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From N.RUSKUC's paper "On Large Subsemigroups and Finiteness conditions of Semigroups", there is a theorem, enter image description here
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 enter image description here

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.

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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.]

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  • $\begingroup$ @Andylang: All clear, then? I don't feel I wrote it up very well, but if you've got the idea, that's the main thing. $\endgroup$ – Tara B Apr 4 '13 at 16:10
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    $\begingroup$ Using the minimal and doing looply can avoid the question we met above.Good answer! $\endgroup$ – Andylang Apr 4 '13 at 16:11

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