    # Normal Subgroups

“I will fart in your general direction,” I muttered in a barely audible and intelligible voice, and I proceeded doing so against what the school etiquette demanded. [···]

The teacher looked at the sky, a cry of anguish for God’s mercy and help that was relentlessly begged, but never answered. “Obviously,” she thought, “it’s God’s revenge for being a committed and militant atheist, proud gay, and communist.” What an unforgiven decision to be a teacher - just three good reasons, June, July, and August!, Anawim, Apocalypse, #justtothepoint.

# Motivation

Let G be an arbitrary group, and let H be a subgroup of G, H ≤ G. A left coset of H in G is a subset of the form aH = {ah | h ∈ H} for some a ∈ G. Analogously, Ha = {ha | h ∈ H} is the right coset of H in G.

Definition. Let G be a group, H ≤ G a subgroup. The set of cosets of a subgroup H in a group G, written or denoted as G/H, is called the quotient set of G by H. G/H = {aH | a ∈ G}. Do the cosets form a group?

If we want the set of cosets to form a group, then given two cosets, say xH and yH, their product need to be a coset, (xH)(yH) = xyH. Let’s see when this happens. Let’s take two representatives, x·n1∈ xH, y·n2 ∈ yH.

(x·n1)·(y·n2) ∈ xyH ⇒ (x·n1)·(y·n2) = x· y·n3
x-1·(x·n1)·(y·n2) = x-1·x· y·n3
n1·y·n2 = y·n3
y-1·n1·y·n2 = y-1·y·n3
y-1·n1·y·n2 = n3
y-1·n1·y = n3·n2-1 ∈ H for any y ∈ G.

Therefore, H ≤ G and we want the set of cosets to form a group, then we need that y-1·n1·y = n3·n2-1 ∈ H for any y ∈ G ↭[n3, n2∈ H, H ≤ G ⇒ n3·n2-1 ∈ H] ∀ y ∈ G, n1 ∈ H, y-1·n1·y ∈ H When this condition is true, we say that H is a normal subgroup of G.

# Definition

Definition. A subgroup H of a group G, H ≤ G, is called a normal subgroup of G if aH = Ha, ∀a ∈ G. That is, a normal subgroup of a group G is one in which the right and left cosets are precisely the same. The usual notation for this relation is H ◁ G. It means ∀a ∈ G, h ∈ H, ∃h’, h’‘∈ H: ah = h’a and ha = ah’’.

Normal Subgroup Test or alternative definition. A subgroup H of a group G is normal in G if and only if is invariant under conjugation by members of the group G. H ◁ G iff gHg-1 ⊆ H, ∀g ∈ G, i.e., ∀g ∈ G, h ∈ H, ghg-1 ∈ H.

Actually, for any arbitrary subgroup H of G, the set ghg-1 will be another subgroup of G, so when H is normal in G, then ghg-1 will be a subgroup of H. In fact, it will be equal to H.

Proof:

⇒) H ◁ G ⇒ ∀a ∈ G, ∀h ∈ H, ∃h’ ∈ H: ah = h’a ⇒ a-1h’a = h ∈ H ⇒ ∀a ∈ G, aHa-1 ⊆ H

⇐) ∀g ∈ G, gHg-1 ⊆ H ⇒ [g = a] aHa-1 ⊆ H ⇒ ∀a ∈ G, aH ⊆ Ha

∀g ∈ G, gHg-1 ⊆ H ⇒ [g = a-1] a-1Ha ⊆ H ⇒ ∀a ∈ G, Ha ⊆ aH ⇒[We have previously demonstrated that aH ⊆ Ha] aH = Ha ⇒ H ◁ G ∎

Theorem. Let G be a group, N ≤ G, the following statements are equivalent.

1. N ◁ G, ∀a∈G, aH = Ha ↭ ∀a ∈ G, h ∈ H, ∃h’, h’‘∈ H: ah = h’a and ha = ah’’.
2. gHg-1 ⊆ H, ∀g ∈ G, i.e., ∀g ∈ G, h ∈ H, ghg-1 ∈ H
3. gHg-1 = H, ∀g ∈ G

Proof. We only need to prove (2) ⇒ (3)

Suppose, gHg-1 ⊆ H, H ⊆ gHg-1?

∀h ∈ H, g ∈ G, ghg-1 ∈ H ⇒ ∃h’ ∈ H: ghg-1 = h’ ⇒ h = g-1h’g = g-1h’(g-1)-1 ⇒ ∀h ∈ H, h = g-1h’(g-1)-1 ∈ g-1H(g-1)-1 ⇒ ∀g ∈ G, H ⊆ g-1H(g-1)-1, and we know (2) g-1H(g-1)-1 ⊆ H ⇒ ∀g ∈ G, g-1H(g-1)-1 = H ↭ ∀g ∈ G, (g-1)-1Hg-1 = H ↭ ∀g∈ G, gHg-1 = H∎

Examples:

• Every subgroup of an Abelian group is normal. It is pretty obvious, ∀a ∈ G, h ∈ H: ah = ha and ha = ah. It will always the case that aH = Ha. Besides, the trivial group {e} and the group itself G are normal, too.

Recall. The index of a subgroup H (H ≤ G) is the number of its distinct left cosets, or equivalently, the number of its distinct right cosets. It is denoted as |G : H|. When G is a finite group, the following equation is satisfied |G : H| = |G|/|H|.

• A subgroup of index 2 is always normal, i.e., |G : H| = 2 ⇒ H ◁ G.

If |G : H| = 2 ⇒ There are only two distinct left cosets, say H and aH (a ∉ H), and two distinct right cosets, say H and Hb (b ∉ H). Since H and aH makes a partition of G ⇒ they are obviously disjoint. The same applies to H and Hb ⇒ aH = Hb, and this coset contains all the elements of G that are not in H, a ∈ G \ H ⇒ aH = Ha = Hb = bH ⇒ H ◁ G.

• Definition. The center of a group, G, is the set of elements that commute with all the elements of G. Z(G) = {g ∈ G | gx = xg, ∀x ∈ G} ≤ G. Futhermore, the center Z(G) of a group is always normal.
• Recall. D3 is the symmetry group of the equilateral triangle. D3 = {e, r (rotation 120° around the center anticlockwise), r2, s, sr, sr2} = ⟨r, s | r3 = s2 = e, rs = sr2⟩.

The subgroup of rotations ⟨r⟩ = {e, r, r2} is a cyclic subgroup of order three. It is also normal, ⟨r⟩ ◁ D3.

Proof:

∀a ∈ G, aH = Ha? We only really need to check this equation for all a ∈ D3 that are not in the subgroup, but even better as D3 = ⟨r, s⟩, we only need to check it for the generator that is not part of the subgroup ⟨r⟩, that is, s.

s⟨r⟩ = {s, sr, sr2}

⟨r⟩s = {s, rs, r2s}, but we know that rs = sr2. Futhermore, recall that rks = srn-k, 1 ≤ k ≤ n-1 ⇒ [n = 3, k = 2] r2s = sr ⇒ s⟨r⟩ = ⟨r⟩s ⇒ ⟨r⟩ ◁ D3. Futhermore, D3/⟨r⟩ = {⟨r⟩, s⟨r⟩} ≋ ℤ2. We can define this isomorphism ⟨r⟩ → 0, and s⟨r⟩ → 1.

⟨r⟩ s⟨r⟩
⟨r⟩ ⟨r⟩ s⟨r⟩
s⟨r⟩ s⟨r⟩ ⟨r⟩ -ss⟨r⟩ = s2⟨r⟩ = e⟨r⟩ = ⟨r⟩-
• Counterexample. Let D3 = {e, r, r2, s, sr, sr2}, the subgroup ⟨s⟩ = {e, s} is not a normal subgroup of D3.

r⟨s⟩ = {r, rs} = {r, sr2} ≠ ⟨s⟩r = {r, sr} because rs ≠ r, rs ≠ sr.

• The symmetric group of order three S3 = {(), (12), (13), (23), (123), (132)} has a normal subgroup of order 3, but its subgroups of order 2 are not normal.

The alternating group of even permutations A3 = {(), (123) -(13)(12)-, (132) -(12)(13)-} is a normal subgroup of S3, A3 ◁ S3, e.g., (12)A3 = {(12), (12)(123), (12)(132)} = {(12), (23), (13)} = A3(12), even though (12)(123) = (23) ≠ (123)(12) = (13). The cosets are A3 = {(), (123), (132)} and (12)A3 = {(12), (23), (13)}. The factor group S3/A3 has the following multiplication table: (and is isomorphic to ℤ2)

A3 (12)A3
A3 A3 (12)A3, e.g., (123)(12) = (13) ∈ (12)A3
(12)A3 (12)A3 A3, e.g., (12)(12)=()

H < S3 = {(1), (12)} is not normal because (123)H = {(123)(1), (123)(12)} = {(123), (13)} ≠ H(123) = {(1)(123), (12)(123)} = {(123), (23)}

• A4 the alternating group of even permutations of S4 = {(), (1, 2, 3), (1, 3, 2), (1, 2, 4), (1, 4, 2), (1, 3, 4), (1, 4, 3), (2, 3, 4), (2, 4, 3), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}. H = {α1, α2, α3, α4} = {(), (12)(34), (13)(24), (14)(23)} is defined as the subgroup comprising the identity element and the three double transpositions, |H| = 4.

H = ()H = (12)(34)H = (13)(24)H = (14)(23)H = H(1) = H(12)(34) = H(13)(24) = H(14)(23)

(123)H = {(123), (134), (243), (142)}. (123)H = (134)H = (243)H = (142)H = H(123) = H(134) = H(243) = H(142)

(123)(12)(34) = (134), (123)(13)(24) = (243), (123)(14)(23) = (142)

(132)H = {(132), (234), (124), (143)}. (132)H = (234)H = (124)H = (143)H = H(132) = H(234) = H(124) = H(143). And therefore, H ◁ A4 ∎ In general, An ◁ Sn

For any subgroup H of G, the set gHg-1 is always another subgroup of G. However, when H is normal, then gHg-1 is a subgroup of H. In fact, it is H itself.

• Consider the general linear group $GL(2, ℝ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ, ad - bc ≠ 0}} \bigr\}$ is defined as the group of invertible 2 x 2 matrices (non-zero determinant) with entries in ℝ and with the group operation being matrix multiplication. The group SL(2, ℝ) of 2 x 2 matrices with determinant 1 is a normal subgroup of GL(2, ℝ), SL(2, ℝ) ◁ GL(2, ℝ).

∀x ∈ GL(2, ℝ), h ∈ SL(2, ℝ), xhx-1 ∈ SL(2, ℝ)? det(xhx-1) =[h ∈ SL(2, ℝ) ⇒ det(h) = 1] det(x)det(h)det(x)-1 = det(x)det(x)-1 = 1 ⇒ xhx-1 ∈ SL(2, ℝ)∎

# Factor Groups

Theorem. If H is a normal subgroup of a group G, H ◁ G, then the set of left (or right) cosets of H in G form a group G/H = {aH | a ∈ G} ≤ G under the operation (aH)(bH) = abH. This group is called the factor or quotient group of G and H of order [G : H]. It comes with a natural group homomorphism Φ: G → G/H such that Φ is onto and Ker(Φ)= H.

Proof:

1. It is well defined, that is, the group multiplication must be independent of the choice of coset representative. Let aH = a’H, bH = b’H. Claim: (aH)(bH) = (a’H)(b’H) ↭ abH = a’b’H. aH = a’H, bH = b’H ⇒[aH = bH ↭ a ∈ bH] ∃h1, h2: a = a’h1, b = b’h2, and therefore, abH = a’h1b’h2H =[aH = H ↭ a ∈ H, h2 ∈ H] a’h1b’H ⇒ [H ◁ G, b’H = Hb’] a’h1Hb’ =[aH = H ↭ a ∈ H, h1 ∈ H] = a’Hb’ =[H ◁ G, b’H = Hb’] a’b’H.
2. eH = H is the identity.
3. a-1H is the inverse of aH, (aH)(a-1H) = aa-1H = eH = H = eH = (a-1a)H = (a-1H)(aH).
4. (aHbH)cH = (ab)HcH = (ab)cH = a(bc)H = aH(bc)H = aH(bHcH)∎

The converse of this theorem is also true, that is, if the operation (aH)(bH) = abH defines a group on the set of left (or right) cosets of H in G, then H is a normal subgroup.

# Examples

• Consider the normal subgroup 3ℤ = {0, ±3, ±6, …} of ℤ. The cosets of 3ℤ in ℤ are:

0 + 3ℤ = 3ℤ = {0, ±3, ±6, ±9, …}
1 + 3ℤ = {1, 4, 7, …; -2, -5, -8, …}
2 + 3ℤ = {2, 5, 8, …; -1, -4, -7, …}

The group ℤ/3ℤ is given by the Cayley table below.

+ 0 + 3ℤ 1 + 3ℤ 2 + 3ℤ
0 + 3ℤ 0 + 3ℤ 1 + 3ℤ 2 + 3ℤ
1 + 3ℤ 1 + 3ℤ 2 + 3ℤ 0 + 3ℤ
2 + 3ℤ 2 + 3ℤ 0 + 3ℤ 1 + 3ℤ

• Consider the normal subgroup 4ℤ = {0, ±4, ±5, …} of ℤ. The cosets of 4ℤ in ℤ are:

0 + 4ℤ = 4ℤ = {0, ±4, ±8, ±12, …}
1 + 4ℤ = {1, 5, 9, 13, …; -3, -7, -11, …}
2 + 4ℤ = {2, 6, 10, 14, …; -2, -6, -10, …}
3 + 4ℤ = {3, 7, 11, 15, …; -1, -5, -9, …}

(1 + 4Z)2 = (1 + 4Z) + (1 + 4Z) = 2 + 4Z
(1 + 4Z)3 = (1 + 4Z) + (1 + 4Z) + (1 + 4ℤ) = 3 + 4Z
(1 + 4Z)4 = (1 + 4Z) + (1 + 4Z) + (1 + 4ℤ) + (1 + 4ℤ) = 4ℤ ⇒ ℤ/4ℤ = {0 + 4ℤ, 1 + 4ℤ, 2 + 4ℤ, 3 + 4ℤ} = ⟨1 + 4ℤ⟩. The factor group is a cyclic group of order 4. The natural homomorphism is Φ: ℤ → ℤ/4ℤ, Φ(m) = m + 4ℤ = r + 4ℤ where m = 4p +r, r is the remainder, 0 ≤ r < 4. Φ is onto and Ker(Φ) = 4ℤ.

There are no other cosets because if k ∈ ℤ, ∃q, r: 0 ≤ r < 4, k = 4q + r ⇒ k ∈ r + 4ℤ for some r, 0 ≤ r < 4.

This group is given by the Cayley table 1.a below. Notice that ℤ/4ℤ ≋ ℤ4. In general, ∀n > 0, nℤ ◁ ℤ. The cosets of ℤ/nℤ are: nℤ = {0, ±n, ±2n, ±3n,…}, 1 + ℤ, 2 + nℤ,…, (n-1) + nℤ. The sum of the coset (k + nℤ) + (l + nℤ) = (k + l) + nℤ. Notice that we have written it additively. Futhermore, ℤ/nℤ ≋ ℤn.

• ℤ/2ℤ = {2ℤ, 1 + 2ℤ} = {e, a} ≋ ℤ2, e.g., e + a = (2ℤ) + (1 +2ℤ) = (0 + 2ℤ) + (1 + 2ℤ) = 1 + 2ℤ = a. a + a = (1 + 2ℤ) + (1 + 2ℤ) = 2 + 2ℤ = 2ℤ = e.
e a
e e a
a a e
Φ: ℤ → ℤ/2ℤ = {e, a}, Φ(m) = m + 2ℤ = $\begin{cases} e, &m~is~ even \\ a, &m~is~ odd \end{cases}$

Φ is onto and Ker(Φ) = {m ∈ ℤ: Φ(m) = e} = {even integers} = 2ℤ.

• G = ℤ18, H = ⟨6⟩ = {0, 6, 12}. Then G/H = {0 + H = H, 1 + H, 2 + H, 3 + H, 4 + H, 5 + H}, e.g., (5 + H) + (4 + H) = 9 + H =[aH = bH ↭ a-1b ∈ H ↭Additive Notation -a + b ∈ H ↭ -3 + 9 ∈ H] 3 + H
• Let’s consider the factor group of the dihedral group of D4 and $K$ = {R0, R180}. $K$ ◁ D4.

$D_4/K$ = {$K, R_{90}K, HK, DK$} Its multiplication table is given in Table 2.3. Observe that the multiplication table for D4 can be simplified into small boxes that are cosets of $K$. We can represent an entire box by a single element of the box! # Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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