Normal Subgroups

A subgroups of is called a normal subgroup if for any aÎ G, aH = Ha.

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:

Every subgroup of an abelian group is normal.

Note3:

The trivial subgroups are also normal.

Note4:

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



Example 1:

Consider the symmetric group 3, à>.

Solution:

From the Table3 it is clear that {p₁, p₂}, {p₁, p₃}, {p₁, p₄} and {p₁, p_5, p₆} are subgroups of 3, à>. The left cosets of
{p₁, p₂} are {p₁, p₂}, {p₃, p₆} and {p₄, p₅}, while the right cosets of {p₁, p₂} are {p₁, p₂}, {p₃, p₅} and {p₄, p₅} and
hence {p₁, p₂} is not a normal subgroup. Similarly, {p₁, p₃}, {p₁, p₄} are not normal subgroups.

But the left and right cosets of {p₁, p₅, p₆} are {p₁, p₅, p₆} and {p₂, p₃, p₄}.

Hence, {p₁, p₅, p₆} is a normal subgroup.



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:

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 ^-1Î H.

Conversely, suppose for all a Î H and x Î G, xax ^-1Î 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 ^-1Hx 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:

Let and D> be groups and g : G ® H be a homomorphism. Then the kernel of g is a normal subgroup.

Proof:

We know that ker(g) is a subgroup of Now for any a Î G, and k Î ker(g),

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:

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:

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 the quotient set of G defined by H. If the subgroup
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:

Let H be a normal subgroup of a group (G, *), then (G/H, Ä ) is a group, called factor group, where aH Ä bH = (a * b)H.
Also, there exists a natural homomorphism from (G, *) onto (G/H, Ä ).

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^-1H such that aH Ä a ^-1H = (a * a ^-1)H = eH.

Thus, for every aH Î G/H, there exists inverse a ^-1H 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):

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:

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: h is well defined.

Suppose aK =bK for some a, b Î G, then a*k₁ = b*k_2, for some k₁, k₂ Î K.

Then, g(a * k₁) = g(a) D g(k₁)
= g(a) D e¢
= g(a)

Also, g(a * k₁) = g(b * k₂)
= g(b) D g(k₂)
= g(b) D e¢
= g(b).

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

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

Hence, the claim.



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))^-1D 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 · on G´ H is given by (g₁ , h₁) · (g₂ , h₂) = (g₁*g₂ , h₁D h₂), for any
(g₁ , h₁) , (g₂ , h₂) Î G´ H.

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 ^-1 , h ^-1).