I'm working on the following problem,

Let $(G,\tau)$ be an abelian topological group, and $H = \bigcap_{e \in U \in \tau}U$ be the intersection of open sets (or equivalently neighbourhoods) of the identity. Prove that $H$ is a subgroup, and that $H = \overline{\{e\}}$. Moreover, conclude that $G/H$ is Hausdorff and that $G$ will be Hausdorff if and only if $H = \{e\}$. Which results hold when $G$ is not abelian?

I have already proved that $H$ is a subgroup and the closure of the identity, without appealing to the commutativity of $G$.

I also have seen the last statement is true, provided that $G/H$ is Hausdorff: if $G$ is Hausdorff then it is $T_1$ which implies $H = \{e\}$, and reciprocally if $H = \{e\}$, then $G \simeq G/H$ which is Hausdorff by hypothesis.

Thus the remaining task is proving that $G/H$ is Hausdorff, and in the case that this fails for non-abelian groups, I should have to show that $G$ is Hausdorff when $H = \{e\}$ via a different argument. So far, I haven't bumped into any difficulties that would indicate the need for commutativity, which makes me suspect that it may not be needed at all. In any case, my approach to see that $G/H$ is Hausdorff was the following: it is equivalent to see that $G/H$ is a topological group which is $T_0$. Since for any $g \in G$, conjugation $c_g$ by $g$ is a homeomorphism,

$$ c_g(H) = \bigcap_{e \in U \in \tau}c_g(U) = \bigcap_{e \in U \in \tau}U = H \quad (\forall g \in G) $$

and so $H$ is normal. This makes $G/H$ be well defined as a group with the operations induced by the projection, which then gives $G/H$ the structure of a topological group.

Thus, it remains to see that $G/H$ is $T_0$, and here is where I've become stuck. I have observed, though, that it suffices to prove the $T_0$ condition for pairs of points of the form $(e,g)$, which then can be generalized via the homeomorphisms given by left multiplication, because separating $h$ from $h'$ can be translated into separating $e$ and $h^{-1}h'$. So, to sum up, I have classes $[e]$ and $[g]$ in $G/H$ and want an open set $U$ so that either $[e] \in U$ and $[g] \not \in U$ or $[g] \in U$ and $[e] \not \in U$.

Could you prove some hints on how to continue from here, in case this approach is correct?

  • $\begingroup$ Note that $H$ could be all of $G$. Also, "the remaining task if", and, in the second to last line, I think you mean $[g] \not \in U$. $\endgroup$ – mathworker21 Sep 23 '18 at 23:20
  • $\begingroup$ Thanks, fixed. As to what you have pointed out, I can't quite see the implications of it. Would you mind elaborating? The case where $H = G$ seems to verify all of what's asked and once again, there's no need for $G$ being abelian in that situation, right? $\endgroup$ – Guido A. Sep 23 '18 at 23:27
  • $\begingroup$ Would you mind making that into an answer? Also, I take from here that I was correct then in assuming that there was no need for $G$ to be abelian? $\endgroup$ – Guido A. Sep 23 '18 at 23:40

Quick note: you can't treat $H$ as an arbitrary subgroup. Consider the example of the indiscrete topology of G and H any nontrivial subgroup. Then G/H is not Hausdorff. The good thing for us is that if $G$ has the indiscrete topology, then $H = G$, so $G/H = \emptyset$ and we're safe.

Proof that $G/\overline{\{e\}}$ is Hausdorff: If $g\not \in \overline{\{e\}}$, then by definition, there is some open set $U \ni g$ of $G$ such that $U \cap H = \emptyset$. Then if $\pi : G \to G/H$ is the obvious thing, $\pi(U)$ will be open in $G/H$ and satisfy $[g] \in \pi(U)$ but $[e] \not \in \pi(U)$.

I did not use that $G$ is Abelian in the above argument (I don't think).

  • $\begingroup$ I'm not seeing why $\pi(U)$ ought to be open. Are you using some additional structure of this particular quotient? $\endgroup$ – Guido A. Sep 23 '18 at 23:46
  • $\begingroup$ Ok, so $G/H$ has the topology provided by the quotient map $\pi$. It's a fact about topological groups that $\pi$ is always an open map. So since $U$ is open, $\pi(U)$ is open. $\endgroup$ – mathworker21 Sep 23 '18 at 23:49
  • $\begingroup$ math.wm.edu/~vinroot/PadicGroups/topgroups.pdf proposition 3.1 here proves this fact. $\endgroup$ – mathworker21 Sep 23 '18 at 23:50
  • $\begingroup$ Great, thanks. I know close to nothing about topological groups, and this was a 'tangential' exercise I was working on, so those notes will be useful. $\endgroup$ – Guido A. Sep 23 '18 at 23:51

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