Photo Credit:@linco
Last lecture I discuss about the introduction of group theory and some basic definition with explanation of the group theory.Now in lecture-2 I will discuss some important theorem and problems of group theory.

Inner Automorphism:

Let G be a group and a be a fixed element of G,then the automorphism f is defined by fa(g)=a-1ga for all g∈G,is called the inner automorphism determined by a.

Example:Identity Mapping.

Problem-1:Let G be a group and f:GG be defind by f(x)=x^-1 ∀ x∈G.Then f is an automorphism on G iff G is abelian.

Solution:
Let G be abelian then for all x,y∈G
xy=yx
Now we have to show that f is automorphism on G.
i)f is one-one:
f(x)=f(y)
-> x−1 =y−1
-> (x−1)−1=(y−1)−1
-> x = y
Hence f is one-one.
ii)f is onto:
f is onto since for x∈G we have x−1 ∈G
Also f(x−1)=f(y−1)
=x
Hence f is onto.
f is automorphism,Since
f(xy)=(xy)
= y−1 x−1
= x−1 y−1
=f(x)f(y)
Thus f is an isomorphism of G onto itself and hence automorphism.

Conversely suppose that f is an automorphism,we have to show that G is abelian group.

Abelian: f(xy)=f(xy)−1

                                   =y−1x−1 
                                   =f(yx)

i.e. xy=yx as f is one-one.Since xy=yx,then G is abelian where x and y are arbitrary elements of G.

Photo Credit:@linco

Problem-2:Consider (z,+) the additive group of integers and define G:zz by Q(x)=-x;for all x∈z then show that Q is automorphism.

Solution:
i)Q is Homomorphism:
For any m,n∈z we have,
Q(m+n)=-(m+n)
=-m-n
=Q(m)+Q(n)
ii)Q is one-one:Q is one-one since
Q(m)=Q(n)
-> -m=-n
-> m=n
iii)Q is onto:
For any m∈z there exists -m∈z such that,
Q(-m)=-(-m)=m
Therefore Q is onto.
Thus Q is isomorphism of z onto itself and so Q is an automorphism of z.

Theorem:Let G be a group,f is an automorphism N be a normal subgroup of G.Prove that f(N) is a normal subgroup of G.

Proof:Let x and y be two elements of f(N).So that x=f(x1) and y=f(x2),where x1,x2∈N,so x1,x2−1∈N
-> f (x1,x2−1)∈f(N)
-> f(x1) . f(x2−1)∈f(N)
-> f(x1).[f(x2)] −1∈f(N)
-> xy−1∈f(N)
Where x,y∈f(N).Therefore f(N) is a subgroup of G.Now let x∈G,h’∈f(N) so that h’=f(h),h∈N.Again x∈G implies x=f(y) for some y∈G as f is an automorphism.
xhx−1=f(y)f(h)[f(y) −1]
=f(y) f(h) f(y−1)
=f(yhy−1) ∈f(N)
As N is a normal subgroup of G y∈G,h∈N,yhy−1∈N.
Here x∈G,h’∈f(N) is a normal subgroup of G.

Photo Credit:@linco

Theorem:

For an abelian group ,the only automorphism is the identity mapping where as for non-abelian group,there exists non-abelian group,there exists non-trival automorphism.

Solution:

Let G be an abelian group and fa be an inner automorphism of G.
Then for all x∈G,we have,
fa(x)=a−1(xa)
= a−1(ax)
=ex
=x
This shows that fa is an identity mapping I on G.On the other hand ,if G is non-abelian then there exists at least two elements a,b∈G such that,
ab≠ba.
Now a,b∈G
-> a-1(ab) ≠a-1(ba)
-> a-1a)b≠a-1ba.
-> b≠a-1ba,where I stands for identity mapping.
Thus in this case fa is different from identity mapping I.Hence the theorem follows.
You can also check:
Theory Of Group: Lecture-1

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