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For any semigroup $S$, let $A$ be a non-empty subset of $S$. Then the subsemigroup generated by $A$ that is, the smallest subsemigroup of $S$ containing $A$, is denoted by $\langle A\rangle$. If there exists a finite subset $A$ of $S$ such that $S=\langle A\rangle$, then $S$ is called a finitely generated semigroup. The rank of a finitely generated semigroup $S$ is defined by $${\rm rank}(S)=\min \{\, |A|:\langle A\rangle =S\}.$$

Let $T$ be a subsemigroup of any finite semigroup $S$ and $T\neq S$. If ${\rm rank}(T)={\rm rank}(S)$, then

  • Is there any relation between $T$ and $S$.
  • Conversly, can we say ${\rm rank}(T)={\rm rank}(S)$ for any relation between $T$ and $S$?
1ENİGMA1
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1 Answers1

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Let $A =\{a,b\}$ and let $W_n$ be the set of nonempty words of length $\leqslant n$ on the alphabet $A$. Define a product $\cdot$ on the set $S = W_n \cup \{0\}$ as follows: $u \cdot 0 = 0 \cdot u = 0$ for all $u \in S$ and, if $u, v \in W_n$ $$ u \cdot v = \begin{cases} uv & \text{if $uv \in W_n$}\\ 0 &\text{otherwise} \end{cases} $$ It is easy to see that $S$ is a finite semigroup of rank $2$, generated by $\{a,b\}$. However, if you take the subsemigroup $T$ consisting of $0$ and all words of length $n$, then its rank is the number of words of length $n$, that is, $2^n$.

On the other hand, if you take $T = \{u,v,0\}$ where $u$ and $v$ are words of length $n$, then $T$ is a subsemigroup of rank $2$ of $T$.

These two examples make me very pessimistic about a reasonable answer to your two questions.

J.-E. Pin
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