Talk:Idempotent Elements form Subsemigroup of Commutative Semigroup

Take a look at Composition of Commuting Idempotent Mappings is Idempotent for what I believe is a somewhat more efficient approach to the manipulations. --Dfeuer (talk) 16:10, 13 March 2013 (UTC)


 * In fact, if you generalize that theorem to semigroups, this one can be proved using that generalization. --Dfeuer (talk) 16:12, 13 March 2013 (UTC)


 * I just generalized that one: Product of Commuting Idempotent Elements is Idempotent. --Dfeuer (talk) 16:22, 13 March 2013 (UTC)


 * Thank you Dfeuer. I think idempotency looks like this: $x^2 = x$. I don't see anything special about $2$ or $1$ though so it could be generalised further for elements that satisfy the equation: $x^m = x^n$ for $m, n \in \mathbb N_{>0}$. I am not very good at generalising in a useful manner though :( --Jshflynn (talk) 18:07, 13 March 2013 (UTC)