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Let $G$ be a real Lie group. What conditions must $G$ satisfy so that the following is true:

For any finite group $\Gamma$ there exist finitely many conjugacy classes of subgroups of $G$ that are isomorphic to $\Gamma$.

I believe that for $G=GL(n,\mathbb{R})$ this is true because: subgroups of $GL(n,\mathbb{R})$ are conjugate if and only if the restriction of the standard representation of $GL(n,\mathbb{R})$ to the subgroups are isomorphic representations. Finite groups have finitely many irreducible representations, and that proves the claim.

I also believe that for $G=SO(n)$ this is true because of this answer: https://mathoverflow.net/a/17074/164084.

It would be enough for my application to know that every real, compact, connected Lie group that has a faithful representation has the property stated above ("For any finite group $\Gamma$...").

This is a crosspost from Stackexchange Mathematics, see here: https://math.stackexchange.com/questions/4219091/which-lie-groups-have-finitely-many-conjugacy-classes-of-subgroups-of-fixed-isom.

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A natural condition is that $G$ has finitely many connected components. One can easily reduce this case to the connected group case, and then to the compact group case, as all maximal compact subgroups in a connected Lie group are conjugated. Then the representation variety $\text{Hom}(\Gamma,G)$ is compact and local rigidity, that is vanishing of $H^1(\Gamma,\mathfrak{g})$, guarantees the finiteness of the number of $G$-orbits. Here $\mathfrak{g}$ denotes the Lie algebra of $G$. The fact that $H^1$ vanishes could be deduced from the fact that every isometric action of $\Gamma$ on $\mathfrak{g}$ has a fixed point, by averaging an orbit.

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    $\begingroup$ If there are finitely many components, it's also true that all maximal compact subgroups are conjugate (this is due to Mostow 1955, extending the connected case done earlier by Iwasawa). $\endgroup$
    – YCor
    Commented Aug 15, 2021 at 16:06
  • $\begingroup$ How beautifully simple. Reading p.152 in 'Andre Weil (1964): Remarks on the Cohomology of Groups' and Lemma 2.1 in 'Tullia Dymarz, Xiangdong Xie: Day's fixed point theorem, group cohomology, and quasi-isometric rigidity' helped me understand the second half of your answer. $\endgroup$
    – user505117
    Commented Aug 16, 2021 at 14:40
  • $\begingroup$ @user505117, indeed, I was very brief, as I do not know your background. If anything is still unclear, I will happily provide more details. $\endgroup$
    – Uri Bader
    Commented Aug 16, 2021 at 16:07
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    $\begingroup$ @user505117 You are right - the word "isometry" in my answer could (and should) be replaced by "affine". Every affine action of a finite group has a fixed point. Given a cocycle $c\in H^1(\Gamma,\frak{g})$, we get an affine action $\phi$ of $\Gamma$ on $\frak{g}$, given by $\phi(\gamma)(X)=\text{Ad}(\gamma)(X)+c(\gamma)$ and for a fixed point $X_0$, $c(\gamma)=X_0-\text{Ad}(\gamma)(X_0)$, thus $c$ is a coboundary. $\endgroup$
    – Uri Bader
    Commented Aug 18, 2021 at 6:56
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    $\begingroup$ However, note that the Killing form on $\frak{g}$ gives an inner product for which the representation $\text{Ad}$ is orthogonal, thus $\phi$ is an isometric action with respect to the induced Eucidean metric on $\frak{g}$. Indeed, this is not needed for my answer, but it gives context to the vanishing of $H^1$ phenomena, as fixed points of isometric actions is a well studied theory. $\endgroup$
    – Uri Bader
    Commented Aug 18, 2021 at 6:56

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