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, $f:X\times\cdots\times X\to X$ be a continuous map and $x\in X$ such that $f(x,\ldots,x)=x$ . Then $f(A\times\cdots\times A)\subseteq A$ , 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 $X\times\cdots\times X$ . Thus (from the definition of product topology and continuous map), there are open subsetes $V_1,\ldots, V_n\subseteq X$ such that each $V_i$ is an open neighbourhood of $x$ and $$f(V_1\times\cdots\times V_n)\subseteq A.$$ 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 $$f(U_1\times\cdots\times U_n)\subseteq A.$$ 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 $$f(A\times\cdots\times A)=f(U_1\times\cdots\times U_n)\subseteq A,$$ which completes the proof. $\square$

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 $$U=\bigcup_{i\in I}\, g_iH.$$

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: $$M:G\times G\to G\mbox{ is such that }M(x,y)=xy;$$ $$\psi:G\to G\mbox{ is such that }\psi(x)=x^{-1};$$ $$\varphi_{g}:G\to G\mbox{ is such that }\varphi_{g}(x)=gxg^{-1}\mbox{ for any }g\in G.$$ 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: $$HH=M(H\times H)\subseteq H;$$ $$H^{-1}=\psi(H)\subseteq H;$$ $$gHg^{-1}=\varphi_{g}(H)\subseteq H\mbox{ for any }g\in G.$$ This shows that $H$ is a normal subgroup, which completes the proof. $\square$

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. $\square$