The simplest way to prove that any left-invariant vector field on a Lie group is complete It's all in the question: I look for the most intuitive proof that the integral curves of any left-invaraint vector field on a Lie group can be extended for all values of "time". I realize  that the argument is always based on the existence of group multiplication; what I look for is the most straightforward proof available. Thanks in advance!
 A: A vector field is complete if and only if its flow $\Phi_t$ exists for all $t \in \Bbb{R}$. So we can show that a left-invariant vector field $X$ on a Lie group $G$ is complete by computing its flow.
The flow $\Phi_t$ of $X$ satisfies the differential equation $\frac{ \rm{d}}{\rm{d}t} \Phi_t(g) |_{t=0} = X_g$ for all $g \in G$. Applying the flow to the identity element, we have $\Phi_t(1) = e^{tX}$, the Lie-theoretic exponential map. Moreover since $X$ is left-invariant, $\Phi_t$ is left-invariant $\Phi_t(g) = g\Phi_t(1) = ge^{tX}.$ So we see that $\Phi_t: G \to G$ is equal to right-multiplication by $e^{tX}$.
A: I do think this statement is true if $G$ is not connected. If $G$ is connected and compact, then you can prove this via Hopf-Rinow theorem using the exponential map. See Terence Tao's article on this in his blog. 
But one still needs to prove the integral curves of a left invariant vector field is the one produced by the exponential map. While this seems standard in the wikipedia definition, it is not entirely clear to me why this is true unless one verify the exponential map of the Lie group coincide with the exponential map of the Riemannian geometry. But then it takes quite a bit of work to prove this "trivial" fact. 
