I found the following statement in my book:
Let $G$ be a Lie group with Lie algebra $\mathfrak g$. Left-invariant foliations are the “same” as Lie subalgebras $\mathfrak h \subset \mathfrak g$ ?
I know that a foliation $\mathscr F$ on $G$ is left-invariant, if for any leaf $L \in F$ and any $g \in G$, we have that the left translate of L: $$g \cdot L := \{g \cdot h: h \in L\}$$ is again a leaf of $\mathscr F$. I guess that by "same" , isomorphic is intended? How do I prove the statement above?
Here is a sketch.
Associated to a foliation $\mathscr{F}$ on $G$ is an involutive distribution: i.e. a distribution $\mathscr{D}$ in $TG$ such that $[\mathscr{D},\mathscr{D}] \subseteq \mathscr{D}$. Conversely, by the Frobenius theorem an involutive distribution gives rise to a foliation. You should check that the left invariance condition on $\mathscr{F}$ is equivalent to the left invariance condition for $\mathscr{D}$, i.e. for all $g,h\in G$ one has $(\ell_h)_*\mathscr{D}_g = \mathscr{D}_{hg}$. Such a distribution is uniquely determined by $\mathscr{D}_e\subseteq T_eG = \mathfrak{g}$, which is thus a vector subspace of $\mathfrak{g}$ with the property of $[\mathscr{D}_e,\mathscr{D}_e]\subseteq \mathscr{D}_e$; so, $\mathscr{D}_e$ is a Lie subalgebra of $\mathfrak{g}$.