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.

EDIT: On re-reading the question, I see that I misread 'at' as 'for' in the first line. This led me to read the first line as a question. Apologies! My answer is retracted.

$hgh^{-1}=kgk^{-1}$ if and only if $k^{-1}h$ commutes with $g$. So your first question is just asking for an infinite abelian subgroup $H$ that is its own centraliser in $G$. There are many examples. For instance, if $G$ is a non-abelian free group and $g\in G$ is not a proper power then $\langle g\rangle$ is its own centraliser. Furthermore, normal subgroups of $G$ are either finite-index or infinitely generated, so $G$ is certainly not normal.

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.