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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!

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  • $\begingroup$ 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. $\endgroup$ – YCor May 2 '20 at 14:18
  • $\begingroup$ @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. $\endgroup$ – user121195 May 2 '20 at 14:22
  • $\begingroup$ @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. $\endgroup$ – user121195 May 2 '20 at 14:27
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    $\begingroup$ It is not a question of "fault", but you do probably have to figure it out for yourself. $\endgroup$ – Geoff Robinson May 2 '20 at 16:25
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    $\begingroup$ 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. $\endgroup$ – Geoff Robinson May 6 '20 at 11:42
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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 $\mathrm{Out}(F(G))$.

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  • $\begingroup$ 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. $\endgroup$ – Santana Afton May 3 '20 at 4:29

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