Let $G$ be a group and $H$ be an abelian and normal subgroup. If $H$ is finite and maximal, prove that $G$ is finite.
What I tried :
Assume $H=\{e,h_2,\cdots,h_{n}\}$. As for each $j$, we have $H\leq C_G(h_j)\leq G$, by maximality, we have $H=C_G(h_j)$ or $h_j\in Z(G)$. WLOG, suppose $\{h_2,\cdots,h_k\}\subset Z(G)$ and for $j\geq k+1$, $C_G(h_j)=H$.
For each $g\in G$, we have $gH=Hg$, i.e. $\{g,gh_2,\cdots,gh_k\}=\{g,h_2g,\cdots,h_kg\}$ and $\{gh_{k+1},\cdots,gh_n\}=\{h_{k+1}g,\cdots,h_ng\}$.
Indeed, for any $g$ and its conjugacy class members, we have assigned one bijection on the set $\{k+1,\cdots,n\}$, that is $h_{k+1}g=gh_{\sigma(k+1)},\cdots,h_{n}g=gh_{\sigma(n)}$
I'm a little mixed up to continue! I think we must prove that $H$ is of finite index and then we're done ...
EDIT: The question has been edited to omit that $H$ is of finite index of $G$. So we first prove that $G/H$ is finite.
Let $H\subset G$ a finite normal maximal subgroup. For all $g\in G-H$ we have $\langle g\rangle H=G$ because $H$ is normal and maximal. The second isomorphism theorem tells us that $$G/H = \langle g\rangle H/H\cong \langle g\rangle/(\langle g\rangle\cap H).$$ Of course the latter is cyclic, so it is either finite or isomorphic to $\Bbb{Z}$. But if it is isomorphic to $\Bbb{Z}$ then $$\langle g\rangle\cong\Bbb{Z}\qquad\text{ and }\qquad\langle g\rangle\cap H=\{1\},$$ from which it follows that $\langle g^2\rangle H$ is a subgroup of $\langle g\rangle H=G$ strictly containing $H$, i.e. satisfying $$H\subsetneq \langle g^2\rangle H\subsetneq G,$$ contradicting the maximality of $H$. This shows that $G/H$ is finite.
By Lagrange's theorem we now have $$|G|=|H|\cdot|G/H|.$$ So $G$ is finite because $H$ and $G/H$ are.
Note that this doesn't use the fact that $H$ is abelian.