While doing some exercises in Lie groups, I see that the Lorentz group $O(1,3)$ has four connected components and $\pi\_0(O(1,3))$ is the Klein four-group $\mathbb{Z}/2 \times \mathbb{Z}/2$. Not only that, but I can find explicit elements representing each connected component $\{1,a,b,ab\}$ which form a subgroup isomorphic to $\pi\_0$.


My question then: is it true that for any Lie group $G$, the group $\pi\_0(G)$ embeds as a subgroup of $G$? I believe it would follow that any such embedding would take distinct elements of $\pi\_0$ to distinct components of $G$.



$\pi_0(G)$ as a subgroup of a Lie group $G$
No. The first example that comes to mind is $\text{Pin}(2) \subset S^3$, given as $S^1 \cup jS^1$. This group has two components, but every element of $jS^1$ squares to $-1 \in S^1$. Thus every element of the non-identity component has order 4. 


So there is no section of the map $\text{Pin}(2) \to \Bbb Z/2$ which sends the non-identity component to 1. 


