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Thompson's normal $p$-complement theorem states that

For an odd prime $p$ dividing the order of a finite group $G$, if $N_G(J(P))$ and $C_G(Z(P))$ has normal $p$-complements, for $P$ a Sylow $p$-subgroup $G$, then so does $G$.

Now, $J(P)$ stands for the Thompson subgroup which is generated by the abelian subgroups of $P$ of maximal rank. How did Thompson know that $J(P)$ plays a central role to prove Frobenius' conjecture? Are there any earlier results on $J(P)$ which gives Thompson the idea that this theorem holds? Or shortly, why did Thompson define $J(P)$?

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    $\begingroup$ Very nice question, but probably it can only be answered by himself or people that have been working closely with him. $\endgroup$ Apr 26 '17 at 8:06
  • $\begingroup$ For clues I would read Gorenstein's paper at the Santa Cruz conference, p17 T10. $\endgroup$ Apr 26 '17 at 8:10
  • $\begingroup$ @ancientmathematician can you give a link or something? Because I wasn't able to find the paper. $\endgroup$
    – Levent
    Apr 29 '17 at 6:59
  • $\begingroup$ books.google.co.uk/… $\endgroup$ Apr 29 '17 at 7:24
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I think the answer is that the definition resulted from experiment and trial and error. Thompson was looking for a single subgroup of $P$ for which the normalizer would determine whwther or not $G$ has a normal $p$-complement.

In the version of his theorem that you state, he does not quite achieve this, because he requires also that $C_G(Z(P))$ has a normal $p$-complement. But there is an improved version of the result in Theorem 3.1, Chapter 8 of Gorenstein's book "Finite Groups", in which the aim is achieved.

The definition of $J(P)$ there is slightly different from the one you give, and probably came later. It is defined as the subgroup of $P$ generated by all abelian subgroups of maximal order (rather than maximal rank), and the theorem, which is attributed to Glauberman and Thompson, states that, for odd primes $p$, $G$ has a normal $p$-complement if and only if $N_G(Z(J(P)))$ does.

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  • $\begingroup$ Thank you for the answer. But how one can notices that the subgroup that he needs to define is $J(P)$ by experiment? Using GAP or some other computer algebra systems, it seems possible. But back then, it seems much more harder. $\endgroup$
    – Levent
    Apr 29 '17 at 7:01
  • $\begingroup$ I have been racking my brains about this, and wish that I'd kept better notes when less ancient. But I do remember a conversation with Glauberman about this very issue, I suppose about 1970. As I recall it (and my memory may be defective) he said the result was very stable in the sense that using any sensible sort of "maximality" led to the result. (I hope I am not ascribing my limited understanding to him, or putting words into his mouth.) $\endgroup$ Apr 29 '17 at 7:29
  • $\begingroup$ Glauberman has an e-mail address that is easy to find so you could consider writing and asking him. $\endgroup$
    – Derek Holt
    Apr 29 '17 at 8:12

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