Let $\Sigma_g$ be the closed orientable surface of genus $g$. There is
no covering map $p\colon\Bbb R^2\backslash \mathbf 0\to \Sigma_g$ so that $p_*\pi_1(\Bbb R^2\backslash \mathbf 0)$ is a normal subgroup
of $\pi_1(\Sigma_g)$ when $g\geq 2$. In other words, the fundamental group of any
closed orientable hyperbolic surface has no normal infinite cyclic subgroup.
Attempt: Suppose, we have a normal covering $\Bbb R^2\backslash \mathbf 0\to \Sigma_g$, where $g\geq 2$. So, there is an infinite cyclic normal subgroup of $\pi_1(\Sigma_g)$. In other words, we have non-trivial elements $a,b\in \pi_1(\Sigma_g)$ such that $bab^{-1}=a^n$ for some integer $n$. Consider the subgroup $G$ of $\pi_1(\Sigma_g)$ generated by $a,b$ and let $X\to \Sigma_g$ be the covering corresponding to the subgroup $G$ of $\pi_1(\Sigma_g)$. Note that $X$ is an orientable surface as $\Sigma_g$ is an orientable surface. Also, $\pi_1(X)=G$.
$\textbf{Case 1:}$ Let $X$ be compact. From classification theory $X\cong \Sigma_h$ for some $h\geq 0$. So, the covering will be finite-fold, say $m$-fold, and then $2-2h=\chi(X)=m\cdot \chi(\Sigma_g)=m(2-2g)$, i.e., $1+m(g-1)=h$, i.e. $h\geq 2$. In other words, $\pi_1(\Sigma_h)=G$ is generated by two elements, in particular, $\Bbb Z^{2h}\cong \frac{G}{[G,G]}$ is also generated by two elements when $h\geq 2$, a contradiction. So, this case is impossible.
$\textbf{Case 2:}$ Let $X$ be non-compact. Hence, $\pi_1(X)=G$ is a free group. Also, from hypothesis, $G$ is generated by $a,b$ such that $bab^{-1}=a^n\implies a=(b^{-1}ab)^n$. ……..
Now, if I show that the problem can be started by assuming that $a$ is not a proper power of some element, then $a=(b^{-1}ab)^n$ gives $n=\pm 1$. When $n=1$ we have $ab=ba$, and in a free group, if two non-trivial elements commute, then they are powers of some common element, i.e., $a=b^{\pm 1}$ as $a$ is not a proper power of any element, and this ends up with giving that every element of $\pi_1(\Sigma_g)$ is a power of $a$, impossible.
But I don't know how to tackle the case $n=-1$, also how to show the
problem can be started by assuming $a$ is not a proper power of any
element.
Best Answer
I'll give an algebraic proof, which is actually quite general. In particular, surface groups of genus $\geq2$ are torsion-free and hyperbolic.
Theorem. Let $G$ be a torsion-free hyperbolic group. If $G$ contains an infinite cyclic normal subgroup, then $G$ is itself cyclic.
Your result follows as surface groups of genus $\geq2$ are non-cyclic.
Proof. We need the following two useful, and standard facts: (1) For all $x\in G$, the centraliser $C_G(x)$ is infinite cyclic. (2) If a torsion-free group $H$ contains a finite-index cyclic subgroup, then $H$ is itself cyclic.
Let $\langle a\rangle$ be an infinite cyclic normal subgroup of $G$. Let $b\in G$ be arbitrary. As $\langle a\rangle\lhd G$, the element $b$ acts on $\langle a\rangle$ as an automorphism, and so $b^2$ centralises $a$. By fact (1), $\langle a, b^{2}\rangle$ is cyclic. Now, $\langle a, b^{2}\rangle$ is normal in $\langle a, b\rangle$, and so infact has index $2$ in this subgroup. Hence, by fact (2), $\langle a, b\rangle$ is cyclic, and in particular $a$ and $b$ commute. As $b$ was arbitrary, $a\in Z(G)$. By fact (1), we have that $G$ is cyclic as required.