Centralizer of a finite normal subgroup has finite index Let $G$ be a (not necessarily finite) group and $N\lhd G$. Show that if $N$ is finite then $C_G(N)$ has finite index in $G$. Further show that if $N$ is finite and $G/N$ is cyclic then the center $Z(G)$ has finite index.
These two are driving me crazy, and I simply cannot figure them out... Any help is appreciated
 A: Since $N$ is normal, $G$ acts on $N$ by conjugation, giving a homomorphism from $G$ to $\rm{Aut}(N)$. The kernel of this map is exactly $C_G(N)$ so since $N$ only has a finite number of automorphisms, the index must be finite.
For the second one, we have $G = N\left< g \right>$ for some $g\in G$ (just take a generator of the quotient). Now, we have that $\left< g \right>$ acts on $N$ by conjugation, so the kernel of the corresponding map has finite index in $\left< g \right>$. But all the elements in this kernel commute with everything in $N$ and also with everything in $\left< g \right>$ and hence with everything in $G$, so the center of $G$ contains this kernel, which means that it must have finite index.
Here is a more general version: Let $G$ be a group with a finite normal subgroup $N$ and an abelian subgroup $H$ such that $G = NH$. Then $Z(G)$ has finite index in $G$. The proof is the same as above, since $H$ acts on $N$ by conjugation and the kernel is contained in $Z(G)$ and has finite index in $H$ which has finite index in $G$.
A: For your first question, use $N/C$ theorem:

$(N/C)$ Theorem: If $G$ is a group and $N\subset G$, then $N_G(N)/C_G(N)$ is isomorphic to a subgroup of $\text{Aut}(N)$.

So since $N$ is normal in the group so the normalizer of it is $G$ and we have $G/C_G(N)\cong S\leq Aut(N)$ which is finite.
