“Some drink at the fountain of knowledge. Others just gargle. You are just a moron stuck to a nose,” I retorted, but my imagination kept running wild, “You do not have a nose, a superlative muzzle owns you, a crooked sundial casting a disproportionate shadow, an elephant dying a terrible death, face up, gasping from breath; the Great Pyramid of Giza or more accurate, all Egyptian pyramids pile together; the king of swordfish, an insult to nature, and a joke to many,” Anawim, Apocalypse, #justtothepoint.

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

- Let G = ℤ
_{8}= {0, 1, 2, 3, 4, 5, 6, 7}, and let H be the subgroup {0, 4} ◁ G. G/H = {H, 1 + H, 2 + H, 3 + H}.

+ | H = {0, 4} | 1 + H = {1, 5} | 2 + H = {2, 6} | 3 + H = {3, 7} |
---|---|---|---|---|

H | H | 1 + H | 2 + H | 3 + H |

1 + H | 1 + H | 2 + H | 3 + H | H |

2 + H | 2 + H | 3 + H | H | 1 + H |

3 + H | 3 + H | H | 1 + H | 2 + H |

To calculate (2 + H) + (3 + H), let’s take two representatives 2 + 7 = 9 =_{mod 8} 1 ∈ 1 + H ⇒ (2 + H) + (3 + H) = 1 + H. There are only two groups of order 4, namely ℤ_{4} (cyclic) and ℤ_{2} x ℤ_{2}. Besides,

- G/H = ⟨1 + H⟩, cyclic, hence G/H ≋ ℤ
_{4}. - Replace H = {0, 4} with 0, 1 + H = {1, 5} with 1, 2 + H = {2, 6} with 2, 3 + H = {3, 7} with 3.

- Let G = ℤ
_{4}x ℤ_{4}, H = ⟨(1, 3)⟩, ℤ_{4}x ℤ_{4}/⟨(1, 3)⟩ = {⟨(1, 3)⟩, (0, 1) + ⟨(1, 3)⟩, (0, 2) + ⟨(1, 3)⟩, (0, 3) + ⟨(1, 3)⟩} where

H = ⟨(1, 3)⟩ = {(0, 0), (1, 3), (2, 2), (3, 1)}

(0, 1) + ⟨(1, 3)⟩ = {(0, 1), (1, 0), (2, 3), (3, 2)}

(0, 2) + ⟨(1, 3)⟩ = {(0, 2), (1, 1), (2, 0), (3, 3)}

(0, 3) + ⟨(1, 3)⟩ = {(0, 3), (1, 2), (2, 1), (3, 0)}

Some calculations: (1, 3) + (1, 3) = (2, 6_{mod 4}) = (2, 2); 3·(1, 3) = (3, 1); 4·(1, 3) = (0, 0) ⇒ H = ⟨(1, 3)⟩ = {(0, 0), (1, 3), (2, 2), (3, 1)}.

Let’s add cosets, e.g., (0, 2) + ⟨(1, 3)⟩ + (0, 1) + ⟨(1, 3)⟩ =[Let’s select two representatives] (0, 2) + (0, 1) + ⟨(1, 3)⟩ = (0, 3) + ⟨(1, 3)⟩

(0, 0) + ⟨(1, 3)⟩ | (0, 1) + ⟨(1, 3)⟩ | (0, 2) + ⟨(1, 3)⟩ | (0, 3) + ⟨(1, 3)⟩ | |
---|---|---|---|---|

(0, 0) + ⟨(1, 3)⟩ | (0, 0) + ⟨(1, 3)⟩ | (0, 1) + ⟨(1, 3)⟩ | (0, 2) + ⟨(1, 3)⟩ | (0, 3) + ⟨(1, 3)⟩ |

(0, 1) + ⟨(1, 3)⟩ | (0, 1) + ⟨(1, 3)⟩ | (0, 2) + ⟨(1, 3)⟩ | (0, 3) + ⟨(1, 3)⟩ | (0, 0) + ⟨(1, 3)⟩ |

(0, 2) + ⟨(1, 3)⟩ | (0, 2) + ⟨(1, 3)⟩ | (0, 3) + ⟨(1, 3)⟩ | (0, 0) + ⟨(1, 3)⟩ | (0, 1) + ⟨(1, 3)⟩ |

(0, 3) + ⟨(1, 3)⟩ | (0, 3) + ⟨(1, 3)⟩ | (0, 2) + ⟨(1, 3)⟩ | (0, 1) + ⟨(1, 3)⟩ | (0, 0) + ⟨(1, 3)⟩ |

Futhermore, the coset |(0, 1) + ⟨(1, 3)⟩| = 4 ⇒ ℤ_{4} x ℤ_{4}/⟨(1, 3)⟩ = {⟨(1, 3)⟩, (0, 1) + ⟨(1, 3)⟩, (0, 2) + ⟨(1, 3)⟩, (0, 3) + ⟨(1, 3)⟩} ≋ ℤ_{4}

- Let G = ℤ
_{4}x ℤ_{8}, and H = ⟨(0, 1)⟩ = {(0, 0), (0, 1), (0, 2), . . . , (0, 7)}. |G : H| = 4·8/4 = 4 ⇒ G/H ≋ ℤ_{4}or ℤ_{2}x ℤ_{2}.

(x, y) + H = (v, w) + H ↭ (x, y) - (v, w) = (x - v, y -w) ∈ H ↭ x = v ⇒[Each coset contains a unique element (x, 0) where x ∈ ℤ_{4})] G/H = {H, (1, 0) + H, (2, 0) + H, (3, 0) + H}. G/H = {H, (1, 0) + H, (2, 0) + H, (3, 0) + H} ≋ ℤ_{4} because there is an isomorphism Φ: ℤ_{4} → G/H, x → (x, 0) + H or, alternatively, we can easily check that |(1, 0) + H | = 4, then G/H = ⟨(1, 0) + H⟩

- Let G = A
_{4}the alternating group of even permutations of S_{4}= {() -α_{1}= 1-,(12)(34) -α_{2}= 2-, (13)(24) -α_{3}= 3-,(14)(23) -4-,(123) -5-, (243)-6-, (142) -7-, (134) -8-, (132) -9, (143) -10-,(234) -11-, (124) -12-}. Let H = {α_{1}, α_{2}, α_{3}, α_{4}} = {(), (12)(34), (13)(24), (14)(23)}, H ◁ A_{4}.

The cosets of H are: H; 5H = {α_{5}, α_{6}, α_{7}, α_{8}}, α_{5}α_{1} = α_{5}; α_{5}α_{2} = (123)(12)(34) = (134) = α_{8}; α_{5}α_{3} = (123)(13)(24) = (243) = α_{6}; α_{5}α_{4} = (123)(14)(23) = (142) = α_{7}. Analogously, 9H = {α_{9}, α_{10}, α_{11}, α_{12}}. Example: (α_{5}α_{1})·(α_{9}α_{1}) = (123)()(132)() = () ⇒ 5H·9H = H, (α_{5}α_{1})·(α_{5}α_{1}) = (123)()(123)() = (132)() ⇒ 5H·5H = 9H, and so on.

We normally use e = () for simplicity’s sake. (α_{5}α_{2})·(α_{9}α_{3}) = (123)(12)(34)(132)(13)(24) = (12)(34) = α_{2} ⇒ 5H·9H = H

- Let G = U(9) = {1, 2, 4, 5, 7, 8}. N = {1, 8}, N ◁ G. |G/N| = |G|/|N| = 6/2 = 3. G/N = {N, 2N, 4N} ≋ ℤ
_{3}where 2N = {2, 7} and 4N = {4, 5}.

2N = {2·1_{mod 9}, 2·8_{mod 9}} = {2, 16_{mod 9}} = {2, 7}, 4N = {4·1_{mod 9}, 4·8_{mod 9}} = {4, 32_{mod 9}} = {4, 5}, 2·5 = 10_{mod 9} = 1 ⇒ 2N·4N = N.

1 | 8 | 2 | 7 | 4 | 5 | |
---|---|---|---|---|---|---|

1 | 1 |
8 |
2 | 7 | 4 | 5 |

8 | 8 |
1 |
7 | 2 | 5 | 4 |

2 | 2 | 7 | 4 | 5 | 8 |
1 |

7 | 7 | 2 | 5 | 4 | 1 |
8 |

4 | 4 | 5 | 8 |
1 |
7 | 2 |

5 | 5 | 4 | 1 |
8 |
2 | 7 |

N = {1, 8} | 2N = {2, 7} | 4N = {4, 5} | |
---|---|---|---|

N | N | 2N | 4N |

2N | 2N | 4N | N |

4N | 4N | N | 2N |

- Let U(20) = {1, 3, 7, 9, 11, 13, 17, 19} be the group of units in ℤ
_{20}. Consider the subgroup H = {1, 19}. U(20)/H = {H, 3H, 7H, 19H}, |U(20)/H| = |U(20)|/|H| = 8/2 = 4 where 3H = {3, 17}, 7H = {7, 13}, and 9H = {9, 11}

3H = {3·1_{mod 20}, 3·19_{mod 20}} = {3, 57_{mod 20}} = {3, 17}, 7N = {7·1_{mod 20}, 7·19_{mod 20}} = {7, 133_{mod 20}} = {7, 13}, 3·7 = 21_{mod 20} = 1 ⇒ 3H·7H = H.

H = {1, 19} | 3H = {3, 17} | 7H = {7, 13} | 9H = {9, 11} | |
---|---|---|---|---|

H = {1, 19} | H | 3H | 7H | 9H |

3H = {3, 17} | 3H | 9H | H | 7H |

7H = {7, 13} | 7H | H | 9H | 3H |

9H = {9, 11} | 9H | 7H | 3H | H |

|3H| = 4 ⇒ U(20)/H = ⟨3H⟩ ≋ ℤ_{4}

- Let G = U(30) = {1, 7, 11, 13, 17, 19, 23, 29}, H = {1, 29},
**H ◁ G**, |G/H| = |G|/|H| = 4.**G/H = {H, 7H, 11H, 17H}**= ⟨7H⟩ where 1H = 29H = {1, 29} = H1 = H29. 11H = 19H = {11, 19} = H11 = H19. 7H = 23H = {7, 23} = H7 = H23. 13H = 17H = {13, 17} = H13 = H17.7·7 = 49 =

_{30}19 ⇒ (7H)^{2}= 11H = {11, 19}. 7·7·7 = 343 =_{30}13 ⇒ (7H)^{3}= 13H. 7·7·7·7 = 2401 =_{30}1 ⇒ (7H)^{4}= H ⇒ |7H| = 4 ⇒ G/H = ⟨7H⟩ ⇒ G/H ≋ ℤ_{4}

H | 7H | 11H | 13H | |
---|---|---|---|---|

H | H | 7H | 11H | 13H |

7H | 7H | 11H | 13H | H |

11H | 11H | 13H | H | 7H |

13H | 13H | H | 7H | 11H |

- Let G = U(32) = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31}, H = {1, 17}. Then, G/H is an Abelian group of order |G|/|H| = 16/2 = 8 ⇒ G is isomorphic to one of the following groups ℤ
_{8}, ℤ_{4}⊕ ℤ_{2}, or ℤ_{2}⊕ ℤ_{2}⊕ ℤ_{2}.

G/H = {1H, 3H, 5H, 7H, 9H, 11H, 13H, 15H} where 1H = {1, 17}, 3H = {3·1, 3·17 = 51_{mod 32} 19} = {3, 19}, 5H = {5·1, 5·17 = 85_{mod 32} 21} = {5, 21}, 7H = {7, 23}, 9H = {9, 25}, 11H = {11, 27}, 13H = {13, 29}, 15H = {15, 31}

(3H)^{2} = (3H)·(3H) = 9H ≠ H, (3H)^{4} = H (3·3·3·3 = 81 = 17 (mod 32)) ⇒ |3H| = 4 and ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} has no elements of order 4 ⇒ G/H ≇ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2}. Since |7H| = |9H| = 2 (7·7 = 49_{mod 32} = 17, 9·9 = 81_{mod 32} = 17) ⇒ G/H has at least two elements of order 2, but ℤ_{8} has only one element of order 2, namely 4 ⇒ G/H ≇ ℤ_{8}, hence G/H ≋ ℤ_{4} ⊕ ℤ_{2}

- Let G = U(32) = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31}, K = {1, 15} ⇒ |G/K| = |G|/|K| = 16/2 = 8 ⇒ G is isomorphic to one of the following groups ℤ
_{8}, ℤ_{4}⊕ ℤ_{2}, or ℤ_{2}⊕ ℤ_{2}⊕ ℤ_{2}. The cosets are: K, 3K = {3, 13}, 5K = {5, 11}, 7K = {7, 9}, 17K = {17, 31}, 19K = {19, 29}, 21K = {21, 27}, and 23K = {23, 25}.

Which of the three Abelian groups of order eight is it? (3K)^{2} = 9K, (3K)^{3} =[3·3·3_{mod 43} = 27] = 21K. (3K)^{4} =[27*3 = 81_{mod 32} = 17] 17K ≠ K ⇒[|G/K| = 8, and so by Lagrange Theorem, the only options are 1, 2, 4 and 8] |3K| = 8 ⇒[G/K is cyclic] G/K ≋ ℤ_{8}.

- The quaternion group Q
_{8}is a non-Abelian group of order eight. ⟨i⟩ = {1, i, -1, -i} ◁ Q_{8}. |Q_{8}/⟨i⟩| = |Q_{8}|/|⟨i⟩| = 8/4 = 2. Q_{8}/⟨i⟩ = {⟨i⟩, j⟨i⟩} where j⟨i⟩ = {j, ji, -j, j(-i)} = {j, -k, -j, k}.

Let’s see the cosets of the group, Q_{8}/⟨i⟩ = {⟨i⟩, j⟨i⟩}. Notice that ⟨i⟩ is the identity of the quotient group. The Cayley table for Q_{8}/⟨i⟩ is exactly equal as ℤ_{2}, and therefore Q_{8}/⟨i⟩ ≋ ℤ_{2}. The same applies with ⟨j⟩ and ⟨k⟩. Figures 1.c. and 1.d.

**The center of the group is Z(Q**= {-1, 1}, then ⟨-1⟩ ◁ Q_{8}) = ⟨-1⟩_{8}. |Q_{8}/⟨-1⟩| = |Q_{8}|/|⟨-1⟩| = 8/2 = 4.

Q_{8}/⟨-1⟩ = {⟨-1⟩, i⟨-1⟩, j⟨-1⟩, k⟨-1⟩} and all elements of this group has order 2 except obviously the identity, that’s why Q_{8}/⟨-1⟩ is isomorphic to ℤ_{2} x ℤ_{2}.

⟨-1⟩ = {-1, 1}, i⟨-1⟩ = {-i, i}, i⟨-1⟩i⟨-1⟩ = i^{2}⟨-1⟩ = -1⟨-1⟩ =[-1 ∈ ⟨-1⟩] ⟨-1⟩

The isomorphism Q_{8}/⟨-1⟩ → ℤ_{2} x ℤ_{2}, is defined by ⟨-1⟩ → (0, 0), i⟨-1⟩ → (1, 0), j⟨-1⟩ → (0, 1), k⟨-1⟩ → (1, 1).

- Consider A
_{4}, the alternating group on 4 letters. A_{4}= {(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234),(243)}. A_{4}has no subgroups of order 6.

Let’s suppose H ≤ G, |H| = 6, then |A_{4}/H| = 12/2 = 2 ⇒ [Let H ≤ G be any subgroup of a finite group G. If |G/H| = |G|/|H|=2 then H ◁ G.] H ◁ G.

|H| = 6, so there are some 3-cycle (there are 8 3-cycles) that are not in H. Let x be a 3-cycle that is not in H, and **consider the cosets H, xH, and x ^{2}H** in A

- H = x
^{2}H ⇒ [x^{3}= 1 as x is a 3-cycle, x^{2}= x^{-1}&& aH = a iff a ∈ H] x^{2}= x^{-1}∈ H ⇒ [H ≤ G, H is closed under inverses as a subgroup] x ∈ H ⊥ - xH = x
^{2}H ⇒ H = x^{-1}x^{2}H = xH ⇒ x ∈ H ⊥

Theorem. Let G be a group, and let be Z(G) the center of G. If G/Z(G) is cyclic, then G is Abelian.

Proof:

Let gZ(G) be a generator of the cyclic group G/Z(G), ⟨gZ(G)⟩ = G/Z(G)

∀a, b ∈ G, ∃i, j ∈ ℤ: aZ(G) = (gZ(G))^{i} = [(aH)(bH)=abH by definition] g^{i}Z(G), and bZ(G) = (gZ(G))^{j} = [(aH)(bH)=abH by definition] g^{j}Z(G)

aZ(G) = g^{i}Z(G), bZ(G) = g^{j}Z(G) ⇒ [A left (or right) coset is uniquely determined or represented by any one of its elements, i.e., aH = bH iff a ∈ bH] a ∈ g^{i}Z(G), b ∈ g^{j}Z(G) ⇒ ∃ x, y ∈ Z(G): a = g^{i}x, b = g^{j}y

ab = (g^{i}x)(g^{j}y) =[Associativity] g^{i}(xg^{j})y [x ∈ Z(G) ⇒ xg^{j} = g^{j}x] g^{i}(g^{j}x)y = (g^{i}g^{j})(xy) =[Trivial exponential rules] (g^{j}g^{i})(xy) =[x ∈ Z(G)] (g^{j}g^{i})(yx) = g^{j}(g^{i}y)x =[y ∈ Z(G)] (g^{j}y)(g^{i}x) = ba ∎

Corollary 1: If G/H is cyclic, H ≤ Z(G) ⇒ G is Abelian.

Proof.

H ≤ Z(G), G/H cyclic ⇒[Claim 🔑] G/Z(G) cyclic ⇒[By the previous theorem] G is Abelian.

Claim 🔑: H ≤ Z(G), G/H cyclic ⇒ G/Z(G) cyclic.

G/H cyclic ⇒ G/H = ⟨aH⟩ ⇒ ∀g ∈ G, ∃n ∈ ℤ: gH = (aH)^{n} = a^{n}H ↭[aH = bH ↭ b^{-1}a ∈ H] g^{-1}a^{n} ∈ H ≤ Z(G) ⇒ g^{-1}a^{n} ∈ Z(G) ↭ gZ(G) = a^{n}Z(G) = (aZ(G))^{n}, so aZ(G) is a generator or G/Z(G)∎

Corollary 2: If G is non-Abelian, then G/Z(G) is not cyclic.

Corollary 3: Let G be a non-Abelian group, |G| = pq, where p and q are primes, then G must have a trivial center.

Proof: Let G be a non-Abelian group ⇒ G/Z(G) is not cyclic.

Z(G) is a normal subgroup of G (Z(G) ◁ G) ⇒[|Z(G)| divides |G| = p·q, p and q are primes] |Z(G)| = 1, p, q or pq.

G is not Abelian ⇒ |Z(G)| ≠ pq. If |Z(G)| = p or q, then |G/Z(G)| is either q or p. Since p and q are primes ⇒ [Every group of prime order is cyclic.] G/Z(G) is a cyclic group ⊥ (G is non-Abelian group, Corollary 2). Therefore, |Z(G)| = 1, G has a trivial center∎

A group of prime order is always cyclic. Let G be a non-trivial group, where |G| is prime. In other words, it has an element g ≠ e, |g| ≠ 1, |g| divides |G| prime ⇒ |g| = |G|, and therefore, G = ⟨g⟩.

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.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn (Abstract Algebra), and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- Andrew Misseldine: College Algebra and Abstract Algebra.