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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.

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    $\begingroup$ Take any finite non-solvable group and view it as a $0$-dimensional Lie group. Its Lie algebra is solvable. $\endgroup$ – Mariano Suárez-Álvarez Jun 7 '16 at 21:37
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    $\begingroup$ (If you do not like zero-dimensional groups for some reason, consider the direct product of my first example with $S^1$) $\endgroup$ – Mariano Suárez-Álvarez Jun 7 '16 at 21:38
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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 $2$.

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  • $\begingroup$ What do you mean by the corresponding Lie group in positive characteristic? $\endgroup$ – Tobias Kildetoft Jun 8 '16 at 9:56
  • $\begingroup$ @TobiasKildetoft: Aha right, here I mean the algebraic group. But perhaps people do not talk about Lie groups in positive characteristic, so maybe this answer will confuse the OP.. $\endgroup$ – spin Jun 9 '16 at 20:07
  • $\begingroup$ I have certainly not heard algebraic groups called Lie groups (and of course one needs to be very careful with "the group corresponding to a Lie algebra" in positive characteristic, as there might be either none or more than one possibility). $\endgroup$ – Tobias Kildetoft Jun 10 '16 at 4:46
  • $\begingroup$ According to Humphrey, there is a bijection $H \mapsto \mathfrak{h} \subset \mathfrak{g}$ for algeraic linear groups. Does this still work for arbitrary (i.e. differential) Lie groups? I think there are Lie sub-groups of $Gl_n$ which are not algebraic. $\endgroup$ – red_trumpet Jun 13 at 13:02
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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.

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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 $i\ge 2$.

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  • $\begingroup$ Do you have a reference for this? $\endgroup$ – red_trumpet Jun 13 at 12:41

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