Talk:Schützenberger group

left and right Schutzenberger groups are isomorphic or antiisomorphic?
"In fact, there are two Schutzenberger groups associated with a given H-class and each is antiisomorphic to the other." I supposte the "two" Schutzenberger groups means the left and right Schutzenberger group for H. Let $$\Gamma_l(H)=\{y_s|sH=H, s \in S^1\}$$ be the left one, $$\Gamma_r(H)=\{z_s|Hs=H, s \in S^1\}$$ the right one.

Then, only this natural isomorphism $$\varphi: \Gamma_r(H) \rightarrow \Gamma_l(H)$$ comes to my mind: $$\varphi(z_s)=y_t \Leftrightarrow ths=h$$. If $$\varphi(z_s) = y_p, \varphi(z_t) = y_q$$, then $$phs=h, qht=h$$, so $$qphst=h$$ meaning that $$\varphi(z_{st})=y_{qp}$$.

Under this mapping, left and right Schutzenberger groups are isomorphic, not antiisomorphic. Moreover, the isomorphism can be used for proving that all Green H-classes in a regular D-class are isomorphic groups, while the antiisomorphism is not directly required. So, I think the article introduction is somewhat confusing. Could someone describe the antiisomorphism explicitly in the article, or add a direct reference? Or is the antiisomorphism just $$(^{-1} \circ \varphi)$$? — Preceding unsigned comment added by Jsedenka (talk • contribs) 12:28, 9 February 2013 (UTC)