### 33rd Putnam 1972

**Problem A2**

Let S be a set with a binary operation * such that (1) a * (a * b) = b for all a, b ∈ S, (2) (a * b) * b = a for all a, b ∈ S. Show that * is commutative. Give an example for which S is not associative.

**Solution**

Consider (a*b)*( (a*b)*b). One can view it as c * (c * b), so that it is b by (1). Or one can consider first (a*b)*b, which is a by (2), so that the expression is (a*b)*a. Hence b = (a*b)*a. Multiplying on the right by a, we get b*a = ((a*b)*a)*a, which is (a*b) by (2). That proves * is commutative.

Take S to have three elements a, b, c. Let a*a = a, b*b = b, c*c = c, a*b = c, b*c = a, c*a = b and assume * is commutative. Then we easily check that the required conditions are met. But c = a*b = (a*a)*b, whereas a*(a*b) = a*c = b. So * is not always associative.

33rd Putnam 1972

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2001