# Why $G/F(G)$ is isomorphic to a subgroup of ${\rm Out}(F(G))$?

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 $${\rm Aut}(F(G))$$.

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! • Often "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. – YCor May 2 at 14:18 • @GeoffRobinson Thanks. I mentioned that in my post, this isomorphism theorem does play a key role in the proof. But I got stuck after using this theorem. – Ebenezer May 2 at 14:22 • @GeoffRobinson$F(G)/Z(F(G))\cong {\rm Inn}(F(G))$and by (1) that$G/Z(F(G))\cong {\rm Aut}(F(G))$. 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. It’s exactly where I got stuck. The isomorphism theorem does bring me very close to the conclusion. But there’s still something in my way. – Ebenezer May 2 at 14:27
• It is not a question of "fault", but you do probably have to figure it out for yourself. – Geoff Robinson May 2 at 16:25
• That's what I explained (or tried to) in the comments before. These are now deleted. It's indeed true that the third isomorphism theorem isn't really necessary. – Geoff Robinson May 6 at 11:42

For (2), there is a general construction at work here. Given a short exact sequence of groups

$$1 \to K \to G \to Q \to 1$$

there is always a well-defined homomorphism $$\varphi: Q\to \mathrm{Out}(K)$$. The idea is to lift elements of $$Q$$ into $$G$$, and have them act on $$K$$ by conjugation. It would be instructive to work out the specifics yourself! Note that $$\ker(\varphi)$$ is the image of $$C_G(K)$$ into $$Q$$.

So, let’s think about the specific situation we’re in. We have a short exact sequence

$$1\to F(G) \to G \to G/F(G) \to 1$$

and a map $$\varphi: G/F(G)\to\mathrm{Out}(F(G))$$. We want to prove this map is injective. Fortunately, the kernel is the image of $$C_G(F(G))$$ into $$G/F(G)$$ ... which we know is trivial since $$G$$ is solvable! (You mentioned $$C_G(F(G)) = Z(F(G)) \subseteq F(G)$$ so it collapses in the quotient).

Thus, the kernel is trivial and so $$G/F(G)$$ naturally sits as a subgroup of $$\mathrm{Out}(F(G))$$.

• I should note that this general construction comes up often in very natural situations. At least, it does with groups I care about in geometry and topology. – Santana Afton May 3 at 4:29