Talk:Element has Idempotent Power in Finite Semigroup

Simplifying the proof

I do not understand the need for the last case disjunction. The argument for $l > 1$ seems to work just as well with $l = 1$. Of course we are then multiplying on both sides by $x^0$ but I don't see why this would be an issue. --A3nm (talk) 11:26, 3 September 2016 (EDT)

$S$ is a semigroup, and the element $x^0 = e$ exists only when $S$ is a monoid. --prime mover (talk) 12:00, 3 September 2016 (EDT)

Argh, got it. (I'm more familiar with monoids so I didn't see the catch.) Thanks, and sorry for the stupid comment then. --A3nm (talk) 12:03, 3 September 2016 (EDT)

No worries. --prime mover (talk) 12:40, 3 September 2016 (EDT)