[proofplan] We show completeness by extending the maximal integral curve through $e$ to all of $\mathbb{R}$. Suppose the maximal interval were bounded above by $q_2 < \infty$. Using left-invariance of $X_\xi$, any short-time integral curve starting at a point $\gamma(t_0)$ close to $q_2$ is obtained by left-translating a short-time integral curve starting at $e$. Pasting this translated curve onto the original past $q_2$ extends the integral curve beyond $q_2$, contradicting maximality. Symmetry handles the lower endpoint, proving $X_\xi$ is complete. The group property $\gamma_\xi(s+t) = \gamma_\xi(s)\gamma_\xi(t)$ follows from left-invariance and uniqueness of integral curves. [/proofplan] [step:Identify the left-invariant vector field $X_\xi$ and its flow near the identity] For $g \in G$, let $L_g: G \to G$, $h \mapsto gh$ be left translation, a diffeomorphism with inverse $L_{g^{-1}}$. Define \begin{align*} X_\xi: G &\to TG \\ g &\mapsto (dL_g)_e(\xi), \end{align*} so that $X_\xi(g) \in T_g G$. The field $X_\xi$ is smooth by smoothness of group multiplication, and it is left-invariant in the sense that for every $g, h \in G$, \begin{align*} (dL_g)_h\bigl(X_\xi(h)\bigr) = (dL_g)_h \circ (dL_h)_e(\xi) = (dL_{gh})_e(\xi) = X_\xi(gh), \end{align*} using the chain rule applied to $L_g \circ L_h = L_{gh}$. By the [existence theorem for flows](/theorems/???), there exists $\delta > 0$ and an open neighbourhood $V$ of $e$ such that the flow $\varphi^{X_\xi}: (-\delta, \delta) \times V \to G$ is defined and smooth, with $\varphi^{X_\xi}_0(g) = g$ and $\frac{d}{dt}\varphi^{X_\xi}_t(g) = X_\xi(\varphi^{X_\xi}_t(g))$. [/step] [step:Define the maximal integral curve through the identity and assume a finite upper endpoint] Let \begin{align*} \gamma: (-q_1, q_2) &\to G \\ t &\mapsto \varphi^{X_\xi}_t(e) \end{align*} be the maximal integral curve of $X_\xi$ with $\gamma(0) = e$, where $0 < q_1, q_2 \le \infty$. We show $q_2 = \infty$; the argument for $q_1$ is identical under $t \mapsto -t$. Suppose for contradiction that $q_2 < \infty$. [/step] [step:Use left-invariance to translate the short-time flow from $e$ to $\gamma(t_0)$] Choose $t_0 \in (-q_1, q_2)$ with $q_2 - t_0 < \delta$, where $\delta > 0$ is the uniform existence time from the local flow statement. Set $g_0 := \gamma(t_0) \in G$. Define the translated curve \begin{align*} \tilde{\gamma}: (t_0 - \delta, t_0 + \delta) &\to G \\ t &\mapsto L_{g_0}\bigl(\varphi^{X_\xi}_{t - t_0}(e)\bigr) = g_0 \cdot \gamma(t - t_0). \end{align*} Then $\tilde\gamma(t_0) = g_0 \cdot e = g_0 = \gamma(t_0)$. We show $\tilde\gamma$ is an integral curve of $X_\xi$. Differentiating and using left-invariance, \begin{align*} \frac{d}{dt}\tilde\gamma(t) &= (dL_{g_0})_{\gamma(t-t_0)}\!\left(\frac{d}{dt}\gamma(t - t_0)\right) \\ &= (dL_{g_0})_{\gamma(t-t_0)}\bigl(X_\xi(\gamma(t - t_0))\bigr) \\ &= X_\xi\bigl(L_{g_0}(\gamma(t - t_0))\bigr) \\ &= X_\xi(\tilde\gamma(t)), \end{align*} where the second equality uses that $\gamma$ is an integral curve of $X_\xi$ and the third is the left-invariance identity from the setup step. [/step] [step:Paste the translated curve onto $\gamma$ to extend beyond $q_2$] On the overlap $(t_0 - \delta, q_2) \cap (t_0 - \delta, t_0 + \delta) = (t_0 - \delta, \min(q_2, t_0 + \delta))$, both $\gamma$ and $\tilde\gamma$ are integral curves of $X_\xi$ and they agree at $t_0$. By the [uniqueness of integral curves](/theorems/???) for smooth vector fields, $\gamma = \tilde\gamma$ on this overlap. Define the pasted curve \begin{align*} \hat\gamma: (-q_1, t_0 + \delta) &\to G \\ t &\mapsto \begin{cases} \gamma(t) & t \in (-q_1, q_2), \\ \tilde\gamma(t) & t \in (t_0 - \delta, t_0 + \delta). \end{cases} \end{align*} The two definitions agree on their overlap, so $\hat\gamma$ is well-defined and smooth, and it is an integral curve of $X_\xi$ through $e$ at time $0$. Since $t_0 + \delta > q_2$ by the choice $q_2 - t_0 < \delta$, the domain $(-q_1, t_0 + \delta)$ strictly contains $(-q_1, q_2)$, contradicting the maximality of $\gamma$. [/step] [step:Conclude completeness of $X_\xi$ and identify $\gamma_\xi$ as a one-parameter subgroup] The contradiction shows $q_2 = \infty$; the mirror argument gives $q_1 = \infty$, so $\gamma$ is defined on all of $\mathbb{R}$. A parallel left-translation argument shows the integral curve through any $g \in G$ is also defined on $\mathbb{R}$: the curve $t \mapsto g \cdot \gamma(t)$ is an integral curve of $X_\xi$ through $g$ by exactly the left-invariance computation above. Therefore $X_\xi$ is complete. Define $\gamma_\xi(t) := \varphi^{X_\xi}_t(e)$ for $t \in \mathbb{R}$. For fixed $s \in \mathbb{R}$, both $t \mapsto \gamma_\xi(s+t)$ and $t \mapsto \gamma_\xi(s)\gamma_\xi(t) = L_{\gamma_\xi(s)}(\gamma_\xi(t))$ are integral curves of $X_\xi$ starting at $\gamma_\xi(s)$ at time $t = 0$: the first by the flow property $\varphi^{X_\xi}_{s+t} = \varphi^{X_\xi}_t \circ \varphi^{X_\xi}_s$ evaluated at $e$, and the second by left-invariance. By uniqueness of integral curves, \begin{align*} \gamma_\xi(s+t) = \gamma_\xi(s)\gamma_\xi(t) \qquad \text{for all } s, t \in \mathbb{R}. \end{align*} In particular $\gamma_\xi(0) = e$ and $\gamma_\xi(-t) = \gamma_\xi(t)^{-1}$, so $\gamma_\xi: \mathbb{R} \to G$ is a smooth group homomorphism, i.e., a one-parameter subgroup. [/step]