factoring a neighborhood of identity in a compact connected Lie group with a closed Lie subgroup Given a closed Lie subgroup of a compact and connected Lie group, it seems plausible to me that there exists such a neighborhood of identity in the original Lie group that it can be written as the product of the aforementioned closed Lie subgroup and the image of the exp map restricted to the complement of the Lie algebra corresponding to the closed Lie subgroup. 
While I suspect it follows from some basic facts in the theory of Lie groups and Lie algebras, I am not sure how to go about proving it. In fact, additional assumptions might be needed to make the statement rigorous. For example, perhaps the complement in the Lie algebra needs be orthogonal with respect to an appropriately invariant inner product? 
Could anyone offer some guidance as to how to make sense of my intuitive and possibly naive thinking? Thanks!
 A: Given $H \subseteq G$ a closed Lie subgroup with corresponding Lie algebras $\mathfrak{h} \subseteq \mathfrak{g}$, let $\mathfrak{k}$ be a vector complement to $\mathfrak{h}$ in $\mathfrak{g}$ (note that $\mathfrak{k}$ is not required to be even a subalgebra), and define $f: \mathfrak{g} \to G$ by 
$$ f(u + v) = \exp(u)\exp(v)$$
where $u \in \mathfrak{h}$ and $v \in \mathfrak{k}$. Then as $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{k}$, the derivative of $f$ at 0 will be the identity map, and thus there are neighborhoods $U$ and $V$ of $0$ in $\mathfrak{h}$ and $\mathfrak{k}$ respectively such that $f: U \times V \to f(U \times V)$ will be a diffeomorphism, and $f(U \times V) = \exp(U)\exp(V) \subseteq G$ will be open. 
Now, this may be the neighborhood you want (since $f(V)$ may intersect $H$), but if you desire, a contradiction argument may be used to show that $V$ may be shrunk to a smaller neighborhood $V'$ of zero such that $f(V') \cap H = 1$. Here, we must use that $\mathfrak{k}$ is a vector complement to $\mathfrak{h}$.
