If a Lie Algebra is solvable, is the corresponding Lie group solvable in the group theoretic sense?

Question

I just started working with Lie Algebras with a professor. The way we defined them is probably the standard way; treat Lie Algebras as tangent spaces at the identity of the Lie Group.

Now, consider the following:

$\mathfrak{g}^{(0)}=\mathfrak{g}$ where $\mathfrak{g}$ is a Lie Algebra and then define $\mathfrak{g}^{(1)} = [\mathfrak{g},\mathfrak{g}] :=\{[x,y] \in \mathfrak{g}:x,y \in \mathfrak{g}\}$. We also have $\mathfrak{g}^{(n+1)}=[\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]$.

We say that a Lie Algebra $\mathfrak{g}$ is solvable if the series $\mathfrak{g} \supseteq \mathfrak{g}^{(1)} \supseteq \mathfrak{g}^{(2)} \supseteq ...$ terminates, i.e. $\exists n \in \mathbb{N}$, such that $\mathfrak{g} \supseteq \mathfrak{g}^{(1)} \supseteq \mathfrak{g}^{(2)} \supseteq ... \supseteq \mathfrak{g}^{(n}) = \{0\}$.

We also say that the corresponding Lie Group is solvable if the Lie Algebra is solvable. Does this imply that the Lie Group is solvable in the group theoretic sense? i.e. $\exists n \in \mathbb{N}$ s.t. $G \trianglerighteq G_1 \trianglerighteq G_2 \trianglerighteq ... \trianglerighteq G_n$ and $G_k /G_{k+1}$ is abelian.

Thank you.

Answer 1

This is true if you assume that the groups involved are connected.

Remark that if ${\cal g}$ is commutative so is $G$ is $G$ is connected, the Lie algebra of $G_i/G_{i+1}$ is ${\cal g}^{(i)}/{\cal g}^{(i+1)}$ which is commutative.

Answer 2

There is a bijective correspondence between connected Lie subgroups of a Lie group and subalgebras of its (finite-dimensional, real) Lie algebra. It follows that $G$ is solvable iff $Lie(G)$ is solvable. We can use the derived series in both cases. That is $G^{(1)}=G$, $G^{(i)}=[G^{(i-1)},G^{(i-1)}]$ for $i\ge 2$.

Answer 3

If you assume connected and characteristic $0$, this is true. Otherwise no.

For disconnected groups, any non-solvable finite group will give a counterexample.

For positive characteristic, consider $G = SL_2(k)$ when $k$ is a field of characteristic $2$.