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 $G/Z(F(G))\cong {\rm Aut}(F(G))$.

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. Any help is welcome. Thanks!