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In the book How groups grow by Avinoam Mann, the author cites the following theorem attributed to Gleason-Montgomery-Zippin.

Theorem 6.4 (Gleason–Montgomery–Zippin: solution of Hilbert’s Fifth Problem [Ka 95, MZ 55]) Let T be a finite dimensional, locally compact, connected and locally connected, homogeneous metric space. Then the group of isometries of T can be given the structure of a Lie group with finitely many components.

Looking at the cited references, I cannot find any statement of the form above. From what I understand, it would suffice to show that $Isom(T)$ is locally Euclidean with finitely many components, but I do not see how that follows from the hypotheses.

Can someone familiar with the literature surrounding Hilbert's fifth problem shed some light on the matter?

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    $\begingroup$ You did not look hard enough, it is in section 6.3, Corollary on page 243 of Montgomery-Zippin. $\endgroup$ Commented Oct 5 at 16:44
  • $\begingroup$ Corollary (from 6.3): If a locally compact group G satisfying (A) acts faithfully and transitively on a locally compact, connected, finite-dimensional space, then G is a Lie group. (A) G is the limit of a countable inverse sequence of Lie groups $\cdots \to G_2\to G_1$ where the maps $G_{i+1}\to G_i$ are continuous with compact kernel. I don't see why the isometry group in the question satisfies (A) nor why it has finitely-many components. $\endgroup$
    – Canno
    Commented Oct 6 at 6:26
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    $\begingroup$ It is a locally compact separable Hausdorff group, hence, it contains an open subgroup which satisfies (A), this is the main result of Montgomery-Zippin. Already this open subgroup acts transitively on $T$. As for finitely many components, take a look at the fiber bundle $G_x\to G\to T$ and note that 𝑇 is connected and $G_x$ is a compact Lie group. I can add more detail if this is not enough. $\endgroup$ Commented Oct 6 at 9:41
  • $\begingroup$ @MoisheKohan, thanks for the comment, more detail would be appreciated (topological groups really aren't my thing). 1) It seems the Corollary applies to an open subgroup of $Isom(T)$ and not the entire group, yet Theorem 6.4 says $Isom(T)$ is a Lie group. 2) Why is $G_x$ a compact Lie group? If you post your reply as an answer, I can accept it. $\endgroup$
    – Canno
    Commented Oct 8 at 17:50
  • $\begingroup$ Ok, I will write a proper answer later. But check Tao's book, see if he includes a proof. It is, by now, a standard argument, used in relation to Gromov's theorem on groups of polynomial growth. $\endgroup$ Commented Oct 8 at 17:56

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