Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

8
ADVANCED ABSTRACT ALGEBRA
Problem 1:
Let G be a non empty set closed under an associative product, which has left indentity e and left inverse for
all elements of g. show that G is a group.
Proof:
Let a ∈G and let b
such that b a = e. Now
b a b = (b a) b = e b = b ...................(i)
such that c b = e
Hence c (b a b) = cb = e from (i)
bgb g
cb a b = e
ab = e
∴ b is also right inverse of a.
Further,
a e = a (b a) = (a b) a = e a = a
Hence e is right identity also
Thus G is a group,
Subgroups
Let H be a non-empty subset of the group G such that
1.
Θ
∈a
≠∃
2. a − 1
∈H ∀a ∈H
We prove that H is a group with the same law of composition as in G.
Proof:
H is closed under multiplication from (1). All elements of H are from G and associative law holds in G,
therefore, multiplication is associative in H also.
Let a ∈ H , then a -1 from (2) and so from (1), a a -1 , i.e. e = a a -1 .
which implies, identity law holds in H, (2) gives inverse law in H. Thus H is a group. H is called a subgroup
of G. Thus a nonempty subset of a group G which is a group under the same law of composition is called a
subgroup G. Note that e, the identity element G is also the identity of H.
A group G is called nontrivial if G (e). A nontrivial group has at teast two subgroups namely G and (e).
Any other subgroup is called a proper subgroup.
Definition:
Let b, a
Problems:
G, b is said to be Conjugate of a
1. Let a ∈G, let C G
(a) = {x
2.
G: x -1 ax = a}
Prove that C G
(a) is a sbgroup of G.
G, if
is a sub group of G.
such that b = x -1 ax.
3. Find the centre of the group GL (2, R) of nonsingular 2 x 2 matrics over real numbers,
Zb