32nd Putnam 1971

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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