Self-homeomorphisms of finite topological spaces This is a problem I came up with today.
Let's call a finite group $G$ "permissible" if there exists a finite set $X$ such that there exist a topology on $X$ for which $\text{Homeo}(X)=G$. The problem is to classify all permissible groups.
I cooked up a proof that this family of groups is generated by taking iterated products and wreath products of symmetric groups, but my proof is clunky (Edit: and wrong. Thanks to Eric Wofsey for pointing this out.). I hope there exists a prettier proof.
 A: Every finite group is the homeomorphism group of some finite $T_0$ space.  This follows fairly easily from the fact that a $T_0$ topology on a finite set is the same thing as a partial order.  Here is a simple construction of a poset with a given automorphism group; there are many others you could give.
Let $G$ be any nontrivial finite group and let $X=G\times\{0,1\}\sqcup G\times G\times\{0,1\}$, ordered as follows.  First, choose some total ordering on $G$, and let that be the ordering of $G\times\{1\}$ as a subset of $X$.  Second, for any $g,h\in G$ declare that $(g,0)\leq (g,h,0)$, $(h,1)\leq (g,h,0)$, $(g,h,0)\leq(g,h,1)$ and $(gh,0)\leq (g,h,1)$.
With this ordering, it is easy to see that any automorphism $\alpha:X\to X$ must fix every point of $G\times\{1\}$ and map $G\times \{0\}$ to itself by some map $\varphi:G\to G$.  It then follows that $\alpha(g,h,0)=(\varphi(g),h,0)$ and $\alpha(g,h,1)=(\varphi(g),h,1)$ for all $g$ and $h$, and thus that $\varphi(gh)=\varphi(g)h$.  In particular, this means $\alpha$ is completely determined by the function $\varphi$, and $\varphi$ in turn is completely determined by $\varphi(1)$ by the formula $\varphi(h)=\varphi(1)h$.  It is now not hard to see that $\alpha\mapsto \varphi(1)$ defines an isomorphism from the automorphism group of $X$ to $G$.
(The idea behind this construction is to encode the action of $G$ on itself by right translation in the poset $X$, so automorphisms of $X$ are the same as automorphisms of $G$ as a a right $G$-set.  The automorphisms of $G$ as a right $G$-set are exactly the left translations, which are isomorphic to $G$.  Here $G\times\{1\}$ should be thought of as the group $G$, $G\times\{0\}$ should be thought of as the underlying set of $G$ which is being acted on by $G\times\{1\}$ by translation, and $G\times G\times \{0,1\}$ is just an auxiliary set that is used to encode that action in the structure of the poset.)
A: First of all, we can obtain the trivial subgroup by choosing an ordering of $X$: $X\approx \{1,...,n\}$, and placing upon it the order topology $\tau=\{\{1,...,i\}\ |\ 1\leq i \leq n\}$. So the trivial subgroup is permissible.
More generally, we may choose a partition $X\approx \coprod_{i=1}^k X_i$, each $X_i$ being a topological space with topology $\tau_i$. Now choose a topology $\tau$ on $X$ given by $\tau=\cup_{i=1}^k \left\{ \left( \cup_{j=1}^{i-1} X_j \right)\cup U | U\in \tau_i \right\}$. This gives $\text{Homeo}(X)=\prod_{i=1}^k \text{Homeo}(X_i)$, so products of permissible subgroups, when they exist, are permissible. In particular products $\prod_{i=1}^k \Sigma_{n_1}$ are permissible.
If we have $p$ copies of a space $X'$, $X=\coprod_{i=1}^p X'$, then we may calculate $\text{Homeo}(X)=\Sigma_n \wr \text{Homeo}(X')$ (the wreath product), so the wreath product of $\Sigma_n$ and a permissible group is again permissible.
Now, I claim that these are all the examples. I.e. the permissible groups are the family generated by the symmetric groups by taking iterated products and wreath products. So a typical member of this family might look like $\Sigma_a\times \left(\Sigma_b\wr (\Sigma_c\times\Sigma_d\times (\Sigma_e\wr \Sigma_f))\right)\times (\Sigma_g\wr \Sigma_h)$. Let us denote this family by $\mathcal{F}$.
First of all, it is sufficient to consider connected spaces. If a space $X$ is not connected, then $\text{Homeo}(X)$ will be a mixture of products and wreath products of the groups of its connected components, depending on their homeomorphism types. Explicitly, grouping the components by homeomorphism type $X=\coprod_{i=1}^k \coprod_{j=1}^{n_i} X_i$, we have $\text{Homeo}(X)=\prod_{i=1}^k \left( \Sigma_{n_i}\wr \text{Homeo}(X_i) \right)$.
Now given a connected finite space $X$ we say an element $x\in X$ has length $l(x)=l$ if $l$ is the length of the maximal chain of proper inclusions of open sets containing $x$. The supremum of $l(x)$ in $X$ is called the length of $X$ and denoted by $l(X)$. We now work by induction on $l(X)$. If $l=0$, then $X$ is indiscrete, so $\text{Homeo}(X)=\Sigma_n$ for some $n$. Assume then for any space $X'$ of length $l(X')\leq k$ we have $\text{Homeo}(X)\in \mathcal{F}$, and let $X$ be a space of length $k+1$. Now the image of an open set $U$ under a homeomorphism $\phi$ inherits the open subsets of $U$, in particular the length of $U$ is preserved. In other words, a homeomorphism will permute the subset of elements in $X$ of maximal length. In other words, $\phi$ acts independently on $\{x\in X \ |\ l(x)=l(X)\}$ and $\{x\in X \ |\ l(x)<l(X)\}$ and the latter subspace has length $k$. In other words, $\text{Homeo}(X)\in \mathcal{F}$. This shows that permissible groups are contained in $\mathcal{F}$, and by the remarks above, any group in $\mathcal{F}$ can be realized by a space, so the reverse inclusion also holds.
