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Let $G$ be a group and $\omega$ be set of all subgroup of $G$.

Since $\omega$ is closed under intersection, it is trivial to check that $\omega$ satisfies to conditions to be a base.

Thus,Let $T$ be topology on $G$ induced by $\omega$.

One trivial observation is that every subgroup of $G$ is open under this topology and every automorphism of $G$ is also continious under this topology since inverse of "subgroups" are "subgroups"and so are their unions which are really interesting for me.

I wonder whether this topology has importance in terms of "Topology" or "Group Theory", any observation or comment is welcome.

mesel
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  • Well, for one thing, every element $x \in G$ has a smallest open neighbourhood: the cyclic group generated by $x$. In particular, a sequence (or net) will converge to $x$ if and only if it is eventually contained in the cyclic group generated by $x$. In fact, I suppose you could use just the cyclic groups as a basis and generate the same topology. – Mike F Feb 24 '14 at 21:21
  • @Mike: Corrallary, $x_n$ goes to $e$ if and only if $x_n=e$ for $n>n_0$ :) – mesel Feb 24 '14 at 21:26
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    Another comment: it seems $G$ will be compact if and only if there is a finite set ${x_1,\ldots,x_n} \subset G$ such that $G = \bigcup_{i=1}^n \langle x_i \rangle$, where $\langle x \rangle$ denotes the cyclic group generated by $x \in G$. – Mike F Feb 24 '14 at 21:26
  • Right, in which case $x_n$ also converges to every other point in $G$ :) – Mike F Feb 24 '14 at 21:30
  • So,every subgroup which are union of finite cylic group are compact which means copmact set must be open.As far as I know,for nice topology, copmact sets are closed?? – mesel Feb 24 '14 at 21:37
  • Well, all subgroups are open regardless of whether they are compact because of how you defined the topology. Other subsets can by compact without being open, for example, any finite set $S \subset G$ with $e \notin S$ is compact, but not open. – Mike F Feb 24 '14 at 21:44
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    By the way, compact sets do not need to be closed if the topology is not Hausdorff, and the topology under discussion here is obviously not Hausdorff. The closure of ${e}$ is the whole group. – Mike F Feb 24 '14 at 21:44
  • relevant to the finite case of this question – Alexander Gruber Feb 25 '14 at 02:59

1 Answers1

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Sorry if my answer comes so late, I hope you're still interested in the question you posed.

I don't think that the topology you propose is of much interest, mainly for the following two reasons:

  1. As Mike F noticed in the comments, the topology is not really well behaved, in the sense that in general it is not Hausdorff ($T_2$). As you might notice, in fact it is not even Kolmogorov ($T_0$), which is really the weakest separation property there is. For example if you take $G=\mathbb{Z}$, then the elements $1$ and $-1$ cannot be separated (in the $T_0$ sense).
  2. The topology behaves badly with respect to the group structure, in the sense that in general multiplication will not be continuous. Indeed if $m:G\times G\to G$ denotes the multiplication, then $m^{-1}(\{e\})=\bigcup_{g\in G}(g,g^{-1})$, which is usually not open (try for example with $G=\mathbb{Z}/2\mathbb{Z}$).
Daniel Robert-Nicoud
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  • Thanks for your answer, I am still interested in the question. I am convined that it has no importance in terms of topology, now I wonder whether it has importance in terms of "Group theory". – mesel Jun 05 '14 at 12:37