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)
 A: 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).

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

A: This very old question keeps getting linked to by other closed questions, and I thought it therefore deserved another answer.
Another option for proving associativity of $Q_8$ is to generalize. The elements of $Q_8$ may be regarded as a collection of "units" for a much larger quaternion algebra $Q$, consisting of all expressions of the form $a + bi + cj + dk$ where $a,b,c,d$ are real numbers, not all equal to zero. The product operation on such expressions is defined in the obvious manner: distribute, simplify, and collect terms, taking care to use the anticommutativity formulas $ij=-ji=k$ when needed, to get
\begin{align*}
(a + bi + cj + dk) (a' + b'i + c'j + d'k) &= (aa' - bb' - cc' - dd') \\ 
&\quad+ (ab' + ba' + cd' - dc') i \\
&\quad+ (ac' + ca' + db' - bd') j \\
&\quad+ (ad' + da' + bc' - cb') k
\end{align*}
And now "all" you have to do is prove that product operation is associative! 
So, you do not escape tedium in this manner.
But, you end up proving something more general and, hence, more valuable, namely associativity of multiplication in $Q$.
