Let $G$ be a connected Lie group, $\mathfrak{g}$ be its Lie algebra. Assume that $\mathfrak{g}$, as a vector space, can be written as the direct sum of its Lie subalgebras $\mathfrak{h}_1$ and $\mathfrak{h}_2$: $\mathfrak{g} = \mathfrak{h}_1 \oplus \mathfrak{h}_2$. Let $H_1$ and $H_2$ be the Lie subgroups of $G$ that correspond to $\mathfrak{h}_1$ and $\mathfrak{h}_2$. The question is: is it true that $H_1 \cdot H_2$ is an open subset of $G$?
I don't know if the statement is true (if not, I'm interested in when it is true), but I'm thinking of the following proof. Let $M = \{h_1 \cdot h_2 \ | \ h_1 \in H_1, \ h_2 \in H_2\}$. For some neighborhoods $V_1 \subseteq \mathfrak{h}_1$ and $V_2 \subseteq \mathfrak{h}_2$ of zero, the exponential map is a diffeo; on the other hand, it is a diffeo from $V_1 \times V_2$ onto a nbd in $G$. If we show that $\exp(V_1)\exp(V_2)$ is an open subset of $\exp(V_1 \times V_2)$, then $M = H_1 \exp(V_1)\exp(V_2) H_2$, so $M$ is an open subset of $G$. However, how can one show that $\exp(V_1)\exp(V_2)$ is open in $\exp(V_1\times V_2)$?
There is no need to think about $\exp(V_1\times V_2)$ here. You just consider the map $(X,Y)\mapsto\exp(X)\exp(Y)$ as a map $\mathfrak h_1\oplus\mathfrak h_2\to G$. Then the tangent map in $(0,0)$ sends $(v,w)$ to $v+w$ and hence is invertible. Thus there are open neighborhoods $V_1$ of $0$ in $\mathfrak h_1$ and $V_2$ of $0$ in $\mathfrak h_2$ such that this map obtains to a diffeomorphism from $V_1\times V_2$ onto an open neighborhood $U$ of $e$ in $G$, which thus is contained in $H_1\cdot H_2$. But then for $g_1\in H_1$ and $g_2\in H_2$, the set $g_1Ug_2$ is an open neighborhood of $g_1g_2$ in $G$ which by construction is contained in $H_1\cdot H_2$ and this shows that $H_1\cdot H_2$ is open.