Are $i,j,k$ commutative?

Question

I am trying to understand quaternions. I read that Hamilton came up with the great equation:

A) $i^2 = j^2 = k^2 = ijk = −1$

In this equation I understand that $i,j,k$ are complex numbers.

Later on, I read that

B) $ij=k$

C) $ji=-k$

So, if $i,j,k$ are complex numbers, and complex number multiplication is commutative, why are these two equations different? I do understand that quaternion multiplication is non-commutative, but I do not understand why multiplying these complex components are also non-commutative.

Could someone please help me understand what is going on here. I (obviously) am not an expert in mathematics, so a simple explanation would be greatly appreciated.

Answer 1

We can extend the complex numbers ($a+bi,\ a,b\in\Bbb R$) with further two imaginary units, named $j$ and $k$, and if we pose those equations, we arrive to the quaternions, where commutativity indeed fails by B) and C).

Answer 2

Indeed commutativity still holds in the quaternions if your complex number only contains i's, j's or k's. I.e. any number's of the form $a + bi + 0j + 0k$ are commutative to each other, similarly $a + 0i + bj + 0k$ is commutative with other numbers of that form also and same with $a + 0i + 0j + bk$. However commutativity doesn't hold over the entire group of quaternions equipped with multiplication as its operation. $\mathbb{Q}_8$ is a non-abelian group in other words.

Answer 3

Suppose $z,z'$ are quaternions. We may write these as $z = a + w$, $z' = a' + w'$ where $a,a'$ are real and $w, w'$ are "purely quarternion" in the sense that if we write out $w$ it is simply some linear combination of $i,j,k$ with no real component. If we identify $i,j,k$ with standard unit vectors in $\mathbb{R}^3$ thus thinking of $w,w'$ as vectors then we have $ww' = -(w \cdot w') + w \times w'$ (where again the first term is the real component, the second term is quaternion one).

$$z z' = (aa' - w \cdot w') + (aw' + a'w + w \times w')$$ $$z' z = (aa' - w \cdot w') + (aw' + a'w + w' \times w)$$ $$ zz' - z'z = w \times w' - w' \times w = 2(w \times w')$$

So commutativity fails precisely when the imaginary quaternion parts aren't colinear.