**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 $G$?
> 
> 
> 



Submanifold of a Lie group whose tangent bundle is invariant under group (left) action
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 $M$).


