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Suppose that, for a connected simply connected real Lie group G, there is a Lie group H such that G = H′ (the commutator subgroup of H). Can H always be chosen to be connected; that is, is there a connected Lie group K such that G = K'?

Such an H is often called an integral for G in general group theory, but that terminology is, perhaps, misleading in the case of Lie groups.

My motivation for the question is that I am trying to answer the following question that was raised in a paper: "Is it true that a Lie algebra is integrable if and only if the corresponding Lie group is?" I believe that I have a proof if the question I have asked has a positive answer.

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  • $\begingroup$ Do you have an example of a connected Lie group not equal to the commutator subgroup of another Lie group (possibly disconnected)? $\endgroup$
    – Moishe Kohan
    Commented Jan 13 at 15:16
  • $\begingroup$ I'm afraid not. I wondered if you took one of the connected components of H (of the identity element?) or the simply connected cover, whether they would have the same commutator subgroup. $\endgroup$
    – David Towers
    Commented Jan 13 at 19:26
  • $\begingroup$ Here's my latest attempt at an answer. Claim: Let $H_0$ be the connected component of the identity of $H$. Then $G=H_0'$. The Lie algebra of H is also the Lie algebra of $H_0$. Now, by (1), the Lie algebra of $H_0'$ is ${\mathfrak h}'$, which is also the Lie algebra of G. I now want to say that $G=H_0'$, but this seems to require that $H_0'$ is simply connected. (1) L. Greenberg, `Commutator groups and algebras'. Journal of Research of the National Bureau of Standards B. Mathematical Sciences 73B (3), (1969), 247-249: nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn3p247 A1b.pdf $\endgroup$
    – David Towers
    Commented Jan 14 at 16:16
  • $\begingroup$ I should say that work on nonassociative algebras and my Lie group theory is rather rusty. $\endgroup$
    – David Towers
    Commented Jan 14 at 16:21
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    $\begingroup$ No, your proof fails in the very first step. Think about $H$ which is the isometry group of the real line and $G$ which is the additive group of real numbers. But you get this $G$ as the derived subgroup of the Heisenberg group. $\endgroup$
    – Moishe Kohan
    Commented Jan 14 at 16:34

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