Compatible relations and normal subgroups

Let G a group and ρ an equivalence relation defined on G, compatible with the operations of the group, i.e. such that

Remembering the definitions of ρ_r, right and ρ_l, left congruence modulo a subgroup, we introduce the following proposition:

7.12.1 Proposition. Let G a group and ρ an equivalence compatible relation defined on G. Let also

1. H is a subgroup of G

2. Indicating with ρ_r and ρ_l, the right and left congruence relations modulo H, it results

   ρ = ρ_r = ρ_l

   That is

   x ρ y   ⇐⇒   xy⁻¹ ∈ H   ⇐⇒   y⁻¹x ∈ H

Proof.

1. We have

   e ρ e  ⇒   e ∈ H

   Moreover ∀x,y ∈ H

   x ρ e,   y ρ e  ⇒   xy ρ ee = e  ⇒   xy ∈ H

   And

   x ρ e,   x⁻¹ ρ x⁻¹  ⇒   e ρ x⁻¹  ⇒   x⁻¹ ∈ H

2. x ρ y and y⁻¹ ρ y⁻¹ imply owing to the compatibility of ρ: xy⁻¹ ρ e, that is x ρ_r y. Conversely, if x ρ_r y then xy⁻¹ ∈ H, that is xy⁻¹ ρ e, from which for the compatibility of ρ: x ρ y. Thus x ρ y ⇒   x ρ_r y. The same argument can be applied to ρ_l.  □

The previous proposition states that given in G an equivalence relation compatible with the operations of the group, the partition in equivalence classes coincides with the partition in right (left) cosets modulo the subgroup H with this latter representing the class of the nuetral element of the group. Moreover left and right cosets coincide i.e. Hx = xH, ∀x ∈ G.

We now ask the converse given a subgroup H of G, when the relations ρ_r, ρ_l modulo H are compatible? It is not generally the case that the two relation are compatible as the following examples shows off:

7.12.2 Example. Let G = 𝓢₃ and H the subgroup H = ⟨(23)⟩.

For ρ_r we have: (13) ρ_r (123), (12) ρ_r (132), (e) ρ_r (23)

For ρ_l we have: (13) ρ_l (132), (12) ρ_l (123), (e) ρ_l (23).

Neither of the two relations are compatible; We've to verify that for example

But this is not the case since:

Observe also that ρ_r ≠ ρ_l. Indeed there exist elements x,y such that xy⁻¹ ∈ H but yx⁻¹ ∉ H. It suffices to take for example x = (132) and y = (12). We have xy⁻¹ = (23) ∈ H, but y⁻¹x = (13) ∉ H.   ■

The following proposition clarify under which additional condition ρ_r and ρ_l are compatible.

7.12.2 Proposition. Let H a subgroup of G, and let ρ_r and ρ_l the right and left congruence relation modulo H. The necessary and sufficient condition for ρ_r (or ρ_l) to be compatile is

Proof. We observe first that by multiplying to the right (or left) by a same element of G two elements still congruent modulo ρ_r (ρ_l). Let ρ_r = ρ_l = ρ and let x ρ x' and y ρ y'. Then we have xy ρ x'y, from which

that is ρ_r (= ρ_l) is compatible.
Conversely suppo that ρ_r is compatible. Then x ρ_r y  ⇐⇒   xy⁻¹ ρ_r e. But then, owing to the compatibility of ρ_r, we have y⁻¹ ρ_r x⁻¹, that is y⁻¹(x⁻¹)⁻¹ = y⁻¹x ∈ H, thus x ρ_r y  ⇐⇒   x ρ_l y and ρ_r = ρ_l.   □

7.12.3 Corollary. Let ρ an equivalence relation compatible defined on G. Then the subset H = {x ∈ G | x ρ e} is a subgroup of G and it holds that ρ = ρ_r = ρ_l modulo H. Conversely, if H is a subgroup of G such that ρ_r = ρ_l, then ρ_r = ρ_l = ρ is a compatible relation.   □

Quotient groups and Normal subgroups

With the equivalence relations defyining right and left cosets 7.8.1 and 7.8.2, we denote the totality of all equivalence classes modulo H by G/H, known as the quotient set of the group G over its subgroup H. This set is the set of cosets G/H = {gH| g ∈ G}. Analogously, the set of right cosets G/H = {Hg | g ∈ G}.

An interesting question is under what conditions on the subgroup H the quotient set G/H becomes a quotient group.

The subgroup H of G such that ρ_r (and thus also ρ_l) is compatible, and for which the right and left cosets coincide are known as normal.

7.12.4 Defintion. A subgroup H of G is said normal in G if ∀x ∈ G

We write H ⊲̲ G. In group theory, (big group)/(subgroup) nearly always means the quotient group, meaning that the subgroup is normal.

Every subgroup of an abelian group is normal, hence all congruence relations modulo a certain subgroup are compatible. In addition every group has two trivial normal subgroups itself and the subgroup reduced to the identity element.

For example, in S₃ the alternating subgroup is a normal subgroup whilst the subgroup {id, (12)} is not.

7.12.5 Example. Let H = A₃ = {(e), (1, 2, 3), (1, 3, 2)} = ⟨(1, 2, 3)⟩

and

G = S₃ = {(e), (1, 2, 3), (1, 3, 2), (1, 2), (1, 3), (2, 3)}.

For x = (1, 2) we have

xH= {(1, 2)(1), (1, 2)(1, 2, 3), (1, 2)(1, 3, 2)} = {(1, 2), (2, 3), (1, 3)}

and

Hx = {(1)(1, 2), (1, 2, 3)(1, 2), (1, 3, 2)(1, 2)} = {(1, 2), (1, 3), (2, 3)}.

Note that the condition xH = Hx is an equality of sets, and it does not require that xh ≠ hx for all h ∈ H. We have xH = Hx, but xh ≠ hx when h = (1, 2, 3) ∈ H. Similar computations show that

(e)H = (1, 2, 3)H = (1, 3, 2)H = {(1), (1, 2, 3), (1, 3, 2)} = H

H(e) = H(1, 2, 3) = H(1, 3, 2) = {(1), (1, 2, 3), (1, 3, 2)} = H

(1, 2)H = (1, 3)H = (2, 3)H = {(1, 2), (1, 3), (2, 3)}

H(1, 2) = H(1, 3) = H(2, 3) = {(1, 2), (1, 3), (2, 3)}.

Thus H is a normal subgroup of G. Additionally we note that G can be expressed as G = H ∪ (1, 2)H.   ■

Remark. In the definition of normal subgroup we have specified normal in G. Indeed a subgroup N may be normal in a subgroup H but not in the whole G. The previous corollary states that all and the only compatible relations defined in a group G are the relations congruent modulo a normal subgroup. If we set

𝓡 := {equivalence relations defined in G, compatible with the operations in G}

𝓝 := {normal subgroups in G}

The mapping

is a bijective function between 𝓡 and 𝓝, whose inverse is

where x ρ y ⇐⇒ xy⁻¹ ∈ N.

The role that ideals play in ring theory is played by normal subroups in group theory. ■

As we did for the case of S_n we introduce the definition of conjiugate elements.

7.12.6 Defintion. Let G a group. Two elements x,y ∈ G are said conjugates if there exists an element g ∈ G such that

Every element of the set gH is equal to some element of Hg, i.e. gh_i = h_jg or h_i = g⁻¹h_jg.

Conjugacy is an equivalence relation, since it is

1. Reflexive: x = e⁻¹xe for any x ∈ G (x is conjugate to itself).

2. Symmetric: if y = g⁻¹xg, then x = (g⁻¹)⁻¹yg⁻¹.

3. Transitive: if y = g⁻¹xg and z = h⁻¹ yh, then z = (gh)⁻¹ x(gh).

Hence, G is the disjoint union of the equivalence classes, which are called conjugacy classes.

Given a subgroup H of G we indicate by xHx⁻¹, the set

It is easy to show that it is still a subgroup of G, known as Conjugated subgroup of H, and may be also indicated with H^x. It coincides with H if and only if H is a normal subroup of G.

We now give a list of equivalence definition of normal subgroup

Proposition 7.12.7. Let N a subgroup of a group G. Then the following propositions are equivalent

1. N is normal in G

2. N^x = N ∀x∈G

3. xnx⁻¹ ∈ N, ∀x ∈ G and ∀n ∈ N

4. N is the uniion of conjugate classes of G.

Proof. b) xNx⁻¹ = N if and only if xN = Nx, for one direction, we multiply on the right by x, and, for the converse we multiply on the right by x⁻¹.
The condition for normality x⁻¹Nx = N for all x ∈ G, says that x⁻¹nx ∈ N for all n ∈ N and all x ∈ G; that is, for every element of N, its entire conjugacy class is contained in N.  □

Definition 7.12.8. Let N be a normal subgroup in the group G. The factor group, or quotient group, G/N

is the set of all left (or right) cosets of N ∈ G. Our first goal is to define an operation on right cosets so that G/N becomes a group. We define it as follows

Example. If G = {e,a,a²} = C₃, and N = {e} then G/{e} = {{e}, {a}, {a²}}.

7.12.9 Proposition. The quotient group G/N, as defined above, is indeed a group.

Proof. First we have to verify that the definition is well-defined. Let Nx = Nx' (that is x' = nx), Ny = Ny' (that is y' = n'y), where n,n' ∈ N, x,y ∈ G: we must prove that Nxy = Nx'y'. Indeed x'y' = nxn'y = nn̅xy. Where the last equality holds because xn' ∈ xN = Nx, and so xn' is equal to n̅x for some n̅ ∈ N. But then Nx'y' = Nxy, as required.

Group axioms, are routine and are left to the reader to verify. The closure law needs no proof, since the operation is well defined. For the associative law, we have

and the right-hand sides are equal, by the associative law for G. Because (aN)(eN) = (ae)N = aN = (ea)N = (eN)(aN), we see that eN = N is the identity element in G/N. Finally the inverse, (a⁻¹N)(aN) = (a⁻¹a)N = eN = (aa⁻¹)N = (a N)(a⁻¹N) shows that a⁻¹N = (aN)⁻¹.  ■

The quotient group comes as the image of a natural homomorphism, which (as in the case of rings) is called the canonical homomorphism. Recall that the elements of G/N are the cosets of N in G.

Proposition 7.12.10 If φ: G → G' is any group homomorphism, then ker φ is a normal subgroup of G.

Proof. We already know that ker φ is a sugroup of G; to verify it is normal notice that

proving that gng⁻¹ ∈ ker φ.   □

We are now going that not only kernel ⇒ Normal, but that the converse is also true leading to kernel ⇐⇒ Normal subgroup. Define a map

Checking that π is a homomorphism is straightforward from the definition of the operation in G/N.

The image of π is G/N, since every coset has the form Ng for some g ∈ G.

For the kernel of π, since the identity element of G/N is the coset e_GN, that is, N itself, we have

Before we had noticed that every kernel (of a group homomorphism) is a normal subgroup; and now we have verified that every normal subgroup is in fact a kernel (of some group homomorphism). We conclude that

in group theory, 'kernel' and 'normal subgroup' are equivalent concepts.

Furthermore, every subgroup in an abelian group is the kernel of some homomorphism: yet another indication that life is simpler in Abelian groups.

7.12.12 Proposition. Every subgroup of index 2 of a group is a normal subgroup

Proof. To see this, suppose H is a subgroup of G such that [G : H] = 2. One coset is H itself: if x happens to be in H, then xH = H and Hx = H ⇒ xH = Hx. Now let a and element of G that is not contained in H. Then the left and right coset decompositions of G by H yield

Accordingly, aH = Ha or aHq⁻¹ = H, i.e. H is a normal subgroup of G.   □