One-parameter subgroups as integral curves

Let $G$ be a Lie group with Lie algebra $\mathfrak g$.

Statement

Fix $X\in\mathfrak g$, and let $X^L$ be the corresponding left-invariant vector field on $G$ (obtained by translating $X\in T_eG$ via left translations ).

1. The integral curve of $X^L$ starting at the identity is the one-parameter subgroup

   $$t\longmapsto \exp(tX),$$

   where $\exp$ is the exponential map . In particular, $\exp(tX)$ solves the ODE $g'(t)=(X^L)_{g(t)}$ with $g(0)=e$.

2. More generally, the integral curve of $X^L$ starting at $g_0\in G$ is

   $$t\longmapsto g_0\,\exp(tX).$$

There is an analogous statement for the right-invariant vector field $X^R$, whose integral curves are $t\mapsto \exp(tX)\,g_0$.

Context

This viewpoint explains why the bracket on $\mathfrak g$ can be recovered from commutators of flows: the Lie bracket is the infinitesimal failure of invariant flows to commute (compare the bracket lemma for left-invariant fields and the structure encoded by the Maurer–Cartan equation ).