A subgroups of is called a

A subgroups

**normal subgroup**if for any aÎ G, aH = Ha.

Note 1:

Note 1:

aH = Ha does not necessarily mean that a * h = h * a for every h Î H. It only means that a * h

_{i}= hj * a for some h

_{i}, hj Î H.

Note2:

Note2:

Every subgroup of an abelian group is normal.

**Note3:**

The trivial subgroups are also normal.

Note4:

Note4:

If H is a normal subgroup, then both the left and right costs of H in G are equal.

Example 1:

Example 1:

Consider the symmetric group

Solution:

Solution:

From the Table3 it is clear that {p

_{1}, p

_{2}}, {p

_{1}, p

_{3}}, {p

_{1}, p

_{4}} and {p

_{1}, p

_{5, }p

_{6}} are subgroups of

{p

_{1}, p

_{2}} are {p

_{1}, p

_{2}}, {p

_{3}, p

_{6}} and {p

_{4}, p

_{5}}, while the right cosets of {p

_{1}, p

_{2}} are {p

_{1}, p

_{2}}, {p

_{3}, p

_{5}} and {p

_{4}, p

_{5}} and

hence {p

_{1}, p

_{2}} is not a normal subgroup. Similarly, {p

_{1}, p

_{3}}, {p

_{1}, p

_{4}} are not normal subgroups.

But the left and right cosets of {p

_{1}, p

_{5}, p

_{6}} are {p

_{1}, p

_{5}, p

_{6}} and {p

_{2}, p

_{3}, p

_{4}}.

Hence, {p

_{1}, p

_{5}, p

_{6}} is a normal subgroup.

Theorem 3.4-1:

Theorem 3.4-1:

A subgroup H is normal in G if and only if for all a Î H and x Î G, xax

^{-1}Î H.

Proof:

Proof:

Given that H is normal, therefore xH = Hx, for all x Î G.

Therefore, xHx

^{-}

^{1}= Hx * x

^{-}

^{1}.

That is, xHx

^{-}

^{1}= H.

Hence, for all aÎ H and x Î G, xax

^{-}

^{1}Î

^{ }H.

Conversely, suppose for all a Î H and x Î G, xax

^{-}

^{1}Î

^{ }H, then xHx

^{-}

^{1 }Í

^{ }H …… (1)

Thus, xH Í Hx, for all x Î G ……(2)

By replacing x by x

^{-}

^{1}in (1), we have x

^{-}

^{1}Hx Í H.

Thus, Hx Í xH, for all x Î G …… (3)

From (2) and (3) we have xH = Hx.

Hence, H is normal.

Theorem 3.4-2:

Theorem 3.4-2:

Let

Proof:

Proof:

We know that ker(g) is a subgroup of

g(a

^{-1 }* k * a) = g(a

^{-1}) D g (k) D g(a).

= g(a

^{-1}) D e

_{H }D g(a).

= [g(a)]

^{-1 }D g(a) = e

_{H }.

Hence, a

^{-1}* k * a Î ker(g).

That is, ker(g) is a normal subgroup of

Theorem 3.4-3:

Theorem 3.4-3:

Let H be a normal subgroup of a group (G, *), then the relation "a º b(mod H)" is a congruence relation, where a º b(mod H)

iff b

^{-1 }* a Î H.

Proof:

Proof:

From Step 1 of the Lagrange’s Theorem, it is clear that the relation a º b(mod H) is an equivalence relation .

Now we shall prove that a º b(mod H) is congruence.

That is, we’ll prove that

if a º a¢ (mod H) and b º b¢ (mod H) then a * b º a¢ * b¢ (mod H).

Let a º a¢ (mod H) and b º b¢ (mod H).

Then by the definition of the relation,

(a¢ )

^{-1 }* a Î H and (b¢ )

^{-1 }* b Î H. Therefore,

(a¢ * b¢ )

^{-1}* (a * b) = ((b¢ )

^{-1}* (a¢ )

^{-1}) * (a * b)

= (b¢ )

^{-1}* ((a¢ )

^{-1}* a) * (b¢ * (b¢ )

^{-1}) * b

= ((b¢ )

^{-1}* ((a¢ )

^{-1}* a) * b¢ ) * (b¢ )

^{-1}* b

Since H is normal and (a¢ )

^{-1}* a Î H, we have (b¢ )

^{-1}* ((a¢ )

^{-1}*a) * b Î H.

Also, (b¢ )

^{-1}* b Î H, we have ((b¢ )

^{-1}* ((a¢ )

^{-1}* a) * b¢ ) * ((b¢ )

^{-1}* b)Î H.

By definition of a º b(mod H), we have a * b º (a¢ * b¢ )(mod H).

Hence, º (mod H) is a congruence relation.

Let G be a group and let H be a subgroup of G, then the relation a º b (mod H) is an equivalence relation. We denote

G/H ={[a] / a Î G}(the set of all equivalence classes defined by the relation "º (mod H)"). As each equivalence class [a] is

nothing but a left coset aH, we have G/H = {aH / a Î G}. G/H is called

**. If the subgroup**

*the quotient set of G defined by H*H is normal, then by Theorem 3.4-3 the relation "º (mod H)" is a congruence relation.

Therefore we can define a binary relation Ä on G/H, by aH Ä bH = (a * b)H.

The operation Ä is well defined since º (mod H) is a congruence relation.

[Suppose aH is equivalent to cH then, aH Ä bH = cH Ä bH, a coset can be replaced by an equivalent coset in the process of

operation. By doing so the result will not be changed and remains the same.]

Theorem 3.4-4:

Theorem 3.4-4:

Let H be a normal subgroup of a group (G, *), then (G/H, Ä ) is a group, called

**, where aH Ä bH = (a * b)H.**

*factor group*Also, there exists a natural homomorphism from (G, *) onto (G/H, Ä ).

Proof:

Proof:

We have already seen that the operation Ä is well defined.

Now, (aH Ä bH) Ä cH = ((a * b)H) *cH

= ((a * b) * c)H

= (a * (b * c))H

= aH Ä (b * c)H

= aH Ä (bH Ä cH)

Thus, Ä is associative.

Further, eH = H acts as an identity element of G/H.

For, H Ä aH = (e * a)H = aH = (a * e)H = aH Ä eH.

Also, for aH Î G/H, there exists a

^{-1}H such that aH Ä a

^{-1}H = (a * a

^{-1})H = eH.

Thus, for every aH Î G/H, there exists inverse a

^{-1}H in G/H.

Hence (G/H, Ä ) is a group.

Define g: G ® G/H , by g(a) = aH.

Then, g(a * b) = (a * b)H

= aH Ä bH

= g(a) Ä g(b)

Thus, g is a homomorphism.

For each aH Î G/H, there exists a Î G such that g(a) = aH.

Thus, g is onto.

Hence, the theorem.

Theorem 3.4-5 (First Isomorphism Theorem):

Theorem 3.4-5 (First Isomorphism Theorem):

Let g be a homomorphism from a group (G , *) to a group (H , D ) and let K be the kernel of g and H¢ Í H be the image set of G under the mapping g. Then G/H is isomorphic to H¢ .

Proof:

Proof:

Since K is a kernel of a homomorphism it must be a normal subgroup of G.

By Theorem 3.4-4, the mapping f from (G , *) into the factor group (G/H , Ä ) defined by f(a) = aK,

for all a Î G is an epimorphism.

Now we define h : G/H ® H¢ such that h(aK) = g(a).

Now we have the following structure of mapping.

Claim 1:

Claim 1:

*h is well defined.*

Suppose aK =bK for some a, b Î G, then a*k

_{1}= b*k

_{2, }for some k

_{1}, k

_{2}Î K.

Then, g(a * k

_{1}) = g(a) D g(k

_{1})

= g(a) D e¢

= g(a)

Also, g(a * k

_{1}) = g(b * k

_{2})

= g(b) D g(k

_{2})

= g(b) D e¢

= g(b).

Thus, g(a) = g(b).

That is, h(aK) = h(bK).

Hence, the claim.

Claim 2:

Claim 2:

*h is a homomorphism.*

h(aK Ä bK) = h((a * b)K)

= g(a * b)

= g(a) D g(b)

= h(aK) D h(bK), for all aK, bK Î G/H.

Hence, h is a homomorphism.

**Claim 3:**

*h is both one-one and onto.*

For y Î H¢ , there exists a Î G such that g(a) = y, since g is onto.

Hence, for each y Î H¢ , there exists a Î G such that y = g(a) = h(aK).

Thus, h is onto.

Suppose h(aK) = h(bK).

Then, g(a) = g(b).

Therefore, (g(b))

^{-1}D

^{ }g(a) = e¢ .

i.e., g(b

^{-1 }* a) = e¢ .

i.e., b

^{-1}* a Î K.

i.e., aK = bK (by definition of the relation a º b(mod K) )

Hence, h is one-one.

Hence, h is an isomorphism from G/H onto H¢ .

*Let (G , *) and (H , D ) be two groups. The direct product of these two groups is the algebraic structure (G*´

*H ,*

**·**

*) in*

which the binary operation

which the binary operation

**·**

*on G*´

*H is given by*

*(g*

_{1}, h_{1})**·**

*(g*

(g´

_{2}, h_{2}) = (g_{1}*g_{2}, h_{1}D_{ }h_{2}), for any(g

_{1}, h_{1}) , (g_{2}, h_{2}) Î G*H.*

It can be shown that (G´

It can be shown that (G

*H ,*

**·**

*) is a group and (e*´

_{G}, e_{H}) is the identity element of (G*H ,*

**·**

*) and inverse of any element*

(g , h) is (g

(g , h) is (g

^{-1}, h^{-1}).
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