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Edit: According to the interesting comment of Tobias Fritz we revise the question.

Assume that $G$ is a Lie group and $M\subseteq G$ is a closed connected smooth submanifold of $G$ containing the neutral element $e\in G$. Assume that for every $m\in M$ we have $D L_m (T_e M)=T_m M$. This means that the differential of left multiplication by $m$ preserves the corresponding tangent space to $M$.

Does this impliy that $M$ is a Lie subgroup of $G$?

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    $\begingroup$ Doesn't your condition hold for $M$ being any open neighbourhood of $e$? $\endgroup$ Sep 5, 2017 at 21:58
  • $\begingroup$ @TobiasFritz thank you for this interesting comment. I revise the question. $\endgroup$ Sep 5, 2017 at 22:04
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    $\begingroup$ You also may want to assume connectedness, since otherwise taking any set of connected components (containing the component of $e$) that is not a group provides another trivial counterexample. $\endgroup$ Sep 5, 2017 at 22:19
  • $\begingroup$ @TobiasFritz yes exactly for example two horizontal line in the plane $\endgroup$ Sep 5, 2017 at 22:21

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The answer is yes.

Any subspace $V\subset\mathfrak g$ in the Lie algebra of $G$ defines a $G$-invariant distribution of planes in $TG$ via the action you have specified: $V_m=DL_m(V)$. This distribution is Frobenius-integrable iff $[V,V]\subset V$, i.e., $V$ is a Lie subalgebra. Note that by $G$-invariance $V$ is integrable at one point iff it is integrable everywhere.

By your assumption, $V$ is integrable, since $M$ is tangent to $V$ at all its points. Now, by the uniqueness of the solutions to ODEs $M$ must coincide with the Lie subgroup $\exp V$ (here we are using closedness and connectedness of $M$).

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  • $\begingroup$ @Yuri Thank you and (+1) for this very interesting answer. $\endgroup$ Sep 6, 2017 at 10:37

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