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We consider a Hausdorff topological group $G$ acting on a topological space $X$ [action simply means a continuous map $G\times X\rightarrow X$ verifying $(gh)(x)=g(h(x))$, and $1(x)=x$].

The set $H=\{g\in G : \forall x\in X, g(x)=x\}$ is a normal subgroup of G. For example, it is closed if singletons are closed in $X$ .

I would like to find a Hausdorff topological group $\mathbf{G}$ and a topological space $\mathbf{X}$ such that $\mathbf{H}$ is not closed.

Note: If we do not require $G$ to be Hausdorff there is a plethora of counterexamples.

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I am assuming that the space $X$ is $T1$ (i.e. points are closed). Consider the orbit map $G\rightarrow X$ taking $g\mapsto gx$. If the action is continuous then this maps is continuous. Then $H$ is the preimage of $x$ which is closed. Am I missing something? –  DBS Jul 29 '13 at 3:35
Of course, if singletons are closed in $X$ then $H$ is closed. But that is not what I am asking is it? In any case I clarified the post above. –  John Jul 29 '13 at 3:41
Okay thanks for the clarification. If you take $G= \mathbb{R}/\mathbb{Z}$ and $H = \mathbb{Q}/\mathbb{Z}$ inside $G$. The $H$ is a normal subgroup of $G$. Take $X= G/H$ with left $G$ action. Then $H$ is not a closed subgroup of $G$ but it is the stabilizer of a point $e.H$. –  DBS Jul 29 '13 at 3:49
Isn't that an answer? –  Kevin Carlson Jul 29 '13 at 4:07
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1 Answer

up vote 3 down vote accepted

Let $G$ be a Hausdorff topological group. Let $N$ be a non-closed normal subgroup of $G$. Let $X = G/N$. $G$ acts on $X$. Let $g$ be an element of $G$. We denote by $G_g$ the set $\{h \in G |\ hgN = gN\}$. Then $G_g = gNg^{-1} = N$. Hence $H = \{g\in G : \forall x\in X, g(x)=x\} = \bigcap \{G_g |\ \forall g \in G\} = N$. Hence $H$ is not closed.

For example, let $G = \mathbb{R}$ and $N = \mathbb{Q}$.

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Thanks, very nice. –  John Jul 29 '13 at 13:44
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