Exponential scales one-parameter subgroups

Fix $a\in \mathbb{R}$ and define ${\delta }_a(u)={\gamma }_X(au)$. By [F4], ${\delta }_a(u+v)={\gamma }_X(a(u+v))={\gamma }_X(au){\gamma }_X(av)={\delta }_a(u){\delta }_a(v),$ so ${\delta }_a$ is a one-parameter subgroup. By [F5], its initial velocity is ${\delta }_a^{\mathrm{\prime }}(0)=a{\gamma }_X^{\mathrm{\prime }}(0)=aX$.

Both ${\delta }_a$ and ${\gamma }_{aX}$ are one-parameter subgroups with initial velocity $aX$. Uniqueness in [F3] therefore gives ${\delta }_a={\gamma }_{aX}$. Evaluating at $u=1$ and applying [F2], ${\gamma }_X(a)={\delta }_a(1)={\gamma }_{aX}(1)={\exp }_G(aX)\mathrm{.}$

For $s,t\in \mathbb{R}$, [F4] and step 2.1 give ${\exp }_G((s+t)X)={\gamma }_X(s+t)={\gamma }_X(s){\gamma }_X(t)={\exp }_G(sX){\exp }_G(tX)\mathrm{.}$

Lie groups are nonempty and boundaryless. If $\dim G=0$, then $X=0$ and all displayed curves are constant; in dimension one the proof is unchanged. Every curve has domain $\mathbb{R}$, so $a,s,t$ may be zero, negative, or any finite values without an endpoint issue. No metric or nondegeneracy condition occurs. The only choice use is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited through [F2] and [F3]; fixing finitely many vectors and scalars adds no choice. The result consists of two identities, not a biconditional.