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I am reading Isaacs' book finte group theory, and I have two questions.

  1. in page 90, there is a Wielandt's theorem (if $G$ has a nilpotent Hall $\pi$-subgroup, then all Hall $\pi$-subgroups of $G$ are conjugate), now I know how to prove it, I want to ask, analogue to Sylow's theorem, if there is some arithmetic property about the number of Hall $\pi$-subgroups of finite group $G$ (or we can assume $G$ is solvable, what results can we get?).
  2. in page 96, we know that theorem 3.23(a) is a generalization of Sylow E-theorem and theorem 3.23(b) is a generalization of Sylow C-theorem. Now I am thinking what we can say about the number of A-invariant Sylow $p$-subgroups of $G$ in theorem 3.23.

Thank you.

Seirios
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user94782
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1 Answers1

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(1) There is a theorem in [Marshall] Hall's The Theory of Groups (9.3.1 in the 1959 edition) a portion of which states that the number of [Philip] Hall $\pi$-subgroups (where only primes dividing $|G|$ occur in $\pi$) of a solvable finite group $G$ is a product of factors each of which (a) is congruent to 1 modulo some member of $\pi$ and (b) divides (the order of) one of the chief factors of $G$. Hall adds that each of these factors is a prime power, but this is superfluous since the chief factors of a finite solvable group are elementary abelian. I was frankly shocked that Isaacs left this out of his text, but I would take that as an indication that this result has not turned out to be very useful over the years.

It's important to note that this gives additional information about the Sylow subgroups of a solvable group. For example, a group of order 315 has a normal subgroup of order 5. Sylow's theorem said the number of Sylow-5 subgroups is congruent to 1 mod 5 and divides 63, so you would think 21 would be possible. But the Philip Hall counting result excludes it, because you can't write 21 as a product of prime powers each of which is congruent to 1 mod 5. It also provides a very fast proof that $A_5$ isn't solvable. $A_5$ has 10 Sylow-3 subgroups, and 10 is not a product of prime powers that are congruent to 1 mod 3. A solvable group of order 60 must have 1 or 4 Sylow-3 subgroups...never 10.

So maybe the reason no text since 1959 has included this result is that it makes homework problems too easy. :)

(2) I can't cite a source, but I'm pretty sure I can prove the usual count conclusions (congruent to 1 mod $p$ and divides the $p'$ part of $|G|$). You can infer from the text that under the hypotheses of 9.23 the number of $A$-invariant Sylow-$p$ subgroups of $G$ is a divisor of the $p'$ part of $G$. This is because if $P$ is one such $A$-invariant Sylow-$P$, then the number of them must be $[C_G(A):N_{C_G(A)}(P)]$ and any Sylow-$p$ subgroup $Q$ of $C_G(A)$ is $A$-invariant and thus can be extended to some $A$-invariant $P$ and thus clearly $Q\le N_{C_G(A)}(P)$, i.e., $N_{C_G(A)}(P)$ contains a Sylow-$p$ subgroup of $C_G(A)$. Also, their number must be congruent to 1 mod $p$ since given an $A$-invariant Sylow-$p$ $P$, $P$ acts by conjugation on all Sylow-$p$ subgroups and takes $A$-invariant ones to $A$-invariant ones, and $P$ is its only fixed point.

C Monsour
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