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)