N. which is a contradiction
in any case we are getting a contradiction.
xy∈N and
Hence our supposition is wrong.
y-1∈ 𝑁
∴ M ⊂N or N⊂ M is obvious.
Hence the theorem.
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Group Theory
Theorem 4: If H is any subgroup of a group G then H -1 = H
Proof: suppose H is a subgroup of group (G,.)
We have to prove that H-1 = H
We prove this by showing that H-1⊂ H and H ⊂ H-1
We know that H-1 = { x-1/x∈H}.
Let x ∈H, since H is a group by inverse law, x-1 ∈H.
∴ H-1⊂ H ( because by definition of H-1, x-1 ∈H-1)
Let x ∈H, since H is a group by inverse law, x-1 ∈H. but by definition of H-1, (x-1)-1 ∈H-1. i.e. x∈H-1.
∴ H ⊂ H-1
Hence H-1 = H.
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BVRICE
Group Theory
NORMAL SUB GROUPS
In this chapter, we consider G is a group under multiplication and H is any subgroup of G.
A sub group H of a group G is said to be a normal subgroup of G if ∀ x ∈ G and ∀ h ∈ H, xhx-1 ∈ H.
Note: 1) From the above definition we conclude that xHx-1 ⊂ G
2) for every group G, {e} and G itself are normal subgroups of G, they are called as improper or trivial
normal subgroups
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