Automorphisms of a finite partition lattice

Question

Is the group of automorphisms of a lattice of partitions of the set $X$, where $|X| = n$, isomorphic to $S_n$? I think it is (for sure it's not 'smaller' than $S_n$), but i can't find any proof of it in papers or books. I did try to prove it myself but it's not that easy, can't think of a counterexample either.

Answer 1

Is the group of automorphisms of a lattice of partitions of the set $X$, where $|X|=n$, isomorphic to $S_n$?

Yes, if $n\neq 2$.

Assume that $X = \{1,2,\ldots,n\}$. Let $\Pi(X)$ be the lattice of partitions of $X$. For each $i\in X$, let $\widehat{i}$ ($i$ hat) be the partition $\{\{i\}, X\setminus\{i\}\}$ of $X$ that has two cells where one is the singleton $\{i\}$ and the other is the complement $X\setminus \{i\}$. Let $\widehat{X} = \{\widehat{1},\widehat{2},\ldots,\widehat{n}\}$, which is a subset of $\Pi(X)$. I will indicate why $\widehat{X}$ is an $\textrm{Aut}(\Pi(X))$-orbit and why the restriction map $\rho\colon \alpha\mapsto \alpha|_{\widehat{X}}$ is an isomorphism of $\textrm{Aut}(\Pi(X))$ onto $\textrm{Sym}(\widehat{X})$.

1. Every automorphism of $\Pi(X)$ permutes the elements of $\widehat{X}$.
   (This says that $\widehat{X}$ is a union of $\textrm{Aut}(\Pi(X))$-orbits, which is enough to guarantee that the restriction map $\rho$ is a homomorphism from $\textrm{Aut}(\Pi(X))$ to $\textrm{Sym}(\widehat{X})$.)
2. Distinct automorphisms of $\Pi(X)$ have distinct restrictions to $\widehat{X}$. ($\rho$ is injective.)
3. Every permutation of $\widehat{X}$ is the restriction of some automorphism of $\Pi(X)$. ($\rho$ is surjective.)

The reason the first claim is true when $n>2$ is that $\widehat{X}$ is definable as the set of partitions of $X$ that are coatoms of $\Pi(X)$ that have exactly $|X|-1$ complementary atoms. (Then we use the fact that lattice automorphisms must map 'coatoms with $|X|-1$ complementary atoms' to other 'coatoms with $|X|-1$ complementary atoms'.) To see that $\widehat{X}$ is exactly the set of partitions of $X$ that are coatoms that have exactly $|X|-1$ complementary atoms, note that the coatoms of $\Pi(X)$ are the partitions of $X$ with $2$ cells, say $\{I,J\}$, while the atoms of $\Pi(X)$ are the partitions that have exactly one nonsingleton cell and that nonsingleton cell has $2$ elements. The number of atoms that are complementary to the coatom $\{I,J\}$ is $|I|\cdot |J|$, since the nonsingleton cell of a complementary atom will have the form $\{i,j\}$ where $i\in I$ and $j\in J$. Now, if $|I|+|J|=|X|$ and $|I|\cdot |J|=|X|-1$, then $|I|$ and $J$ are the roots of the polynomial $f(t)=t^2-|X|\cdot t + (|X|-1)$, hence $|I|$ and $|J|$ are the numbers $1$ and $|X|-1$ in some order. This is enough to show that a coatom of $\Pi(X)$ has exactly $|X|-1$ complementary atoms iff the coatom is a member of $\widehat{X}$.

The reason the second claim is true when $n>2$ is that $\widehat{X}$ is a lattice-theoretic generating set for $\Pi(X)$. (Every atom in $\Pi(X)$ is a meet of elements of $\widehat{X}$, and every element of $\Pi(X)$ is a join of atoms. Thus $\bigvee \bigwedge \widehat{X} = \Pi(X)$.) Since two lattice automorphisms that agree on a lattice-theoretic generating set are equal, distinct automorphisms of $\Pi(X)$ will have distinct restrictions to $\widehat{X}$

The reason the third claim is true when $n>2$ is that any permutation $\pi$ of the set $X$ induces an automorphism $\bar{\pi}$ on $\Pi(X)$ whose action on $\widehat{X}$ mirrors the action of $\pi$ on $X$: $\bar{\pi}(\widehat{i})=\widehat{\pi(i)}$. For example, if $X=\{1,2,3,4\}$ and $\pi=(1\;2)(3\;4)\in \textrm{Sym}(X)$, then $\bar{\pi}$ is the automorphism of $\Pi(X)$ that maps a partition $\{I,J,K\}$ to $\{\pi(I),\pi(J),\pi(K)\}$ and in particular it maps $\widehat{i} = \{\{i\},X\setminus \{i\}\}$ to $\{\{\pi(i)\},X\setminus \{\pi(i)\}\}$. Thus $\bar{\pi}|_{\widehat{X}}=(\widehat{1}\;\widehat{2})(\widehat{3}\;\widehat{4})$.