# 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. $\mathrm{\square}$

Title | characterization of Alexandroff groups |
---|---|

Canonical name | CharacterizationOfAlexandroffGroups |

Date of creation | 2013-03-22 18:45:43 |

Last modified on | 2013-03-22 18:45:43 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 22A05 |