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Let $G$ be a Lie group and $\mathfrak{g}$ be the lie algebra of $G$. We know that for any $X\in\mathfrak{g}$ there exists an unique $\alpha_X:(\mathbb{R},+)\longrightarrow (G,\cdot)$ one-parameter subgroup such that $X_{\alpha_X(t)}=\alpha_X'(t)$ for all $t\in\mathbb{R}$.

If the exponential map is defined by: $$\begin{array}{rcll} \text{exp}:&\mathfrak{g}&\longrightarrow & G\\ &X&\longmapsto& \text{exp}(X)=\alpha_X(1) \end{array}$$ How can I prove the following statement?


If $X\in\mathfrak{g}$ then $\text{exp}(tX)=\alpha_X(t)$ for all $t\in\mathbb{R}$?

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Fix $s\in\Bbb R$. By uniqueness of solutions to ODEs we find $\alpha_{sX}(t)=\alpha_X(st)$ for all $t\in\Bbb R$. In particular, taking $t=1$, we find $\alpha_{sX}(1)=\alpha_X(s)$. By the definition of $\exp$, this means $$\exp(sX)=\alpha_X(s),\quad\forall s\in\Bbb R.$$

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