characterization of Alexandroff groups

Topological group $G$ is called Alexandroff if $G$ is an Alexandroff space as a topological space. For example every finite topological group is Alexandroff. We wish to characterize them. First recall, that if $A$ is a subset of a topological space, then $A^o$ denotes an intersection of all open neighbourhoods of $A$.

Lemma. Let $X$ be an Alexandroff space, $f:X\times \mathrm{\cdots }\times X\to X$ be a continuous map and $x\in X$ such that $f(x,\mathrm{\dots },x)=x$. Then $f(A\times \mathrm{\cdots }\times A)\subseteq A$, where $A={\{x\}}^o$.

Proof. Let $A={\{x\}}^o$. Of course $A$ is open (because $X$ is Alexandroff). Therefore $f^{-1}(A)$ is open in $X\times \mathrm{\cdots }\times X$. Thus (from the definition of product topology and continuous map), there are open subsetes $V_1,\mathrm{\dots },V_n\subseteq X$ such that each $V_i$ is an open neighbourhood of $x$ and

$$f(V_1\times \mathrm{\cdots }\times V_n)\subseteq A.$$

Now let $U_i=V_i\cap A$. Of course $x\in U_i$, so $U_i$ is nonempty and $U_i$ is open. Furthermore $U_i\subseteq V_i$ and thus

$$f(U_1\times \mathrm{\cdots }\times U_n)\subseteq A.$$

On the other hand $U_i\subseteq A$ and $U_i$ is open neighbourhood of $x$. Thus $U_i=A$, because $A$ is minimal open neighbourhood of $x$. Therefore

$$f(A\times \mathrm{\cdots }\times A)=f(U_1\times \mathrm{\cdots }\times U_n)\subseteq A,$$

which completes the proof. $\mathrm{\square }$

Proposition. Let $G$ be an Alexandroff group. Then there exists open, normal subgroup $H$ of $G$ such that for every open subset $U\subseteq G$ there exist ${\{g_i\}}_{i\in I}\subseteq G$ such that

$$U=\bigcup _{i\in I}{g}_{i}H.$$

Proof. Let $H={\{e\}}^o$ be an intersection of all open neighbourhoods of the identity $e\in G$. Let $U$ be an open subset of $G$. If $g\in U$, then $g^{-1}U$ is an open neighbourhood of $e$. Thus $H\subseteq g^{-1}U$ and therefore $gH\subseteq U$. Thus

$$U=\bigcup _{g\in U}gH.$$

To complete the proof we need to show that $H$ is normal subgroup of $G$. Consider the following mappings:

$$M:G\times G\to G\text{ is such that }M(x,y)=xy;$$

$$\psi :G\to G\text{ is such that }\psi (x)=x^{-1};$$

$${\phi }_g:G\to G\text{ is such that }{\phi }_g(x)=gx{g}^{-1}\text{ for any }g\in G.$$

Of course each of them is continuous (because $G$ is a topological group). Furthermore each of them satisfies Lemma’s assumptions (for $x=e$). Thus we have:

$$HH=M(H\times H)\subseteq H;$$

$$H^{-1}=\psi (H)\subseteq H;$$

$$gH{g}^{-1}={\phi }_g(H)\subseteq H\text{ for any }g\in G.$$

This shows that $H$ is a normal subgroup, which completes the proof. $\mathrm{\square }$

Corollary. Let $G$ be a topological group such that $G$ is finite and simple. Then $G$ is either discrete or antidiscrete.

Proof. Of course finite topological groups are Alexandroff. Since $G$ is simple, then there are only two normal subgroups of $G$, namely the trivial group and entire $G$. Therfore (due to proposition) the topology on $G$ is ,,generated” by either the trivial group or entire $G$. In the first case we gain the discrete topology and in the second the antidiscrete topology. $\mathrm{\square }$