I know two facts and I’ve managed to figure out how to prove one, but the other one is still a little confusing.
Let $G$ be a finite solvable group and $F(G)$ is the Fitting subgroup of $G$.
(1) $G/Z(F(G))$ is isomorphic to a subgroup of ${\rm Aut}(F(G))$;
(2) $G/F(G)$ is isomorphic to a subgroup of ${\rm Out}(F(G))$.
Proof of (1):
$F(G)$ is normal in $G$, so $G=N_G(F(G))$. Since $G$ is solvable, $Z(F(G))=C_G(F(G))$. $F(G)$ is a characteristic subgroup of $G$ and $Z(F(G))$ is a characteristic subgroup of $F(G)$, therefore $Z(F(G))$ is characteristic and normal in $G$, and $G/Z(F(G))$ is hence well-defined. By the $N/C$ theorem, $G/Z(F(G))=N_G(F(G))/C_G(F(G))$ is isomorphic to a subgroup of ${\rm Aut}(F(G))$.
About (2), I asked a question and have got some ideas.
I know that $F(G)/Z(F(G))\cong {\rm Inn}(F(G))$ and by (1) that $G/Z(F(G))$ is isomorphic to a subgroup of ${\rm Aut}(F(G))$.
So by the third isomorphism theorem, we have $G/F(G) \cong G/Z(F(G)) \big/ F(G)/Z(F(G))$.
If it is true that, say, “if $A\cong M$ and $B\cong N$ where $B\trianglelefteq A$ and $N\trianglelefteq M$ then $A/B\cong M/N$”, then we’re done. However, it is not true in general. I believe that I ignored something important. So what should I do next? It really seems very close. It’s quite obvious to think in an intuitive way that $G/F(G)\cong G/Z(F(G)\big/ F(G)/Z(F(G))$ is isomorphic to a subgroup of ${\rm Aut}(G)/{\rm Inn}(G)$ since the $G/Z(F(G))$ is isomorphic to a subgroup of ${\rm Aut}(G)$ and $F(G)/Z(F(G))\cong {\rm Inn}(G)$. But it’s not sufficient in a proof. I think there’s still something missing.
Let me just make my question clear. I want to take an example. Assume that $A$ is a subgroup of $C$ and $B\trianglelefteq A$. Also, assume that $N\trianglelefteq M$. If $A\cong M$ and $B\cong N$, then it is Not true in general that $A/B\cong M/N$. So in the case that we were talking about, $C={\rm Aut}(F(G))$, $B={\rm Inn}(F(G))$, $M=G/Z(F(G))$, $N=F/Z(F(G))$, it’s just the same: $M$ is isomorphic to a subgroup of $C$, namely $A$, and $N\cong B$. But we don’t have $A/B\cong M/N$ in general. I want to know how to prove it in this specific case.
Any help is welcome. Thanks!
Since $F(G)$ is normal in $G$, $G$ acts by conjugation on $F(G)$. This gives you a map $G\to \mathrm{Aut}(G)$. The kernel of this map is $C_G(F(G))=Z(F(G))$ as you note, so the map factors to an embedding $G/Z(F(G))\to \mathrm{Aut}(G)$.
Now, the image of $F(G)$ under this embedding, is the image of $F/Z(F(G))$ under the map that sends an element to the conjugation map by that element; this is the group of inner automorphisms of $F(G)$ (not just "isomorphic to it", but exactly it), so $$\frac{G/Z(F(G))}{F(G)/Z(F(G))} \cong \frac{G}{F(G)}$$ embeds into $\mathrm{Aut}(G)/\mathrm{Inn}(F(G)) = \mathrm{Out}(F(G))$. That is all you need.