This question concerns a statement in a short paper by S. P. Wang titled “A note on free subgroups in linear groups" from 1981. The main result of this paper is the following theorem.


**Theorem** (Wang, 1981): For every field $k$ of characteristic 0 and subgroup $\Gamma$ of $GL(n, k)$, the group $\Gamma$ either has a nonabelian free subgroup or possesses a normal solvable subgroup of index < $\lambda(n)$.


To prove this theorem, Wang needs the following statement:


“It is easily shown that in any Lie group with finitely many connected components, there is a finite subgroup which meets every component.”


Please give me a proof of this statement.



In any Lie group with finitely many connected components, does there exist a finite subgroup which meets every component?
An immediate consequence of Theorem 3.1(ii) of Ch. XV of Hochschild's book "The structure of Lie groups" is that in such a Lie group, maximal compact subgroups meet every connected component (and are all conjugate to each other by part (iii)). So your question thereby reduces to the case of compact Lie groups.


Consider a compact Lie group $K$, and let $T$ be a maximal torus in $K$. I claim that the normalizer $N\_K(T)$ (whose identity component is $T$) maps onto $\pi\_0(K)$, so we could then replace $K$ with $N\_K(T)$ to reduce to the case when $K^0$ is a torus. To prove the claim, for any $\gamma \in \pi\_0(K)$ and $k \in K$ lifting $\gamma$, the conjugate $kTk^{-1}$ is a maximal torus in $K^0$, so it is $K^0$-conjugate to $T$. That enables us to change $k$ by left multiplication against an element of $K^0$ so that $kTk^{-1}=T$, thereby verifying that $N\_K(T) \rightarrow \pi\_0(K)$ is surjective.


Now we may assume $K^0$ is a torus $T$, so we seek to show that if $G$ is an extension of a finite group $\Gamma$ by a torus $T$ then $G$ contains a finite subgroup mapping onto $\Gamma$. Since $T$ is commutative, it is a $\Gamma$-module via $G$-conjugation on $T$ (since $G/T = \Gamma$). This enables us to define ${\rm{H}}^2(\Gamma, T)$, and the isomorphism class of $G$ as an extension of $\Gamma$ by $T$ is classified by a class in this cohomology group (with $T$ encoding the analytic structure on $G$). Since this cohomology group is killed by the size $n$ of $\Gamma$, we see that ${\rm{H}}^2(\Gamma, T[n]) \rightarrow {\rm{H}}^2(\Gamma, T)$ is surjective (consider the $\Gamma$-cohomology sequence attached to the exact sequence $1 \rightarrow T[n] \rightarrow T \stackrel{n}{\rightarrow} T \rightarrow 1$). Hence, we get a finite group $E$ that is an extension of $\Gamma$ by $T[n]$ such that its pushout along the inclusion $T[n] \rightarrow T$ is $G$. That identifies $E$ as a finite subgroup of $G$ mapping onto $\Gamma = \pi\_0(G)$.


QED


**EDIT**: I should also note that the analogue for linear algebraic groups is much more elementary to prove insofar as it avoids the hard work on maximal compact subgroups in the presence of disconnectedness, and perhaps this suffices for the motivating application (which was not explained much in the question posted). Namely, if $G$ is a linear algebraic group over a field $F$ of characteristic 0 then we claim there is a finite $F$-subgroup in $G$ that meets all connected components of $G\_{\overline{F}}$. Indeed, if $T$ is a maximal $F$-torus in $G$ then $N\_G(T)$ maps onto the finite etale $F$-group $G/G^0$ by a calculation with $\overline{F}$-points exactly as in the compact case above, so we reduce to the case when $G^0$ is a torus. 


If $F$ is algebraically closed then we can carry out the degree-2 group cohomology argument as in the compact case above to conclude. For general $F$ we can proceed similarly by working with the abelian category of $\Gamma$-modules for the etale topology on the category of finite type $F$-schemes, with $\Gamma$ a finite $F$-group, and considering the sheafified "standard cochain" complex for such $\Gamma$.


