# if we set topology on a group like that, is it important?

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.

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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
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
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

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 $G=\mathbb{Z}/2\mathbb{Z}$).