Yes, it is true. Let us show that, in a compact Lie group $G$ with $G^0$ nonabelian, finite subgroups that are not contained in any proper subgroup of positive dimension have bounded cardinal.

By contradiction, let finite subgroups $G_n$ have cardinal tending to infinity, and not contained in any proper subgroup of positive dimension.

Let $B$ be a neighborhood of 0 in the Lie algebra of $G$ such that $\exp|_B$ is a homeomorphism onto its image, with inverse denoted by $\log$ (we will let $B$ vary, but can assume it is contained in a single such neighborhood $B_0$ once and for all, hence the function $\log$ will not depend on $B$).

Define $T_n^B=\exp\big(\mathrm{span}\big(\log(G_n\cap\exp(B))\big)\big)$.

The first point is that for each $B\subset B_0$ and for $n$ large enough, $G_n\cap\exp(B)$ consists of commuting elements. This is close to Jordan's theorem and is a particular case of Zassenhaus' theorem (see for instance Theorem 1.3 here, with in mind that connected nilpotent subgroups of compact groups are abelian). We can extract once and for all to suppose that this holds for all $n$.

The next point is that $T_n^B$ is a torus. For an arbitrary commutative subset, it might have been dense in a torus. But the point is that elements of finite subgroups are well-behaved. More precisely: for this point, we can suppose that $G$ is embedded in the unitary group $\mathrm{U}(d)$. Every element of $G$ can be diagonalized with eigenvalues $e^{i\pi k_1/m},\dots,e^{i\pi k_d/m}$. We can arrange $B_0$ so that it avoids elements with eigenvalues $-1$ and define the log accordingly, so assume all $k_i$ to be in $]-m,m[$. Hence the log of such an element will be, in the same diagonal basis, the matrix with eigenvalues $i\pi k_1/m,\dots i\pi k_d/m$, and the exp of its span will be the 1-dimensional torus of diagonal matrices with eigenvalues $\theta^{k_1},\dots,\theta^{k_d}$ for $\theta$ in the unit circle. Now when we consider all $G_n\cap \exp(B)$, we just consider the torus generated by several such (commuting) tori.

Now consider $\liminf_n \dim(T_n^B)$. This is positive (because $G_n$ is infinite, so that $G_n\cap\exp(B)$ is never reduced to $\{1\}$. This liming decreases when $B$ decreases, and hence we can find $B_1$ for which its value, say $c>0$, is minimal. After extracting once again, we an suppose that $\lim_n\dim(T_n^{B_1})=c$. Define $T_n=T_n^{B_1}$. Hence, for every neighborhood $B\subset B_1$, there exists $n_0$ (depending on $B$) such that for all $n\ge n_0$ we have $T_n^B=T_n$ (because $T_n^B\subset T_n$ and these are tori of the same dimension).

This observation allows to deduce that $T_n$ is, for large $n$, normalized by $G_n$. Indeed, for $g_n\in G_n$, we have $$g_nT_n^Bg_n^{-1}=\exp(\mathrm{span}\log(G_n\cap \exp(g_nBg_n^{-1})));$$
since $1$ has a basis of conjugation-invariant neighborhoods, we can choose $B$ such that $\bigcup_g gBg^{-1}$ is contained in $B_1$. We deduce that $g_nT_ng_n^{-1}\subseteq T_n$ for large $n$, hence this is an equality.

Then $G_nT_n$ is a positive-dimensional closed subgroup containing $G_n$. By assumption, it equals $G$, and this is a contradiction since $G^0$ is non-abelian.

unlabeledsummands, the normalizer of the centralizer of the abelian subgroup (which is what preserves the labeled decomp). There are three possibilities for this Lie group. (1) All of $O(n)$: the abelian group is contained in the center and the whole group is at most size $2F(n)$. (2) Positive dimensional but not everything—not primitive. (3) Discrete: $S_n\ltimes O(1)^n$—finitely many cases. $\endgroup$9more comments