# Solvable group determined by the action on its Fitting subgroup and the isoclass of some of its Sylows

Let QP be finite p-groups, H ≤ Aut(Q).

Is it really true that there is at most one p-solvable group G such that $Q \unlhd G$, $C_G(Q) \leq Q$, P is a Sylow p-subgroup of G, and the map from G to Aut(Q) surjects onto H?

I think this is true (up to an isomorphism of G restricting to the identity on Q), but I am worried about a consequence:$\newcommand{\Aut}{\operatorname{Aut}}\newcommand{\Fit}{\operatorname{Fit}}$

Let F be a finite nilpotent group, let H be a subgroup of Aut(F), and for each prime p dividing the order of F, let FpEp be a Sylow p-subgroup of F contained in some finite p-group.

Say that G is a model of $(H,E)$ if G is solvable, $F \unlhd G$, $C_G(F) \leq F$, the homomorphism from G to $\Aut(F)$ is surjective onto H, and each Ep is a Sylow p-subgroup of G.

Is it really true that given F, H, and E there is at most one (up to an isomorphism restricting to the identity on F) model of $(H,E)$?

In such a model, F is the largest nilpotent normal subgroup of G, and a theorem of Fitting guarantees that $G/Z(F) \cong H$, so of course $G/Z(F)$ is uniquely determined by H. I had no idea G itself could be uniquely recovered if only ones knows the Sylow subgroups (assuming I am not wrong).

If it is false, I would appreciate an example where F = Q is a p-group.

If it is true, I would appreciate an older reference than 21st century topology, as surely this is “Fitting's other theorem.”

A special case is clear: if F has vanishing second cohomology as an H-module, then G must be the semi-direct product. In particular, if the orders of F and H are coprime, then of course G is uniquely determined.

On the other hand, if the second cohomology does not vanish, then while multiple extensions can arise, in the examples I've seen, each extension is uniquely identified by its Sylow subgroups that intersect the Fitting subgroup non-trivially.

I am not sure I understand how knowing the Sylows, but not knowing how they interact is sufficient to know the group. A good answer (if it is true) might begin “but Jack we do know how they interact, because…”.

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@Geoff: Thanks. This is part of the "a constrained fusion system has a unique model theorem". ams.org/mathscinet-getitem?mr=2167090 is given as the standard reference. I removed most of the bit about H coming from a fusion system on P, and modeled it after I.4.9c in AKO ams.org/mathscinet-getitem?mr=2848834 – Jack Schmidt Jan 7 '12 at 7:46

That's an intersting question! I decided to look for an example with $Q$ elementary abelian. The restriction map $H^2(G/Q,Q) \rightarrow H^2(P/Q,Q)$ is injective, but inequivalent extensions can be isomorphic as groups, so it seems possible that you could have inequivalent extensions for which the groups were non-isomorphic, but the restrictions to $P$ were isomorphic.
Anyway, I think GAP (or Magma) SmallGroup(486,149) and SmallGroup(486,150) provide a counterexample to your conjecture. I want $Q$ to be elementary abelian of order 27, and these groups have more than one such normal subgroup. I chose $H$ in SmallGroup(486,149) to be generated by generators 4,5,6 and $H$ in SmallGroup(486,150) to be generated by generators 3,5,6. These are self-centralizing elementary abelian normal subgroups of order 27. The quotient groups $G/H$ are isomorphic as are the Sylow 3-subgroups of the two groups.