# $\pi_0(G)$ as a subgroup of a Lie group $G$

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 $$G$$.

• Not sure about the general case , but $\{\pm I_2\}\subset\mathrm{GL}_2(\mathbb{R})$ lie in the same component. There are two distinct questions: the existence of sections of $G\to\pi_0(G)$ and the existence of subgoups of $G$ isomorphic to $\pi_0(G)$. Jan 30, 2020 at 2:48
• In the case where $G$ is the normaliser of a maximal torus in a reductive Lie group, your question is asking about a lift of the normaliser of the Weyl group, which does not exist in general. (The easiest example is when the ambient group is $\operatorname{SL}_2$, in which case the normaliser is a non-split extension of $\operatorname{GL}_1$ by $\mathbb Z/2\mathbb Z$.) @MikeMiller's example is the compact form of this example. Jan 30, 2020 at 3:04
• See also mathoverflow.net/questions/150949/… (already served up by MO as a related question), which offers a weaker version of this property. Jan 30, 2020 at 3:06
• The claim "I believe..." is not true. Consider the quaternion group $Q$ (of order 8) and $S$ the 1-dimensional circle group; let $z_Q$ and $z_S$ be their unique elements of order $2$ and $z=(z_Q,z_S)\in Q\times S$. Define $G=(Q\times S)/\langle z\rangle$. Then the quotient $C_2\times C_2$ indeed embeds into $G$, but cannot be mapped as a section (i.e., for each embedding $C_2\times C_2\to G$ there exists two distinct elements mapping into the same component).
– YCor
Jan 30, 2020 at 8:43

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.