Originally posted by adam warlock
Yes, yes I'm assuming that the * defined in the set is commutative but I think it is a fair assumption.
a*a*...a[k-1]*a[k]*a[k+1]*...*a[n-1]*a[n]=a[k] the things is if we can swap a[k] all the way through the end of the product, or front, to have a*a*...a[k-1]*a[k+1]*...*a[n-1]*a[n]*a[k]=a[k]. If so the result follows. So what really at ...[text shortened]... oesn't use the assumption of commutability of a[k]. If so just tell me and I'll think harder.
I have more of a sketch proof (I know, I know...but i'm not sure how correct it is). I came across a similar problem in one of my modules-Finite Maths-where we were asked to prove that in a finite field F there exists an integer n
such that 1+1+...+1=0 (where there are n 1s). I couldn't think how to tackle it so i showed it (roughly) for a general finite set, although under a non-commutative (but closed!) operation (which, in retrospect, is silly as a field is commutative by definition).
I really can't fault my logic, and it is actually quite a natural thing to assume in a finite set under a closed operation. However, I couldn't fault my logic when i made an elementary mistake the other day, so I could be completly wrong...