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Let $G$ be any group. Is there a topological space $(X,\tau)$ such that the automorphism group $\textrm{Aut}(X,\tau)$ is isomorphic to $G$?

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    $\begingroup$ At least for finite groups it is true. $\endgroup$ – Dune Nov 25 '14 at 8:13
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    $\begingroup$ According to the thread mathoverflow.net/questions/37356/… on MathOverflow, the answer is yes. $\endgroup$ – Jeremy Rickard Nov 25 '14 at 8:36
  • $\begingroup$ @JeremyRickard I don't understand; the topological automorphisms of a graph is a big group, much bigger than the group of graph theoretic automorphisms. $\endgroup$ – user98602 Nov 25 '14 at 18:02
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    $\begingroup$ @MikeMiller The thread I pointed to contains a reference to a construction for topological spaces as well as one for graphs: de Groot, J. (1959), Groups represented by homeomorphism groups, Mathematische Annalen 138 $\endgroup$ – Jeremy Rickard Nov 25 '14 at 19:29
  • $\begingroup$ @JeremyRickard Ah, thanks. Sorry, don't know how I missed that. $\endgroup$ – user98602 Nov 25 '14 at 19:30
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As pointed out in the comments, this has been answered by Tony Huynh on MathOverflow. In

de Groot, J. ($1959$), Groups represented by homeomorphism groups, Mathematische Annalen $138$

the author shows that:

"for every group $G$ one can find a complete, connected, locally connected metric space $M$ of any positive dimension such that $G \cong A(M)$"

where $A(M)$ denotes the autohomeomorphism group of $M$.

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