Take $G$ to be a non-abelian free group and $h\in G$ not a proper power. Then $H=\langle h\rangle$ is its own centraliser. In particular, for $g\in G\smallsetminus H$,
$h^{m}gh^{-m}\neq h^ngh^{-n}$
if $m\neq n$, so $(g^{h^n})$ is certainly infinite.
Greenberg's Theorem asserts that normal subgroups of $G$ are either of finite index or infinitely generated, so $H$ is certainly not normal.
EDIT: I just noticed your condition on Cartan subalgebras for the second question. I have nothing to say about that!
FURTHER EDIT:
Any abelian subgroup equal to its own centraliser would work have worked (eg the maximal $\mathbb{Z}$ in the infinite dihedral group). I gave the free group example because $H$ isn't normal.