I remember being asked by a student many years ago if I could think of an example of an algebraic structure that was commutative but not associative. Introductory undergraduate algebra courses abound with examples in the other direction: matrix multiplication, multiplication of quaternions, in fact “most” groups. To recall: if *S* is a set and ∗ is a binary operation on *S* (i.e. an operation that picks up any two elements of *S* and outputs another element of *S*), then ∗ is called **associative** if, for all *x*, *y*, *z* ∈ *S*,

*x*∗(*y*∗*z*) = (*x*∗*y*)∗*z*;

and ∗ is called **commutative** if, for all *x*, *y* ∈ *S*,

*x*∗*y* = *y*∗*x*.

I scratched my head for a little while and then realized that the game of **rock-paper-scissors** provides a nice example of what my student sought.

The idea is this. Consider the set *S* consisting of the three moves in a game of rock-paper-scissors: *r* for rock, *p* for paper, and *s* for scissors. Now define the product *m*_{1}∗*m*_{2} of two moves *m*_{1}, *m*_{2} ∈ *S* to be whichever of the moves would win according to the usual rule that rock beats scissors, scissors beats paper, and paper beats rock; a rock-rock draw counts as a win for rock, and so on. The multiplication table for (*S*, ∗) is

∗ | r |
p |
s |
---|---|---|---|

r |
r |
p |
r |

p |
p |
p |
s |

s |
r |
s |
s |

This multiplication table is symmetric about the diagonal, so commutativity holds. However, associativity fails:

*r*∗(*p*∗*s*) = *r*∗*s* = *r* ≠ *s* = *p*∗*s* = (*r*∗*p*)∗*s*.

Simple! In general, starting with any finite “cyclically ordered” set, one could proceed just as above to define a commutative but non-associative magma.