This is somehow related. A simply-connected Lie group $G$ has uncountably many infinite cyclic subgroups. To see this simply observe that the exponential map at the identity gives a group homomorphism $\mathbb{R}\to G$ defined by $t\to\operatorname{exp}(tX)$ for each $X\in\mathfrak{g}$. This homorpphism is surjective because $G$ is simply connected. Then resticting it to $\mathbb{Z}\subset\mathbb{R}$ gives an infinite cyclic subgroup of $G$ for each vector $X$.

Maybe this was obvious but I just realised it, found it surprising, and tought this could be a good spot to leave it.

Also I think this argument could be used to answer the easy case: if $G$ is non-compact then at least one geodesic is non-compact, hence the map is surjectve and the claim follows.

This is somehow related. A simply-connected Lie group $G$ has uncountably many infinite cyclic subgroups. To see this simply observe that the exponential map at the identity gives a group homomorphism $\mathbb{R}\to G$ defined by $t\to\operatorname{exp}(tX)$ for each $X\in\mathfrak{g}$. This homorpphism is injective because $G$ is simply connected. Then resticting it to $\mathbb{Z}\subset\mathbb{R}$ gives an infinite cyclic subgroup of $G$ for each vector $X$.

Maybe this was obvious but I just realised it, found it surprising, and tought this could be a good spot to leave it.

Also I think this argument could be used to answer the easy case: if $G$ is non-compact then at least one geodesic is non-compact, hence the map is surjectve and the claim follows.