Group actions on manifolds - exponential map Let $M$ be a smooth manifold. Suppose $K$ is a Lie group (with Lie algebra $\mathfrak{k}$) acting EDIT: TRANSITIVELY on $M$ from the left and $G$ is a Lie group (with Lie algebra $\mathfrak{g}$) acting on $M$ from the right. Suppose further these actions commute - ie $k(pg)=(kp)g=kpg$. Fix a point $x\in M$. Suppose $\phi(t)$ is some smooth curve in $M$ such that $\phi(0)=x$.
Is it possible to find $X\in \mathfrak{k}$ and $A\in\mathfrak{g}$ such that
\begin{align*}
\frac{d}{dt}\bigg|_{t=0}\exp(tX)x\exp(tA)=\phi^{\prime}(0)??
\end{align*}
And if so, how could you prove such a fact?
 A: Yes, it is always possible. In fact, you can always take $A=0\in\mathfrak g$.
Fix $x\in M$, and consider the smooth map $F\colon K\to M$ given by $F(k) = kx$. Because $K$ acts transitively, $F$ is surjective. Moreover, $F$ is equivariant with respect to the (transitive) left actions of $K$ on $K$ and $M$: For all $k,k'\in K$, 
$$
k'F(k) = k'(kx) = (k'k)x = F(k'k).
$$
The equivariant rank theorem (Theorem 7.25 in my Introduction to Smooth Manifolds) shows that $F$ has constant rank, and because it's surjective, the global rank theorem (Theorem 4.14 in ISM) shows that it's a smooth submersion. This means that $dF_e\colon T_eK \to T_xM$ is surjective (where $e$ is the identity of $K$). Identifying $T_eK$ with $\mathfrak k$, we can choose $X\in \mathfrak k$ such that $dF_e(X) = \phi'(0)$. With $A=0\in \mathfrak g$, it then follows that 
\begin{align*}
\left.\frac d {dt}\right|_{t=0}\exp(tX)x\exp(tA) &= 
\left.\frac d {dt}\right|_{t=0}F(\exp(tX)) 
=
dF_e (d(\exp)_e(X)) = 
dF_e (X) = \phi'(0).
\end{align*}
(In the second-to-last equation, I used the fact that $d(\exp)_e$ is the identity map on $\mathfrak k$.)
A: What if the action of $G$ and $K$ are both trivial? E.g. $k\cdot x=x=x\cdot g$ for all $k$ and $g$? Then your expression $\mathrm{exp}(tX)\,x\, \mathrm{exp}(tA)$ is identically $x$, and so has trivial derivative. Hence you can't hit $\phi'(0)$ in general. 
