Finitely generated semigroup gets the finitely generated subsemigroup?

Question

From N.RUSKUC's paper "On Large Subsemigroups and Finiteness conditions of Semigroups", there is a theorem,
Here large subsemigroup means $S$\ $T$ is finite. In this side "=>" of the proof in the paper, suppose $S$ is finitely generated by the set $A$. Then the set

generats $T$.

My question is how can I get that $X$ generates $T$. Of course from the definition of $X$, I know $T$ contains $X$. So the point may be how to get that for any element in T, this element can be presented by X.
Thanks for your assistance.

Answer 1

Let $t\in T$ with $t = a_1\ldots a_n$ for $a_i\in A$. I'm going to use $U$ for $S^1\setminus T$. To express $t$ as a product of elements of $X$:

1. Let $k$ be minimal such that $a_1\ldots a_k\in U$ (where the empty product represents $1$) and $a_1\ldots a_{k+1}\in T$. (Certainly $k$ exists, since $t\in T$.) Set $s_1 = a_1\ldots a_k$.
2. If $a_{k+2}\ldots a_n\in U$, then set $s_2 = a_{k+2}\ldots a_n$ and we have $t = s_1 a_{k+1} s_2\in X$. Otherwise, replace $t$ by $a_{k+2}\ldots a_n$ and return to 1.

We keep going through that loop as long as possible, obtaining $t = s_1 b_1 s_2 b_2 \ldots s_m b_m s_{m+1}$, where $s_i\in U$, $b_i\in A$ and $s_i b_i\in T$ for all $i$ and $s_m b_m s_{m+1}\in T$ (for if $s_m b_m s_{m+1}$ were not in $T$, we would have already stopped the process earlier).

[I'll probably come back and try to improve the exposition of this later, but for now I wanted to just quickly record the idea.]