**Problem**: (2001 Putnam - A1) Consider a set $ S $ and a binary operation $ * $, i.e., for each $ a,b \in S $, $ a*b \in S $. Assume $ (a*b)*a = b $ for all $ a, b \in S $. Prove that $ a*(b*a) = b $ for all $ a, b \in S $.

**Solution**: If we make the substitution $ a = b*a $, we get

$ ((b*a)*b)*(b*a) = b $.

However, since $ (b*a)*b = a $ (by switching $ a $ and $ b $), we thus have

$ a*(b*a) = b $,

as desired. QED.

--------------------

Comment: Not an extremely hard problem, but making the right substitution took some time to find. Playing around with the given equality is pretty much the only option in this problem.

--------------------

Practice Problem: Find a set $ S $ (preferably nontrivial) and operation $ * $ satisfying the conditions of the problem above.

The set of reflections of an arbitrary geometric figure with the lines of reflection passing through a common point, under composition.

ReplyDeleteAlternately, the subset of S_n containing only 2-cycles.

... I think.