For two days I've been trying to show the following: Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and consider the smooth distribution $$F=\{F_p=DR_p(e)\mathfrak{h}; p\in G\},$$ where $\mathfrak{h}\subseteq \mathfrak{g}$ is a Lie subalgebra and $R_p:G\rightarrow G$, $a\mapsto ap$, is the right translation. Show that $F$ is involutive.
My definition of an involutive distribution is: Let $X, Y\in \mathfrak{X}(G)$ be smooth fields such that $X(p), Y(p)\in F_p$ for all $p\in G$. We say $F$ is involutive if $[X(p), Y(p)]\in F_p$ for all $p\in G$.
Can anyone help me?
What I thought up till now is: Suppose $\mathfrak{h}$ is a Lie subalgebra. If $F$ were not involutive there would exist smooth fields $X, Y\in \mathfrak{X}(G)$ such that $$X(p), Y(p)\in F_p$$ for all $p\in G$ but $$[X(q), Y(q)]\not\in F_q$$ for some $q\in G$ . In particular, $X(e), Y(e)\in F_e=\mathfrak{h}$, hence, $$[X(e), Y(e)]\in \mathfrak{h}$$ for $\mathfrak{h}$ is a Lie subalgebra. Now I'd like to conclude $$[X(q), Y(q)]=DR_q(e)h$$ for some $h\in \mathfrak{h}$ for getting a contradiction. I'm conjecturing $h=[X(e), Y(e)]$, so that it would suffice proving $$DR_q(e)[X(e), Y(e)]=[X(q), Y(q)],$$ but I don't know if that is true.
Fix a basis $x_1,\dots,x_m$ of $\frak{h}$, and consider the right invariant vector fields $X_1,\dots X_m$ extending this basis of $\frak{h} \subset \frak{g}$. Every vector field that is tangent to your distribution is a $C^{\infty}(G)$-linear combination of those vector fields. Thus, to show integrability, you only need to show that the brackets $[X_i,X_j]$ are still tangent to the distribution. This follows from the fact that the lie algebra of right-invariant vector fields and the Lie algebra $T_e G$ are isomorphic under the map that takes a right invariant vector field and gives you its value at the identity of $G$. Concretely this means that $[X_i,X_j]=$ the right invariant vector field extending $[x_i,x_j]\in\frak{h}$.