HMMT 二月 2004 · 团队赛 · 第 11 题

HMMT February 2004 — Team Round — Problem 11

[40] Prove that there exists an element a ∈ T such that the equation a ? b = a holds for all b ∈ T .

解析

Prove that there exists an element a ∈ T such that the equation a ? b = a holds for all b ∈ T . Solution: Choose a whose image contains as few elements as possible — we know we can do this, since T , being a subset of S , is finite. We claim that this a works. Indeed, suppose c is in the image of a . Then, for any d in the image of c , a ? ( c ? d ) = ( a ? c ) ? d = c ? d = d , so d is also in the image of a . So the image of c is contained in the image of a . But a was chosen to have image as small as possible, so the two images must coincide. In particular, a ? a = a is in the image of c . So a = c ? a = a ? c = c. This argument shows that a is the only element of the image of a , which gives what we wanted. Alternative Solution: This can also be solved without using Problem 10: The product of any two elements of T is also in T , since commutativity and associativity give ( b ? b ) ? ( c ? c ) = ( b ? c ) ? ( b ? c ) for b, c ∈ S . Then let a , a , . . . , a be all the 1 2 n elements of T , and put a = a ? a ? · · · ? a ; this value does not depend on the ordering 1 2 n of the elements. If b ∈ T , then a = c ? b , where c is the ? -product of all elements of T different from b , and consequently a ? b = ( c ? b ) ? b = c ? ( b ? b ) = c ? b = a .