Talk:Aperiodic semigroup

Reference for the cited 'well-known' theorem, please.

Charles Matthews 21:18, 17 June 2006 (UTC)[reply]

If I'm not mistaken, a Grothendieck group can only be formed for a commutative monoid. Shouldn't the Grothendieck group of an aperiodic commutative monoid be trivial? Of course, that also makes it cyclic, but the result seems rather irrelevant to me. Yours, Huon 09:39, 19 June 2006 (UTC)[reply]

Let me add that I have read (and fully understood) the paper previously cited as a reference for the supposedly "well-known" theorem. There is simply nothing in there about Grothendieck groups. Pascal.Tesson 02:51, 6 July 2006 (UTC)[reply]

My OR gripe

"Aperiodic" is a bit of a misnomer because the period of their mongenic subsemigroups is exactly one rather than having no period, cf. def of period at monogenic semigroup. Tijfo098 (talk) 03:58, 12 October 2012 (UTC)[reply]

You'll love the "infinite cyclic group" then :-) --192.75.48.8 (talk) 20:17, 11 March 2015 (UTC)[reply]

Result about epigroups

The article currently says "In terms of Green's relations, a finite semigroup is aperiodic if and only if its H-relation is trivial. These two characterizations extend to group-bound semigroups.[citation needed]"

I was only able to find this in Higgins (1994) "Theorem 1.2.12 Let S be a group-bound semigroup in which every subgroup is trivial. Then H is trivial on S." Which is not quite as powerful. Tijfo098 (talk) 08:07, 12 October 2012 (UTC)[reply]