### 32nd Putnam 1971

**Problem B1**

S is a set with a binary operation * such that (1) a * a = a for all a ∈ S, and (2) (a * b) * c = (b * c) * a for all a, b, c ∈ S. Show that * is associative and commutative.

**Solution**

b*a = (b*a)(b*a) using (1)

= ( (b*a)*b )*a using (2) twice

But (b*a)*b = (b*b)*a using (2) twice

= b*a using (1)

So b*a = (b*a)*a

= (a*a)*b using (2) twice

= a*b

That shows that * is commutative. (2) now gives immediately that * is associative, because (a*b)*c = (b*c)*a = a*(b*c).

32nd Putnam 1971

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2001