# Which Lie groups have finitely many conjugacy classes of subgroups of fixed isomorphism type?

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.

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.
• @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. Aug 18, 2021 at 6:56
• 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. Aug 18, 2021 at 6:56