I'm currently trying to understand the following part of a script (translated from German to English). It is the first part where quaternions get introduced, so I don't know anything about them except what is written there:
Quaternions are an expansion of the concept of complex numbers on structures with four (instead of two) components. A quaterion $h$ can be written as a vector or in the form of $h = h_0 + ih_1 + j h_2 + > kh_3$, where $i$, $j$ and $k$ are related to the $i$ in complex numbers. Accordingly $h_0$ is often called real part and $h_1, h_2, > h_3$ are called imaginary part of a quaternion.
For $i$, $j$ and $k$ the following rules are applied:
$$i^2 = j^2 = k^2 = -1$$ $$ijk=-1$$
From these rules follows:
$\begin{align} ij &= k\\ ji &= -k\\ jk &= i\\ kj &= -i\\ ki &= j\\ ik &= -j \end{align}$
I don't understand why these rules follow.
My thoughts
$ij = k$ and $jk = i$
$$\begin{align} ijk &= -1\\ \Leftrightarrow ijk \cdot (k^3) &= (-1) \cdot (k^3)\\ \Leftrightarrow ij \cdot (-1) \cdot (-1) &= (-1) \cdot (-1) \cdot k\\ \Leftrightarrow ij &= k \end{align}$$
You get $jk = i$ with the same idea.
$-j = ik$ and $-k = ji$
$$\begin{align} ij &= k\\ \Leftrightarrow iij &= ik\\ \Leftrightarrow -j &= ik \end{align}$$
you get $-k = ji$ with the same idea.
Question
Why is $ki=j$ and $kj = -i$?
First I thought I could proof it like this:
$\begin{align} -k &= ji\\ \Leftrightarrow (-1) \cdot ki &= jii\\ \Leftrightarrow (-1) \cdot ki &= j \cdot (-1)\\ \Leftrightarrow ki &= (-1) \cdot j \cdot (-1)\\ \stackrel{*}{\Leftrightarrow} ki &= j \end{align}$
But for the last transformation * I would need commutativity. Obviously, $ij \neq ji$ so how can I know that $(-1) \cdot j = j \cdot (-1)$?
it's very trivial. $$(-1)j=(j^2)j=j(j^2)=j(-1).$$ Because $.$ operator in $Q_8$ is associative.