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Let $S, T$ be two semigroups. In the following all semigroups are supposed to be finite. We write $S \prec T$ if there exists a surjective semigroup morphism from a subsemigroup of $T$ onto $S$. A class of finite semigroups $\mathcal V$ is defined by a set of equations if exactly the semigroup in $\mathcal V$ fulfill those equations, similar for a class of finite monoids. Let $\mathcal W$ be a class of monoid, we denote by $\mathcal W_S$ the class of semigroups $$ \mathcal W_S = \{ S \mbox{ is a semigroup} \mid S \prec M, M \in \mathcal W \} $$ where the monoid $M$ is considered as a semigroup. Let $R_n$ denote the class of monoid defined by $$ (xy)^n x = (xy)^n. $$ Then the class of semigroup $(R_n)_S$ is defined by the equations $$ x^{n+1} = x^n, \quad (xy)^n x = (xy)^n. $$ Denote by $V_n$ the class of semigroups defined by the single equation $$ (xy)^n x = (xy)^n. $$ Show that $$ (R_n)_S \subseteq V_n \subseteq (R_{n+1})_S. $$ Hint: Utilize the free semigroup over two generators and its subsemigroup of sequenezes of length at least $k$ for suitable $k$.

This is exercise V.3.2 from Samuel Eilenberg, Automata, Machines and Languages, Volume B.

The inclusion $(R_n)_S \subseteq V_n$ is obvious, for properness we can look at the cyclic semigroup $S = \{x,x^2, \ldots, x^{n+1}\}$ with $x^{n+2} = x^{n+1}$. But for the other inclusion I have no idea? I see that if we set $y = x$ we get $V_n \subseteq (R_{2n})_S$, but that is not enough.

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  • $\begingroup$ Call your example $S_{n+1}$. Then $S_{2n}$, where $x^{2n+1}=x^{2n}$, seems to be in $V_n$, but it does not satisfy $x^{n+2}=x^{n+1}$ which indeed seems to be implied by being in $(R_{n+1}) _S$. $\endgroup$ – Berci Jun 26 '18 at 13:20
  • $\begingroup$ Yes, so the exercise is wrong? I also could not make much sense out of the hint given... $\endgroup$ – StefanH Jun 26 '18 at 13:22
  • $\begingroup$ Based on our current understanding, I can imagine it's only a typo, meaning $(R_{2n})_S$ on the right hand side. But I'm not sure. $\endgroup$ – Berci Jun 26 '18 at 13:29
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It is indeed a typo. The second inclusion fails for $n = 2$. Take the semigroup $\{a, a^2, a^3, a^4\}$ with $a^4 = a^5$. Then it satisfies the equation $(xy)^2x = (xy)^2$ but it does not satisfy the equation $x^4 = x^3$.

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