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Let $G$ be a group and $H\leq G$. Let $B$ be a sub-semigroup of $H$ such that $B\subseteq B^1(B\setminus B^2)$ where $B^{1}:=B\cup\{1\}$, $B^2:=BB$ (we have $B^2\subseteq B$). Now, put $$ \tau :=\{ A\subseteq G: BA\subseteq A\} \; ,\; \tau' :=\{ HD\cup B^1E: \; D,E\subseteq G \; \&\; E\cap BE=\emptyset\}. $$ We have $\tau'\subseteq \tau$. For if $A=HD\cup B^1E$, for some $D,E\subseteq G$, then $$BA=B(HD)\cup B(B^1E)=HD\cup BE\subseteq HD\cup B^1E=A,$$ (since $B$ is a sub-semigroup of $H$).

Now, my question is whether or not $\tau\subseteq \tau'$ (i.e., if $BA\subseteq A$, then $A=HD\cup B^1E$ for some $D,E\subseteq G$ such that $E\cap BE=\emptyset$).

Note that $G,H,B$, and the empty set are always members of $\tau \cap \tau'$, since $BB\subseteq B$, $B^1(B\setminus B^2)=B$, $B(B\setminus B^2)\cap (B\setminus B^2)=\emptyset$, $HG=G$, and $X\emptyset=\emptyset$, for all $X\subseteq G$.

Also, see https://mathoverflow.net/questions/187125/lower-periodic-subsets-of-groups-and-semigroups

M.H.Hooshmand
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1 Answers1

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Let $G = H = \mathbf Z^2$ and $B = \{(x,0) : x > 0\}$. Then $\tau$ consists of all subsets $A$ of $\mathbf Z^2$ such that $(x,y) \in A \implies (x+1,y) \in A$, while $\tau'$ consists of the subsets which are either (a) all of $\mathbf Z^2$ or (b) a set of the form $\{(x,y) : x \ge f(y), y \in S\}$. In particular the $x$-axis is in $\tau$ but not $\tau'$.

Possibly you will respond to the above example by requiring that $\langle B \rangle = H$, but that won't help, as the following example shows.

Let $G = H = \mathbf Z^2$ and $B = \{(x,y) : x, y \ge 0, x+y>0\}$. Note $B \setminus B^2 = \{(1,0),(0,1)\}$, so the hypothesis $B = B^1(B\setminus B^2)$ holds. Now $\tau$ consists of all "northeast-closed" subsets of $\mathbf Z^2$, while $\tau'$ consists of all subsets which are either (a) all of $\mathbf Z^2$ or (b) the set of elements to the northeast of some subset $E \subset \mathbf Z^2$, no element of which is northeast of another. Now the upper half plane is in $\tau \setminus \tau'$.

Sean Eberhard
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  • I understand that $\mathbb{N}$ here is really the copy ${(n, 0) : n\in\mathbb{N}}\subseteq \mathbb{Z}^2$, correct? – Jakobian Jul 20 '23 at 11:15
  • @Jakobian Right. I edited to make that clearer. – Sean Eberhard Jul 20 '23 at 12:06
  • @ Sean Eberhard. Thanks for your time and attention. (1) Whate are $f$ and $S$ in the 3rd line of your answer? (2) Do you have a counter-example satisfying the condition $(H^\setminus B)^{-1}\subseteq B$ (sorry for forgetting this condition), where $H^=H\setminus {1}$ and $X^{-1}$ denotes the set of all inverses of elements of $X$? – M.H.Hooshmand Jul 26 '23 at 18:12
  • (if $H$ is a proper subgroup of $G$, that is more desirable) – M.H.Hooshmand Jul 26 '23 at 18:22
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    @M.H.Hooshmand (1) $f$ and $S$ are arbitrary. For any subset $S \subset \mathbf Z$ and function $f : S \to \mathbf Z$ we get a set of that form. (2) I would phrase this as $H = B \cup B^{-1} \cup {1}$. Will think about it. By the way I don't think there is any reason to consider $G \ne H$ in your question. – Sean Eberhard Jul 27 '23 at 08:43
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    @M.H.Hooshmand The hypotheses that $H = B \cup B^{-1} \cup {1}$ and $B \subset B^1 (B \setminus B^2)$ forces $H \cong \mathbf Z$ and $B \cong \mathbf N$. For let $x, y \in B \setminus B^2$. Since $x \notin B^2$ we have $xy^{-1} \notin B$, and similarly $yx^{-1} \notin B$, so we must have $xy^{-1} = 1$. Therefore $B \setminus B^2$ is a singleton and the hypothesis $B \subset B^1 (B \setminus B^2)$ then implies $B \cong \mathbf N$. So, it's true in this case, because there is only one example. – Sean Eberhard Jul 27 '23 at 09:13
  • @Sean Eberhard. I meant an example from another group, anyway thanks for your helpful comments. – M.H.Hooshmand Jul 27 '23 at 12:17
  • Everybody, also see: https://math.stackexchange.com/questions/4742981/semigroups-with-the-property-s-s1s-setminus-s2 – M.H.Hooshmand Jul 27 '23 at 12:19