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!

"X is isomorphic a subgroup of Y"is poor way of starting"this natural map from X to Y is an isomorphism", which in case is not easy to state can better be stated as"there is a natural injective homomorphism from X to Y". By poor way, I mean that when one applies such a statement, one needs to refer to some given natural map, not the bare existence of such a map. $\endgroup$ – YCor May 2 '20 at 14:1810more comments