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Here is a generalization of some of @SalvoTringali's ideas. My basic feeling is that probably this conjecture boils down to the case of meet semilattices and hence isn't so far from the original union-closed sets conjecture. I'm hoping this approach will allow for induction.

I'll give two slight improvements on @SalvoTringali's answer. If $s\in S$, let $H_s=\{t\in S\mid sS^1= tS^1\}$ (the $\mathcal H$-class of $s$) and $I_s=sS^1\setminus H_s$. Here $S^1$ means $S$ with an adjoined identity. Then standard finite semigroup theory tells you $I_s$ is an ideal, and that either $H_s$ is a group or $H_sH_s\subseteq I_s$. The element $s$ is (von Neumannn) regular if $H_s$ is a group and otherwise is called null.

I will use $s^{\omega}$ to denote the idempotent positive power of $s$ and $s^{\omega+1}$ for $s^{\omega}s$.

The first claim shows that if $sS^1$ is a maximal principal ideal with $H_s$ a nontrivial group, then the conjecture is true. This generalizes Salvo's Proposition 1.

Claim 1. Suppose that $|S|\geq 4$ and $S$ has a maximal principal ideal $sS^1$ with $H_s$ a nontrivial group and $s$ not idempotent.

Proof. Let $e$ be the the identity of $H_s$ and let $s'$ be the inverse of $s$ in $H_s$. Then for any $z\in S$ we have that $ez = s'(sz)$. Assume now that $z\neq e,s,s^2$. Since $sS^1$ was maximal, we cannot have $s'z\in\{e,s\}$ unless $z\in H_s$, in which case $z=s,s^2$, which we excluded. There are $n-3$ choices of $z$, more than enough to meet the desired bound. []

The next claim generalizes Salvo's Propositions 2 and 3. It is trickier.

Claim 2. Suppose that there is a maximal principal ideal $sS^1$ with $H_s$ not a group. Then there are $a\neq b\in S$ such that $az=bz$ for all $z\in S\setminus \{a,b\}$.

Proof. Our assumption implies we can find a minimal principal ideal $tS^1$ subject to the property that if $tS^1\subseteq uS^1$, then $H_u$ is not a group. Indeed, our assumption gives us at least one principal ideal with this property and hence there is a minimal such by finiteness. Then since $H_tH_t\subseteq I_t$, we must have that $t^k\notin H_t$ for all $k>1$.

We claim that $I_t$ is a monoid with identity $t^{\omega}$. For if $uS^1$ is a maximal principal ideal of $S$ contained in $I_t$, then by choice of $t$, we must have that $H_u$ is a group. Write $u=tx$ with $x\in S^1$. Choose $k>1$ such that $u^k=u^{\omega}$ and $t^k=t^{\omega}$. Then $u=u^ku = t^kx^ku=t^{\omega}t^kx^ku=t^{\omega}u$. Now if $v\in I_s$ is any element, then $v\in uS^1$ for some maximal principal left ideal of $S$ contained in $I_s$, and so if $v=uy$, then $t^{\omega}v=t^{\omega}uy=uy=v$ by the previous case.

I claim that $tz=t^{\omega+1}z$ for all $z\in S\setminus \{t,t^{\omega+1}\}$. First we observe that if $z\in S$, then $tz\in I_t$. Indeed, if $tz\in H_t$, then $tzS^1=tS^1$ and so $tzy=t$ for some $y\in S^1$. Then $t(zy)^{\omega}=t$ and $tS^1\subseteq (zy)^{\omega}S^1$. But $(zy)^{\omega}$ is regular, so this contradicts the choice of $t$. Thus $tz\in I_t$, and so by the above claim $tz = t^\omega(tz) = t^{\omega+1}z$. []

So we are left with the case that each maximal $\mathcal H$-class is an idempotent. Now one should try to do some sort of induction to get to the case each element is an idempotent.

Here is a generalization of some of @SalvoTringali's ideas. My basic feeling is that probably this conjecture boils down to the case of meet semilattices and hence isn't so far from the original union-closed sets conjecture. I'm hoping this approach will allow for induction.

I'll give two slight improvements on @SalvoTringali's answer. If $s\in S$, let $H_s=\{t\in S\mid sS^1= tS^1\}$ (the $\mathcal H$-class of $s$) and $I_s=sS^1\setminus H_s$. Here $S^1$ means $S$ with an adjoined identity. Then standard finite semigroup theory tells you $I_s$ is an ideal, and that either $H_s$ is a group or $H_sH_s\subseteq I_s$. The element $s$ is (von Neumannn) regular if $H_s$ is a group and otherwise is called null.

I will use $s^{\omega}$ to denote the idempotent positive power of $s$ and $s^{\omega+1}$ for $s^{\omega}s$.

The first claim shows that if $sS^1$ is a maximal principal ideal with $H_s$ a nontrivial group, then the conjecture is true. This generalizes Salvo's Proposition 1.

This claim has a gap. I’ll leave it in case I can fix it. Claim 2 seems ok.

Claim 1. Suppose that $|S|\geq 4$ and $S$ has a maximal principal ideal $sS^1$ with $H_s$ a nontrivial group and $s$ not idempotent.

Proof. Let $e$ be the the identity of $H_s$ and let $s'$ be the inverse of $s$ in $H_s$. Then for any $z\in S$ we have that $ez = s'(sz)$. Assume now that $z\neq e,s,s^2$. Since $sS^1$ was maximal, we cannot have $s'z\in\{e,s\}$ unless $z\in H_s$, in which case $z=s,s^2$, which we excluded. There are $n-3$ choices of $z$, more than enough to meet the desired bound. []

The next claim generalizes Salvo's Propositions 2 and 3. It is trickier.

Claim 2. Suppose that there is a maximal principal ideal $sS^1$ with $H_s$ not a group. Then there are $a\neq b\in S$ such that $az=bz$ for all $z\in S\setminus \{a,b\}$.

Proof. Our assumption implies we can find a minimal principal ideal $tS^1$ subject to the property that if $tS^1\subseteq uS^1$, then $H_u$ is not a group. Indeed, our assumption gives us at least one principal ideal with this property and hence there is a minimal such by finiteness. Then since $H_tH_t\subseteq I_t$, we must have that $t^k\notin H_t$ for all $k>1$.

We claim that $I_t$ is a monoid with identity $t^{\omega}$. For if $uS^1$ is a maximal principal ideal of $S$ contained in $I_t$, then by choice of $t$, we must have that $H_u$ is a group. Write $u=tx$ with $x\in S^1$. Choose $k>1$ such that $u^k=u^{\omega}$ and $t^k=t^{\omega}$. Then $u=u^ku = t^kx^ku=t^{\omega}t^kx^ku=t^{\omega}u$. Now if $v\in I_s$ is any element, then $v\in uS^1$ for some maximal principal left ideal of $S$ contained in $I_s$, and so if $v=uy$, then $t^{\omega}v=t^{\omega}uy=uy=v$ by the previous case.

I claim that $tz=t^{\omega+1}z$ for all $z\in S\setminus \{t,t^{\omega+1}\}$. First we observe that if $z\in S$, then $tz\in I_t$. Indeed, if $tz\in H_t$, then $tzS^1=tS^1$ and so $tzy=t$ for some $y\in S^1$. Then $t(zy)^{\omega}=t$ and $tS^1\subseteq (zy)^{\omega}S^1$. But $(zy)^{\omega}$ is regular, so this contradicts the choice of $t$. Thus $tz\in I_t$, and so by the above claim $tz = t^\omega(tz) = t^{\omega+1}z$. []

So we are left with the case that each maximal $\mathcal H$-class is an idempotent. Now one should try to do some sort of induction to get to the case each element is an idempotent.

Here is a generalization of some of @SalvoTringali's ideas. My basic feeling is that probably this conjecture boils down to the case of meet semilattices and hence isn't so far from the original union-closed sets conjecture. I'm hoping this approach will allow for induction.

I'll give two slight improvements on @SalvoTringali's answer. If $s\in S$, let $H_s=\{t\in S\mid sS^1= tS^1\}$ (the $\mathcal H$-class of $s$) and $I_s=sS^1\setminus H_s$. Here $S^1$ means $S$ with an adjoined identity. Then standard finite semigroup theory tells you $I_s$ is an ideal, and that either $H_s$ is a group or $H_sH_s\subseteq I_s$. The element $s$ is (von Neumannn) regular if $H_s$ is a group and otherwise is called null.

I will use $s^{\omega}$ to denote the idempotent positive power of $s$ and $s^{\omega+1}$ for $s^{\omega}s$.

The first claim shows that if $sS^1$ is a maximal principal ideal with $H_s$ a nontrivial group, then the conjecture is true. This generalizes Salvo's Proposition 1.

Claim 1. Suppose that $|S|\geq 4$ and $S$ has a maximal principal ideal $sS^1$ with $H_s$ a group. Then the conjecture is true.

Proof. Let $e$ be the the identity of $H_s$ and let $s'$ be the inverse of $s$ in $H_s$. Then for any $z\in S$ we have that $ez = s'(sz)$. Assume now that $z\neq e,s,s^2$. Since $sS^1$ was maximal, we cannot have $s'z\in\{e,s\}$ unless $z\in H_s$, in which case $z=s,s^2$, which we excluded. There are $n-3$ choices of $z$, more than enough to meet the desired bound. []

The next claim generalizes Salvo's Propositions 2 and 3. It is trickier.

Claim 2. Suppose that there is a maximal principal ideal $sS^1$ with $H_s$ not a group. Then there are $a\neq b\in S$ such that $az=bz$ for all $z\in S\setminus \{a,b\}$.

Proof. Our assumption implies we can find a minimal principal ideal $tS^1$ subject to the property that if $tS^1\subseteq uS^1$, then $H_u$ is not a group. Indeed, our assumption gives us at least one principal ideal with this property and hence there is a minimal such by finiteness. Then since $H_tH_t\subseteq I_t$, we must have that $t^k\notin H_t$ for all $k>1$.

We claim that $I_t$ is a monoid with identity $t^{\omega}$. For if $uS^1$ is a maximal principal ideal of $S$ contained in $I_t$, then by choice of $t$, we must have that $H_u$ is a group. Write $u=tx$ with $x\in S^1$. Choose $k>1$ such that $u^k=u^{\omega}$ and $t^k=t^{\omega}$. Then $u=u^ku = t^kx^ku=t^{\omega}t^kx^ku=t^{\omega}u$. Now if $v\in I_s$ is any element, then $v\in uS^1$ for some maximal principal left ideal of $S$ contained in $I_s$, and so if $v=uy$, then $t^{\omega}v=t^{\omega}uy=uy=v$ by the previous case.

I claim that $tz=t^{\omega+1}z$ for all $z\in S\setminus \{t,t^{\omega+1}\}$. First we observe that if $z\in S$, then $tz\in I_t$. Indeed, if $tz\in H_t$, then $tzS^1=tS^1$ and so $tzy=t$ for some $y\in S^1$. Then $t(zy)^{\omega}=t$ and $tS^1\subseteq (zy)^{\omega}S^1$. But $(zy)^{\omega}$ is regular, so this contradicts the choice of $t$. Thus $tz\in I_t$, and so by the above claim $tz = t^\omega(tz) = t^{\omega+1}z$. []

So we are left with the case that each maximal $\mathcal H$-class is an idempotent. Now one should try to do some sort of induction to get to the case each element is an idempotent.

Here is a generalization of some of @SalvoTringali's ideas. My basic feeling is that probably this conjecture boils down to the case of meet semilattices and hence isn't so far from the original union-closed sets conjecture. I'm hoping this approach will allow for induction.

I'll give two slight improvements on @SalvoTringali's answer. If $s\in S$, let $H_s=\{t\in S\mid sS^1= tS^1\}$ (the $\mathcal H$-class of $s$) and $I_s=sS^1\setminus H_s$. Here $S^1$ means $S$ with an adjoined identity. Then standard finite semigroup theory tells you $I_s$ is an ideal, and that either $H_s$ is a group or $H_sH_s\subseteq I_s$. The element $s$ is (von Neumannn) regular if $H_s$ is a group and otherwise is called null.

I will use $s^{\omega}$ to denote the idempotent positive power of $s$ and $s^{\omega+1}$ for $s^{\omega}s$.

The first claim shows that if $sS^1$ is a maximal principal ideal with $H_s$ a nontrivial group, then the conjecture is true. This generalizes Salvo's Proposition 1.

Claim 1. Suppose that $|S|\geq 4$ and $S$ has a maximal principal ideal $sS^1$ with $H_s$ a nontrivial group and $s$ not idempotent.

Proof. Let $e$ be the the identity of $H_s$ and let $s'$ be the inverse of $s$ in $H_s$. Then for any $z\in S$ we have that $ez = s'(sz)$. Assume now that $z\neq e,s,s^2$. Since $sS^1$ was maximal, we cannot have $s'z\in\{e,s\}$ unless $z\in H_s$, in which case $z=s,s^2$, which we excluded. There are $n-3$ choices of $z$, more than enough to meet the desired bound. []

The next claim generalizes Salvo's Propositions 2 and 3. It is trickier.

Claim 2. Suppose that there is a maximal principal ideal $sS^1$ with $H_s$ not a group. Then there are $a\neq b\in S$ such that $az=bz$ for all $z\in S\setminus \{a,b\}$.

Proof. Our assumption implies we can find a minimal principal ideal $tS^1$ subject to the property that if $tS^1\subseteq uS^1$, then $H_u$ is not a group. Indeed, our assumption gives us at least one principal ideal with this property and hence there is a minimal such by finiteness. Then since $H_tH_t\subseteq I_t$, we must have that $t^k\notin H_t$ for all $k>1$.

We claim that $I_t$ is a monoid with identity $t^{\omega}$. For if $uS^1$ is a maximal principal ideal of $S$ contained in $I_t$, then by choice of $t$, we must have that $H_u$ is a group. Write $u=tx$ with $x\in S^1$. Choose $k>1$ such that $u^k=u^{\omega}$ and $t^k=t^{\omega}$. Then $u=u^ku = t^kx^ku=t^{\omega}t^kx^ku=t^{\omega}u$. Now if $v\in I_s$ is any element, then $v\in uS^1$ for some maximal principal left ideal of $S$ contained in $I_s$, and so if $v=uy$, then $t^{\omega}v=t^{\omega}uy=uy=v$ by the previous case.

I claim that $tz=t^{\omega+1}z$ for all $z\in S\setminus \{t,t^{\omega+1}\}$. First we observe that if $z\in S$, then $tz\in I_t$. Indeed, if $tz\in H_t$, then $tzS^1=tS^1$ and so $tzy=t$ for some $y\in S^1$. Then $t(zy)^{\omega}=t$ and $tS^1\subseteq (zy)^{\omega}S^1$. But $(zy)^{\omega}$ is regular, so this contradicts the choice of $t$. Thus $tz\in I_t$, and so by the above claim $tz = t^\omega(tz) = t^{\omega+1}z$. []

So we are left with the case that each maximal $\mathcal H$-class is an idempotent. Now one should try to do some sort of induction to get to the case each element is an idempotent.

Here is a generalization of some of @SalvoTringali's ideas. My basic feeling is that probably this conjecture boils down to the case of meet semilattices and hence isn't so far from the original union-closed sets conjecture. I'm hoping this approach will allow for induction.

I'll give two slight improvements on @SalvoTringali's answer. If $s\in S$, let $H_s=\{t\in S\mid sS^1= tS^1\}$ (the $\mathcal H$-class of $s$) and $I_s=sS^1\setminus H_s$. Here $S^1$ means $S$ with an adjoined identity. Then standard finite semigroup theory tells you $I_s$ is an ideal, and that either $H_s$ is a group or $H_sH_s\subseteq I_s$. The element $s$ is (von Neumannn) regular if $H_s$ is a group and otherwise is called null.

I will use $s^{\omega}$ to denote the idempotent positive power of $s$ and $s^{\omega+1}$ for $s^{\omega}s$.

The first claim shows that if $sS^1$ is a maximal principal ideal with $H_s$ a nontrivial group, then the conjecture is true. This generalizes Salvo's Proposition 1.

Claim 1. Suppose that $|S|\geq 4$ and $S$ has a maximal principal ideal $sS^1$ with $H_s$ a group. Then the conjecture is true.

Proof. Let $e$ be the the identity of $H_s$ and let $s'$ be the inverse of $s$ in $H_s$. Then for any $z\in S$ we have that $ez = s'(sz)$. Assume now that $z\neq e,s,s^2$. Since $sS^1$ was maximal, we cannot have $s'z\in\{e,s\}$ unless $z\in H_s$, in which case $z=s,s^2$, which we excluded. There are $n-3$ choices of $z$, more than enough to meet the desired bound. []

The next claim generalizes Salvo's Propositions 2 and 3. It is trickier.

Claim 2. Suppose that there is a maximal principal ideal $sS^1$ with $H_s$ not a group. Then there are $a\neq b\in S$ such that $az=bz$ for all $z\in S\setminus \{a,b\}$.

Proof. Our assumption implies we can find a minimal principal ideal $tS^1$ subject to the property that if $tS^1\subseteq uS^1$, then $H_u$ is not a group. Indeed, our assumption gives us at least one principal ideal with this property and hence there is a minimal such by finiteness. Then since $H_tH_t\subseteq I_t$, we must have that $t^k\notin H_t$ for all $k>1$.

We claim that $I_t$ is a monoid with identity $t^{\omega}$. For if $uS^1$ is a maximal principal ideal of $S$ contained in $I_t$, then by choice of $t$, we must have that $H_u$ is a group. Write $u=tx$ with $x\in S^1$. Choose $k>1$ such that $u^k=u^{\omega}$ and $t^k=t^{\omega}$. Then $u=u^ku = t^kx^ku=t^{\omega}t^kx^ku=t^{\omega}u$. Now if $v\in I_s$ is any element, then $v\in uS^1$ for some maximal principal left ideal of $S$ contained in $I_s$, and so if $v=uy$, then $t^{\omega}v=t^{\omega}uy=uy=v$ by the previous case.

I claim that $tz=t^{\omega+1}z$ for all $z\in S\setminus \{t,t^{\omega+1}\}$. First we observe that if $z\in S$, then $tz\in I_s$. Indeed, if $tz\in H_s$, then $tzS^1=tS^1$ and so $tzy=t$ for some $y\in S^1$. Then $t(zy)^{\omega}=t$ and $tS^1\subseteq (zy)^{\omega}S^1$. But $(zy)^{\omega}$ is regular, so this contradicts the choice of $t$. Thus $tz\in I_s$, and so by the above claim $tz = t^\omega(tz) = t^{\omega+1}z$. []

So we are left with the case that each maximal $\mathcal H$-class is an idempotent. Now one should try to do some sort of induction to get to the case each element is an idempotent.

Here is a generalization of some of @SalvoTringali's ideas. My basic feeling is that probably this conjecture boils down to the case of meet semilattices and hence isn't so far from the original union-closed sets conjecture. I'm hoping this approach will allow for induction.

I'll give two slight improvements on @SalvoTringali's answer. If $s\in S$, let $H_s=\{t\in S\mid sS^1= tS^1\}$ (the $\mathcal H$-class of $s$) and $I_s=sS^1\setminus H_s$. Here $S^1$ means $S$ with an adjoined identity. Then standard finite semigroup theory tells you $I_s$ is an ideal, and that either $H_s$ is a group or $H_sH_s\subseteq I_s$. The element $s$ is (von Neumannn) regular if $H_s$ is a group and otherwise is called null.

I will use $s^{\omega}$ to denote the idempotent positive power of $s$ and $s^{\omega+1}$ for $s^{\omega}s$.

The first claim shows that if $sS^1$ is a maximal principal ideal with $H_s$ a nontrivial group, then the conjecture is true. This generalizes Salvo's Proposition 1.

Claim 1. Suppose that $|S|\geq 4$ and $S$ has a maximal principal ideal $sS^1$ with $H_s$ a group. Then the conjecture is true.

Proof. Let $e$ be the the identity of $H_s$ and let $s'$ be the inverse of $s$ in $H_s$. Then for any $z\in S$ we have that $ez = s'(sz)$. Assume now that $z\neq e,s,s^2$. Since $sS^1$ was maximal, we cannot have $s'z\in\{e,s\}$ unless $z\in H_s$, in which case $z=s,s^2$, which we excluded. There are $n-3$ choices of $z$, more than enough to meet the desired bound. []

The next claim generalizes Salvo's Propositions 2 and 3. It is trickier.

Claim 2. Suppose that there is a maximal principal ideal $sS^1$ with $H_s$ not a group. Then there are $a\neq b\in S$ such that $az=bz$ for all $z\in S\setminus \{a,b\}$.

Proof. Our assumption implies we can find a minimal principal ideal $tS^1$ subject to the property that if $tS^1\subseteq uS^1$, then $H_u$ is not a group. Indeed, our assumption gives us at least one principal ideal with this property and hence there is a minimal such by finiteness. Then since $H_tH_t\subseteq I_t$, we must have that $t^k\notin H_t$ for all $k>1$.

We claim that $I_t$ is a monoid with identity $t^{\omega}$. For if $uS^1$ is a maximal principal ideal of $S$ contained in $I_t$, then by choice of $t$, we must have that $H_u$ is a group. Write $u=tx$ with $x\in S^1$. Choose $k>1$ such that $u^k=u^{\omega}$ and $t^k=t^{\omega}$. Then $u=u^ku = t^kx^ku=t^{\omega}t^kx^ku=t^{\omega}u$. Now if $v\in I_s$ is any element, then $v\in uS^1$ for some maximal principal left ideal of $S$ contained in $I_s$, and so if $v=uy$, then $t^{\omega}v=t^{\omega}uy=uy=v$ by the previous case.

I claim that $tz=t^{\omega+1}z$ for all $z\in S\setminus \{t,t^{\omega+1}\}$. First we observe that if $z\in S$, then $tz\in I_t$. Indeed, if $tz\in H_t$, then $tzS^1=tS^1$ and so $tzy=t$ for some $y\in S^1$. Then $t(zy)^{\omega}=t$ and $tS^1\subseteq (zy)^{\omega}S^1$. But $(zy)^{\omega}$ is regular, so this contradicts the choice of $t$. Thus $tz\in I_t$, and so by the above claim $tz = t^\omega(tz) = t^{\omega+1}z$. []

So we are left with the case that each maximal $\mathcal H$-class is an idempotent. Now one should try to do some sort of induction to get to the case each element is an idempotent.