Leaves of a Lie subalgebra distribution

The map $t\mapsto I+t{E}_{12}$ is a one-parameter subgroup because [given] $E_{12}^2=0$, so $(I+s{E}_{12})(I+t{E}_{12})=I+(s+t)E_{12}\mathrm{.}$ Hence $H$ is a connected immersed Lie subgroup with tangent space $T_{I}H=\mathfrak{h}$.

For any $g\in G$, the left coset $gH$ has tangent line [given] $d(L_g{)}_I(\mathfrak{h})$ at the point $g$. Thus each coset is an integral manifold of the distribution $\mathcal{D}$.

For each $g\in G$, the curve $c_g(t)=g(I+t{E}_{12})$ has image $gH$ and [given] derivative $c_g^{\mathrm{\prime }}(t)=g{E}_{12}=d(L_{c_g(t)}{)}_I(E_{12}),$ so it is an integral curve of the nowhere-zero rank-$1$ distribution $\mathcal{D}$. Hence any connected integral manifold through $g$ is locally an open piece of that same integral curve and is contained in $gH$. Since step 1.2 shows that $gH$ itself is a connected integral manifold through $g$, [L1] identifies $gH$ as the leaf through $g$. Varying $g$ gives all leaves.