“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 Mn(ℝ) of all n x n matrices with entries in ℝ is not a group because not every matrix has an inverse. GLn(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 GL2(ℂ). Then, the set Q8 = {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 GL2(ℂ), Q8 ≤ GL2(ℂ).
Q8 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 Q8 is a non-Abelian group with eight elements, Q8 = {1, -1, i, -i, j, -j, k, -k}. It is given by the group representation ⟨-1, i, j, k: (-1⟩2 = 1, i2 = j2 = k2 = ijk = -1⟩ = ⟨i, j: i4 = 1, i2 = j2, ij = j-1i⟩ Figure 1.b.
If you follow the arrow clockwise on the diagram (1.b), you always get a positive sign,
ij = j-1i ↔ ijk = j-1ik ↭[ij=k, kk =-1] -1 = j-1ik =[Obviously, j-1·j = 1, -1 ∈ Z(G)] j-1(-j).
Recall ⟨a⟩ = {an | n ∈ ℤ}. Figure 1.d. is the subgroup lattice of Q8. 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}.
Q8 = ⟨i, j: i4 = 1, i2 = j2, ij = j-1i⟩
⟨i⟩ = {1, i, -1, -i}. j⟨i⟩ = {j, -k, -j, k} = ⟨i⟩j = {j, k, -j, -k} ⇒ ⟨i⟩ ◁ Q8. Observe that we do not need to check the elements of ⟨i⟩ = (1, i, -1, -i), just the generators of the group Q8, i.e, i and j.
Let’s see the cosets of the group, Q8/⟨i⟩ = {⟨i⟩, j⟨i⟩}. Notice that ⟨i⟩ is the identity of the quotient group. The Cayley table for Q8/⟨i⟩ is exactly equal as ℤ2 (there is only one group of order 2 up to isomorphism), and therefore Q8/⟨i⟩ ≋ ℤ2. 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.
Q8/⟨-1⟩ = {⟨-1⟩, i⟨-1⟩, j⟨-1⟩, k⟨-1⟩} and all elements of this group has order 2 except obviously the identity, that’s why Q8/⟨-1⟩ is isomorphic to ℤ2 x ℤ2.
i⟨-1⟩i⟨-1⟩ = i2⟨-1⟩ = -1⟨-1⟩ = ⟨-1⟩
The isomorphism Q8/⟨-1⟩ → ℤ2 x ℤ2, 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.