Does the quotient of an algebraic group by its neutral component always split?

Question

Let $G$ be a complex algebraic group. Let $G^\circ\subseteq G$ be the connected component of $G$ which contains the neutral element $1\in G$. Then, it can be shown that $G^\circ\trianglelefteq G$ is a normal subgroup and the left (or right) cosets of $G^\circ$ are the connected (and irreducible) components of $G$. In particular, $G/G^\circ$ is a finite group.

I am wondering whether there exists a subgroup $Q\subseteq G$ with $Q\simeq G/G^\circ$ and $G=G^\circ Q$. This is not true for groups in general, but the counterexamples I know are no counterexamples for this particular case.

Answer 1

Here's a counterexample with $G^\circ$ a 1-torus $G_m$: consider the cyclic group $C_4$ of order 4 with generator $u$. Let it act on $G_m$ by $u\cdot x=x^{-1}$; let $z$ denote the unique element of order 2 in $G_m$. Mod out the semidirect product $G_m\rtimes C_4$ by the central element of order two $u^{-1}z$. The resulting group $G$ has $G/G^\circ$ of order 2, but $G\smallsetminus G^\circ$ consists of elements of order 4 and thus the extension is not split: indeed, for $x\in G_m$ we have $(ux)^2=uxux=u^2(u^{-1}xu)x=u^2=z$ which has order exactly 2.

Answer 2

Just for my own notes: Yves Cornulier's example is actually a very standard group (as well as a very clear example). We just take the standard surjection from the “$N$” of the $(B,N)$-pair to the Weyl group. In many cases this does not split. The specific example takes the large group to be SL2.

Let $K$ be a field of characteristic not 2, and let: $$ X=\operatorname{SL}_2(K), \quad T=\left\{ \left[\begin{matrix} t & 0 \\ 0 & 1/t \end{matrix}\right] ~\middle|~ t \in K^\times \right\}, \quad N_X(T) = T \cup Tw_0, \quad w_0 = \left[\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right]$$

Then set $G=N_X(T)$ and note $G_0 = T$, but $$N_X(T) \setminus T = \left\{ \left[\begin{matrix} 0 & t \\ -1/t & 0 \end{matrix}\right] ~\middle|~ t \in K^\times \right\}$$ consists entirely of elements that square to $\left[\begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix}\right]$ and so there is no section of $G \twoheadrightarrow G/G_0 = W$.

This lack of a section was particularly annoying to me when studying GL first and then SL, since the Weyl group of GL splits (it is the subgroup of permutation matrices), while the Weyl subgroup of SL must be viewed in the generic way as $N_X(T)/T = G/G_0$ and one gets that $w_0^2 \in Z(X)$ but $w_0^2 \neq 1$ so that one has lots of annoying signs running around if one tries to stick with matrices instead of abstract quotient groups.