“No code is perfect, so one man’s crappy or buggy software is another man’s job. If everything works just right, you’d be out of your job and starve.” “Are you suggesting that coconuts migrate, bugs evaporate, and decontextualized news are not fake?” I asked.

“Not at all. Coconuts could be carried, bugs debugged, and fake news, debunked. You are the best thing I’ve ever done with my life. OK, let me see it,” Dad retorted, a proud smile shining on his face, Anawim, Apocalypse, #justtothepoint.

The set M_{n}(ℝ) of all n x n matrices with entries in ℝ is not a group because not every matrix has an inverse. GL_{n}(ℝ), known as the general linear group, is the set of n x n invertible matrices, together with the operation of matrix multiplication. This does form a group because the product of two invertible matrices is again invertible, and the inverse of an invertible matrix is invertible, with identity matrix as the identity element of the group.

Let 1 = $(\begin{smallmatrix}1 & 0\\ 0 & 1\end{smallmatrix})$, I = $(\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix})$, J = $(\begin{smallmatrix}0 & i\\ i & 0\end{smallmatrix})$, K = $(\begin{smallmatrix}i & 0\\ 0 & -i\end{smallmatrix})$ be matrices in GL_{2}(ℂ). Then, the set Q_{8} = {1, -1, I, -I, J, -J, K, -K} forms a group under matrix multiplication. It is known as the quaternion group. It is a subgroup of GL_{2}(ℂ), Q_{8} ≤ GL_{2}(ℂ).

Q_{8} is a non-Abelian group, e.g., IJ (K) ≠ JI (-K) or KJ (-I) ≠ JK (I).

KJ = $(\begin{smallmatrix}i & 0\\ 0 & -i\end{smallmatrix})(\begin{smallmatrix}0 & i\\ i & 0\end{smallmatrix})=(\begin{smallmatrix}0 & -1\\ 1 & 0\end{smallmatrix})=-I,~ JK=(\begin{smallmatrix}0 & i\\ i & 0\end{smallmatrix})(\begin{smallmatrix}i & 0\\ 0 & -i\end{smallmatrix})=(\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix})=I.$

1 | -1 | I | -I | J | -J | K | -K | |
---|---|---|---|---|---|---|---|---|

1 | 1 | -1 | I | -I | J | -J | K | -K |

1 | -1 | 1 | -I | I | -J | J | -K | K |

I | I | -I | -1 | 1 | K | -K | -J | J |

-I | -I | I | 1 | -1 | -K | K | J | -J |

J | J | -J | -K | K | -1 | 1 | I | -I |

-J | -J | J | K | -K | 1 | -1 | -I | I |

K | K | -K | J | -J | -I | I | -1 | 1 |

-K | -K | K | -J | J | I | -I | 1 | -1 |

Alternative definition. The quaternion group Q_{8} is a non-Abelian group with eight elements, Q_{8} = {1, -1, i, -i, j, -j, k, -k}. It is given by the group representation ⟨-1, i, j, k: (-1⟩^{2} = 1, i^{2} = j^{2} = k^{2} = ijk = -1⟩ = ⟨i, j: i^{4} = 1, i^{2} = j^{2}, ij = j^{-1}i⟩ Figure 1.b.

If you follow the arrow clockwise on the diagram (1.b), you always get a positive sign,

- ij = k, ji = -k
- jk = i, kj = -i
- ki = j, ik = -j
ij = j

^{-1}i ↔ ijk = j^{-1}ik ↭[ij=k, kk =-1] -1 = j^{-1}ik =[Obviously, j^{-1}·j = 1, -1 ∈ Z(G)] j^{-1}(-j).

Recall ⟨a⟩ = {a^{n} | n ∈ ℤ}. Figure 1.d. is the **subgroup lattice of Q _{8}**. There are three subgroups of order 4,, namely ⟨i⟩ = {1, i, -1, -i}; ⟨j⟩ = {1, j, -1, -j}; and ⟨k⟩ = {1, k, -1, -k}), one subgroup of order 2, that is, ⟨-1⟩ = {1, -1}, and the trivial subgroup of order 1 {1}.

- It is
**not Abelian**, ij = k ≠ ji = -k - There are, up to isomorphism,
**only five groups of order 8**: non-Abelian groups (Q_{8}, the quaternion group; D_{8}, the dihedral group of order 8), and Abelian groups (ℤ_{8}-cyclic group of order 8-, ℤ_{4}x ℤ_{2}, ℤ_{2}x ℤ_{2}x ℤ_{2}). - Q
_{8}is not Abelian, but**every proper subgroup, namely ⟨i⟩, ⟨j⟩, and ⟨k⟩, is normal**.

Q_{8} = ⟨i, j: i^{4} = 1, i^{2} = j^{2}, ij = j^{-1}i⟩

⟨i⟩ = {1, i, -1, -i}. j⟨i⟩ = {j, -k, -j, k} = ⟨i⟩j = {j, k, -j, -k} ⇒ ⟨i⟩ ◁ Q_{8}. Observe that we do not need to check the elements of ⟨i⟩ = (1, i, -1, -i), just the generators of the group Q_{8}, i.e, i and j.

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} (there is only one group of order 2 up to isomorphism), and therefore Q_{8}/⟨i⟩ ≋ ℤ_{2}. The same applies with ⟨j⟩ and ⟨k⟩. Figure 1.c.

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.

**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}.

i⟨-1⟩i⟨-1⟩ = i^{2}⟨-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).

The quaternions form a non-commutative division ring.

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.