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


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)

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

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).

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

  • $\begingroup$ 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 $\endgroup$ – Jack Schmidt May 24 '13 at 20:32

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