Exercise 2, § 7

Let G be a Lie group, $\phi : U \to G$ an exponential mapping with the properties of Proposition 3, V a neighbourhood of 0 in U such that $\mathbf{Z}V \subset V$ and $\psi : V \to G$ a tangent mapping at 0 to $\phi$ such that $\psi(nx) = \psi(x)^n$ for all $n \in V$ and all $n \in \mathbf{Z}$. If $p > 0$, then $\psi = \phi | V$. (Give L(G) a norm. Let $x \in V$. Then $p^n x$ tends to 0 as n tends to $+\infty$, hence there exists $\alpha_n > 0$ such that $\alpha_n$ tends to 0 and $\| (\phi^{-1} \circ \psi)(p^n x) - p^n x \| \leq \alpha_n \| p^n x \|$. But $\psi(p^n x) = \psi(x)^{p^n}$, whence $\| (\phi^{-1} \circ \psi)(x) - x \| \leq \alpha_n \| x \|$.)