I am trying to prove this: Let $G$ be a finite-by-solvable group, i.e. $G$ has a normal subgroup $N$ that is finite with $G/N$ solvable. Prove that $G$ is solvable-by-finite, i.e., $G$ has a solvable normal subgroup $K$ such that $G/K$ is finite.
My attempt:
There exists a natural map $\psi$ from $G$ to Aut$N$ that takes $g\in G$ to the automorphism of $N$ that’s ‘conjugation by g’. Now, the kernel $K$ of this $\psi$ is $C_G(N)$, the centralizer of $N \in G$. What is $K\cap N$ then? Considering $N$ is finite, we have $G/K$ is also finite. I want to show that $K$ is solvable but I am stuck here. Any help would be appreciated.
You're proof is nearly done. You ask
but this is what you are missing!
Specifically, $K\cap N=Z(N)$ (in particular, it is solvable). By one of the isomorphism theorems, we have $K/Z(N)=K/(K\cap N)\cong KN/N$, and so $K/Z(N)$ embeds into the solvable group $G/N$, and so $K/Z(N)$ is solvable. As solvable-by-solvable groups are solvable, it follows that $K$ is solvable, as required.