# Quaternion group associativity

Consider the eight objects $\pm 1, \pm i, \pm j, \pm k$ with multiplication rules:

$ij=k,jk=i,ki=j,ji=-k,kj=-i,ik=-j,i^2=j^2=k^2=-1$,

where the minus signs behave as expected and $1$ and $-1$ multiply as expected. Show that these objects form a group containing exactly one involution.

Well, it is easy to determine closure from the definition. $1$ is clearly the identity, and the inverses can also be determined $(i,-i),(j,-j),(k,-k)$, $-1$ with itself, and $1$ with itself. The only involution is $-1$. Is there an easy way to check associativity of this group? There are too many possible combinations $(ab)c=a(bc)$ to check directly. ($8^3$ possible combinations)

• First idea: negatives work as expected, so only need to check associativity of {i,j,k} which is 3^3 or 3^2 after symmetry. – Jack Schmidt May 24 '13 at 20:14
• second idea: find matrices that multiply as requested, then you know the multiplication is associative. – Jack Schmidt May 24 '13 at 20:15
• @Jack : If this was an answer I'd upvote it. You should just copy paste it down there. – Patrick Da Silva May 24 '13 at 20:21
• You probably intended a minus sign in $i^2=\dots=-1$. – Andreas Blass May 24 '13 at 20:32
• My answer assumes $i^2=j^2=k^2=-1$, since otherwise $\pm i,\pm j, \pm k$ are all also involutions. – Jack Schmidt May 24 '13 at 20:35

## 1 Answer

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).

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

• The matrix method involves 16 nonzero products, so is faster, unless of course you don't guess the correct matrices on your first try. :P – Jack Schmidt May 24 '13 at 20:32