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Is there any theorem which says anything like $SN_G(D)=G$ where $S$ is a $p$-Sylow subgroup and $D$ is the intersection it has with some other subgroup?

I know the first that will probably spring to mind is the Frattini argument, but the thing is in the proof where I saw this the author is using this to show that $\cap_{g \in G}S^g$ is normal in $G$ (to get a contradiction as we assumed $G$ was simple) but then surely we can't use the Frattini argument as it would require $S$ to be normal.

Let G be a group of order $p^nq$ $n \geq 0$ p and q are primes Then G is not simple.

Edit: To add more clarity here is the full argument of the section I'm stuck on .

Proof: Wlog assume $p \neq q$ Suppose G is not simple

( NOTE: I'm going to skip the parts for when n = 1 and where $n>1$ but has trivial intersection as they are separate cases and their arguments don't play any role in the part of the proof I'm interested in.)

So suppose $n>1$ and pick any two Sylow subgroups $S \neq P$ s.t. $P\cap S=D$ is largest possible.

Then $D<S \Rightarrow D<N_S(D)<N_G(D)$

and also

$D<S \Rightarrow D<N_P(D)<N_G(D)$

It follows that $N_G(D)$ can not be a p group since if so it lies in some P-sylow subgroup T and Then $T\cap P$ $\geq N_G(D)$$\geq N_P(D)>D$ $\Rightarrow T=P$ But then similarly $S\cap P=S\cap T<D \Rightarrow \Leftarrow$

Thus a q sylow subgroup Q lives in $N_G(D)$ Then $SN_G(D)=G$ and so if we pick any g it will be of the form $g=sx,s\in S, x\in N_G(D)$

Then $S^g=S^{sx}=S^x\geq D^x=D$ and thus D lies in in every P sylow subgroup of G $\Rightarrow 1<D \leq \cap_{g\in G} S^g$ is normal in G which is a contradiction.

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    $\begingroup$ Why don't you write down the argument where you saw this used? $\endgroup$ – the_fox Dec 24 '18 at 2:04
  • $\begingroup$ @the_fox I added the full argument there $\endgroup$ – excalibirr Dec 24 '18 at 14:49
  • $\begingroup$ Is this from Isaacs' FGT book? $\endgroup$ – the_fox Dec 25 '18 at 2:05
  • $\begingroup$ @the_fox no It's just from lecture notes $\endgroup$ – excalibirr Dec 25 '18 at 2:27
  • $\begingroup$ Just so it is more clear, let me post as an answer Isaacs' proof from his book. You can then comment on which part of it you don't understand. (I find it a little difficult to parse what you wrote.) $\endgroup$ – the_fox Dec 25 '18 at 2:32
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So, this is how Isaacs proves the theorem you mentioned. Strategy and notation look very similar to the proof in your notes, so this is likely where your teacher took it from.


enter image description here

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  • $\begingroup$ This is great thank you so much, This does look quite similar alright but I can't read it properly right now I cant concentrate very well I've been studying for the past 15 hours XP. I'll give it a thorough read tomorrow though and let you know where I get confused. happy holidays by the way ..I'm quite relieved to have finally gotten an answer(and a detailed one at that) for this problem :) $\endgroup$ – excalibirr Dec 25 '18 at 2:41
  • $\begingroup$ Merry Christmas to you too :) I'm guessing that you have a little trouble with the last paragraph, but let me know when you have had a chance to look at it carefully. $\endgroup$ – the_fox Dec 25 '18 at 2:46
  • $\begingroup$ Just got a chance to really look over this just now . It seems like the only major difference between the proof in my notes and this one were (i) he doesn't assume that G is simple and aim for contradiction. (ii) rather than using the normaliser he just uses the Q group which lives in it . $\endgroup$ – excalibirr Dec 26 '18 at 20:46
  • $\begingroup$ My only questions are (a) we can easily start this proof by assuming that G is not simple and aim for contradiction without it interrupting any of the steps right ? ( I mean at the end we could just say so D is a subgroup of the core of S which is normal (FALSE) and then say so G must be simple because everything we've tried is false otherwise .) (b) I'm a tad confused as to why the fact $D<S \cap N \Subset s \cap R$ implies that S=R . is it because the intersection of S and N lies in S and therefore R must be S ? $\endgroup$ – excalibirr Dec 26 '18 at 20:47
  • $\begingroup$ I did not quite understand (a). The goal is to prove that $G$ is not simple. Do you want to argue instead by contradiction? For (b) $S=R$ follows from the maximal choice of $D$ in the beginning. If $S \neq R$ then we have found two Sylow $p$-subgroups with intersection greater than is possible. $\endgroup$ – the_fox Dec 27 '18 at 0:53

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