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Quotient posets of the Boolean lattice $B_n$ have interesting properties and are for example discussed in chapter 5 of Stanley's book on algebraic combinatorics.

$B_n/G$ for a subgroup $G$ of the symmetric group $S_n$ (that acts naturally on $B_n$) is defined as the poset of orbits under the natural order (that is one orbit $a$ is $\geq$ another orbit $b$ if and only if there exists elements $a' \in a$ and $b' \in b$ such that $a' \geq b'$).

The posets $B_n /G$ are graded of rank $n$, rank-symmetric, rank-unimodal, and Sperner. See theorem 5.8. in the book of Stanley. One might ask whether one can add some more conditions to have a characterisation of them by natural properties (not mentioning the Boolean lattice in this characterisation).

Question: Is there a characterisation when a given bounded poset $P$ is isomorphic to $B_n /G$ for a subgroup $G$ of the symmetric group?

Can this at least be done when $G$ is assumed to be cyclic?

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    $\begingroup$ My initial reaction was "because any poset automorphism is a boolean algebra automorphism, won't this poset quotient be equally a boolean algebra quotient?". But of course this is wrong. For instance, quotienting $\mathbb 2 \times \mathbb 2$ by the the obvious involution, we obtain the ordinal $\mathbb 3$, which is not a boolean algebra. $\endgroup$
    – Tim Campion
    Commented Aug 23, 2020 at 15:22
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    $\begingroup$ There are numerous necessary conditions, but I doubt whether there is a useful necessary and sufficient condition. For instance, if $k<n/2$ then there must be a set of disjoint saturated chains going from every element of rank $k$ to every element of rank $n-k$. See math.mit.edu/~rstan/pubs/pubfiles/60.pdf. Another more trivial necessary condition is that the number of elements of rank $k$ cannot exceed ${n\choose k}$. $\endgroup$
    – Richard Stanley
    Commented Aug 24, 2020 at 2:11
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    $\begingroup$ @Mare: this is not known to me and looks impossibly difficult. $\endgroup$
    – Richard Stanley
    Commented Aug 24, 2020 at 13:35
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    $\begingroup$ What happens if we extend the set to also include quotients of the form $L_n(q)/G$ where $L_n(q)$ is the lattice of linear subspaces of $\mathbb F_q^n$ and $G$ is some subgroup of $GL_n(\mathbb F_q)$. Is there any chance this can make the classification easier? $\endgroup$
    – Gjergji Zaimi
    Commented Aug 24, 2020 at 16:20
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    $\begingroup$ A further property of $B_n/G$: if $n\geq 12$ and $B_n/G$ has exactly one element of rank 6, then $B_n/G$ is a chain. This is a deep result requiring the classification of finite simple groups. See math.stackexchange.com/questions/388488/…. $\endgroup$
    – Richard Stanley
    Commented Aug 24, 2020 at 16:29

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