4
$\begingroup$

Theorem: Suppose G is a finite group with Sylow p-subgroup P. Then the following are equivalent:

  • The set K of elements of G of order relatively prime to p (the p′-elements) form a subgroup
  • If A and Ag are both subsets of P, then there is some x in CG(A) such that xg is in P

That the first implies the second is a silly trick: $(a^{-1} a^g)^{-1} = g^{-1} g^a \in P \cap K = 1$ for any $a \in P$ and $g \in K$ such that $a^g \in P$.

The second implies the first is not too hard (Frobenius normal p-complement theorem), but I'm trying to use this as a first example, and so don't want to have any prerequisites outside a very gentle undergraduate group theory course. Most of the rest of the talk is just using Sylow's theorem.

Is there a very low-tech, short, few-preliminaries proof that "absolutely no fusion" implies a normal p-complement?

I would be ok with assuming P is abelian, so that we get:

  • If A and Ag are both subsets of P, then g in CG(A).

I'm also fine with assuming p = 2 so that "relatively prime to p" shortens to "odd".

I don't think using the transfer is appropriate, as it won't be used again, and the whole point of this proof is to motivate learning something else.

For a "no" answer to my question, it would be sufficient to convince me that transfer is needed (and nice if you can suggest a special case where it wouldn't be needed, other than P of order 2).

  • 0
    You can do it with characters, but I don't suppose you would accept that as a "few preliminaries" proof.2012-06-18
  • 0
    @Geoff: I know that no one in the target audience knows anything about characters, so it won't work for this. I do like that proof myself, and in general how characters and transfer can often accomplish the same tasks. One version is Isaacs's CT 8.22 (Brauer–Suzuki) http://books.google.com/books?id=MeE7BFXwQroC&pg=PA1372012-06-18
  • 0
    I am confused. Isn't $A_4$ a counterexample, with $P=V_4$, and $A$ being any subgroup of order 2? The set of odd order elements is certainly a group. Moreover, any $A^g$ is in $P$, since $P$ is normal. But $C_G(A)=1$, so there is no $x$ such that $xg\in P$ if $g$ is an element of order 3.2012-06-18
  • 0
    @Alex: (1,2,3) and (2,3,4) has a product of order 2, so the odd order elements are not a group. A4 satisfies both hypotheses with p=3, but neither with p=2.2012-06-18
  • 0
    I agree about characters. Argument for $|P| = 2,$ works with little modification when $P$ is a cyclic $2$-group. Consider determinant of regular representation. A generator of $P$ acts with determinant $-1.$2012-06-18
  • 2
    It's not hard to see that what you are asking for in the case $P$ abelian is equivalent to Burnside's Transfer Theorem. (If $A,A^g \subseteq P$ then $P$ and $P^{g^{-1}}$ are conjugate in $C_G(A)$, so $P^{hg}=P$ with $h \in C_G(A)$ and then the hypothesis of BTT hives $hg \in C_G(P)$, so $g \in C_G(A)$ and we have your hypothesis.) So you are really just asking for a transfer-free proof of BTT.2012-06-18
  • 1
    @Derek: Thanks! I agree that when P is abelian my version and BTT are easily equivalent, so yes I am apparently asking for a transfer-free proof of the transfer theorem. I'm going to keep looking for such a published proof, but am I right in thinking there probably hasn't been one?2012-06-18
  • 1
    I have never come across such a proof!2012-06-18
  • 0
    Sorry, Jack, I was being silly. I knew I shouldn't be posting at 3 am. Thanks!2012-06-19
  • 1
    I can't even do the special case of a self-normalizing Sylow subgroup of order $p$ without BTT. You can show by counting that you have the approrpiate number of $p'$-elements, but you still need to show that they form a subgroup.2012-06-19
  • 1
    @Derek: That is exactly the problem (no sylow intersections means the hypothesis is basically useless). The more extreme that is, the easier the character theory argument gets (until we get a Frobenius group with complement P and kernel O_{p'}(G)), but it's still a far cry from the unsophisticated level I'm looking for. I plan on writing a negative answer in the next few days, indicating the character theory and transfer proofs as the canonical alternatives. One book I have has a survey of 5 or 6 BTT proofs (all using transfer, but in different ways).2012-06-19

0 Answers 0