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 $T$ and $T'$ are subsemigroups of any finite semigroup $S$ and $T\cup T'$ be disjoint union and subsemigroup of $S$.
If we know the rank of $T$ and $T'$, then can we determine the rank of $T\cup T'$ ?
Best Answer
As pointed out in the comments, your condition is rather peculiar. Nevertheless, even under this strong assumption, nothing interesting can be said about the rank of $T \cup T'$, as shown by the following two extreme cases:
Case 1. Let $S= \{1, \ldots, n\}$ under the product defined by $i * j = j$ and let $T$ and $T'$ be two disjoint subsets of $S$. Then $T$, $T'$ and $T \cup T'$ are subsemigroups of $S$. Then $\operatorname{rank}(T) = |T|$ and $\operatorname{rank}(T') = |T'|$. Moreover $$\quad\operatorname{rank}(T \cup T') = |T \cup T'| = \operatorname{rank}(T) + \operatorname{rank}(T'). $$ Case 2. Let $A = \{a, b\}$ and let $S = \{u \in A^* \mid 0 < |u| \leqslant n\} \cup \{0_a, 0_b\}$ under the product defined by $$ u * v = \begin{cases} uv &\text{if $|uv| \leqslant n$},\\ 0_a &\text{if $u = 0_a$ or if $|uv| > n$ and the first letter of $u$ is $a$},\\ 0_b &\text{if $u = 0_b$ or if $|uv| > n$ and the first letter of $u$ is $b$}. \\ \end{cases} $$ Let now \begin{align} T_a &= \{au \mid u \in A^* \text{ and } 0 \leqslant |u| < n\} \cup \{0_a\}\\ T_b &= \{bu \mid u \in A^* \text{ and } 0 \leqslant |u| < n\} \cup \{0_b\} \end{align} Then $T_a$ and $T_b$ are disjoint subsemigroups of $S$ and $T_a \cup T_b = S$. Since $S$ is generated by $A$, one gets $\operatorname{rank}(T_a \cup T_b) = 2$. On the other hand, if I am not wrong, $\operatorname{rank}(T_a) = \operatorname{rank}(T_b) = n$ (for instance, $\{a, ab, ab^2, \ldots, ab^{n-1}\}$ generates $T_a$ and no smaller set is generating).