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.