# 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, $f:X\times\cdots\times X\to X$ 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. $\square$

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:

 $M:G\times G\to G\mbox{ is such that }M(x,y)=xy;$
 $\psi:G\to G\mbox{ is such that }\psi(x)=x^{-1};$
 $\varphi_{g}:G\to G\mbox{ is such that }\varphi_{g}(x)=gxg^{-1}\mbox{ for any }% g\in G.$

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. $\square$

Title characterization of Alexandroff groups CharacterizationOfAlexandroffGroups 2013-03-22 18:45:43 2013-03-22 18:45:43 joking (16130) joking (16130) 4 joking (16130) Theorem msc 22A05