Quaternion group associativity

Solution 1:

One idea that saves some work, but still might make your eyes cross is to realize that negatives work as expected, so you don't need to test associativity with 1, -1, or any of the negative i,j,k. That leaves only $\{i,j,k\}^3$ which is 27 tests, each requiring 4 very easy multiplications, 108 easy lookups. If you notice that $i \mapsto j \mapsto k$ is an automorphism, then you can say that $a=i$ in $a(bc) \stackrel{?}{=} (ab)c$ reducing it to 9 checks, 36 easy multiplications.

However, a method with more fringe benefits is to find matrices that multiply as expected. It is easiest if you know complex numbers, since $i$ and $j$ are both like $\sqrt{-1}$, but they don't commute.

Hopefully you can discover the matrices by yourself, but maybe it is tricky without a lot of experience. Here are matrices that work ($\sqrt{-1}$ is any fixed square root of $-1$ in a field of characteristic not 2, say $\mathbb{C}$ for instance; take 1 to be the identity matrix, and if $a$ corresponds to the matrix $A$, then $-a$ corresponds to $-A$, so that negatives work just like expected).

$$i = \begin{bmatrix} \sqrt{-1} & 0 \\ 0 & -\sqrt{-1} \end{bmatrix}, \qquad j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \qquad k = \begin{bmatrix} 0 & \sqrt{-1} \\ \sqrt{-1} & 0 \end{bmatrix}$$