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Let $G$ be a Lie group with a bi-invariant metric. Then the exponential map at the identity in the Riemannian sense, $\operatorname{exp}(v)$, is the same as the exponential map in the Lie sense, $e^v$. In particular $1$-parameter subgroups $L=e^\ell$ corresponding to lines $\ell\subset T_eG$ are geodesics, but this may fail if only right (or left) invariant.

I would like to have an example of this fact, to know if the result still holds if I restrict to a central subgroup $N\subset G$, and have a better reference than this (unreferenced) wiki article.

The question regarding the central subgroup is justified by the fact that any invariant metric on an abelian Lie group is automatically bi-invariant.

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    $\begingroup$ In general, computing geodesics is hard. But, in the simple case of left invariant metrics on $SO(3)$ or $SU(2)$, it has been done in special cases: see, e.g. arxiv.org/pdf/1504.05472v3.pdf where formula (4) gives an explicit formula for geodesics on $S^3$ under the Berger metric (which is left $S^3$-invariant and right $S^1$ invariant.) $\endgroup$ – Jason DeVito Feb 23 at 17:43
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You can really take your favorite nonabelian Lie group and any metric that isn't bi-invariant. The simplest example is $$ G = \left\{ \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} \mid a,b \in \mathbb{R} \right\} .$$

This group acts simply transitively by isometries on the hyperbolic plane $\mathbb{H}^2 = \{x+iy \mid y >0 \}$, and admits a (unique) left-invariant Riemannian metric making the orbit map an isometry. For that metric, the geodesics in $G$ correspond to the geodesics in the hyperbolic plane.

The one-parameter subgroups need not correspond to Riemannian geodesics. For example, the curve $$ c(t) = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} \cdot i = t + i $$ is a horocycle.

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