Is the fundamental group of a path connected Lie group generated by one parameter groups Let $G$ be a path connected Matrix Lie Group with identity element $I \in G$. Does there always exist a set of generators $[\gamma_i]_{i \in I}$ of $\pi_1(G)$ such that each homotopy class $[\gamma_i]$ contains a loop of the form $f(t) = e^{tA}$, for some $A$ in the Lie algebra of $G$? My idea was to show that any homotopy class of loops based at $I$ contains some subgroup isomorphic to a circle, but I'm not sure if this works.
 A: This proposition is true in the case when the universal covering group of $G$, call it $\pi:\tilde{G}\to G$, is compact.
From homotopy theory, we know that $\pi_1(G,e)$ is in bijective correspondence with paths connecting $\tilde{e}\in \tilde{G}$ to $x\in\pi^{-1}(e)$. Given $[\gamma]\in \pi_1(G,e)$, there is a path $\tilde{\gamma}: I\to \tilde{G}$ with $\tilde{\gamma}(0)=\tilde{e}$ and $\pi\circ\tilde{\gamma}=\gamma$ and for all loops $\beta\sim \gamma$, $\tilde{\beta}(1)=\tilde{\gamma}(1)$.
Since $\tilde{G}$ is compact, the exponential map $\mathrm{exp}: \tilde{\mathfrak{g}}\to \tilde{G}$ is surjective. Given $x\in \pi^{-1}(e)$, there is an $A_x\in \tilde{\mathfrak{g}}$ such that $\mathrm{exp}(A_x)=x$ and $\exp(tA_x)$ gives us a path connecting $e$ to $x$. This means that $\pi\circ \exp(tA_x)$ defines a loop at $e\in G$ and all homotopy classes are given by $\{\pi\circ \exp(tA_x)| x\in \pi^{-1}(e)\}$.
All that remains to be shown is that $\gamma(t)=\pi\circ \exp(tA_x)$ coincides with the exponential map on $G$. Differentiating this path gives $\frac{d\gamma}{dt}(t)=T_{\exp(tA_x)}\pi(T_eL_{\exp(tA_x)}A_x)=T_e(\pi\circ L_{\exp(tA_x)})(A_x)$. Since $\pi$ is a group homomorphism $\pi\circ L_{\exp(tA_x)}=L_{\pi(\exp(tA_x))}\circ \pi$ and we have \begin{align}\frac{d\gamma}{dt}(t)&=T_{e}(L_{\pi(\exp(tA_x))}\circ \pi)(A_x)\\
&=T_{\tilde{e}}L_{\pi(\exp(tA_x))}T_e\pi(A_x)
\end{align}
Since the exponential of $tA$ is the unique solution to $\frac{d\gamma}{dt}(t)=T_{\tilde{e}}L_{\gamma(t)} A$, we conclude that $\gamma(t)=\exp(tB_x)$ for $T_e\pi(A_x)=B_x$. Hence, $$\pi_1(G,e)=\{ [\exp(tB_x)]| x\in \pi^{-1}(e)\}$$
