Connected subgroup determined by its Lie algebra

Let $G$ be a Lie group. If $H_1,H_2\subseteq G$ are connected immersed Lie subgroups and

\mathrm{Lie}(H_1)=\mathrm{Lie}(H_2)\subseteq \mathfrak g,

then $H_1=H_2$ .

Equivalently, every Lie subalgebra $\mathfrak h\subseteq\mathfrak g$ determines at most one connected immersed Lie subgroup of $G$ with Lie algebra $\mathfrak h$ .

Why this is true

One convenient way to see the mechanism is via the exponential map. A connected Lie subgroup $H$ with Lie algebra $\mathfrak h$ is generated by $\exp(U\cap\mathfrak h)$ for any sufficiently small neighborhood $U$ of $0$ . Thus two connected subgroups with the same $\mathfrak h$ are generated by the same one-parameter subgroups.

This uniqueness is one half of the Lie correspondence between connected subgroups and subalgebras; existence typically uses integrability of left-invariant distributions or the third theorem of Lie.