Exponential map in the Lie vs. Riemannian category 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.

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.
