Let $G$ be a finite group and assume it has more than one Sylow $p$-subgroup.

It is known that order of intersection of two Sylow p-subgroups may change depending on the pairs of Sylow p-subgroups.

I wonder whether there is a condition which guarantees that intersection of any two Sylow $p$-subgroups has the same order.

Thanks for your help.

  • $\begingroup$ Well, if the Sylow subgroups have order a prime then all the intersections have one single element... $\endgroup$ – DonAntonio Nov 16 '13 at 15:36
  • $\begingroup$ If the number of Sylow p-subgroups is at most 2... $\endgroup$ – azimut Nov 16 '13 at 16:03
  • 1
    $\begingroup$ number of the Sylow p-subgroup can not be 2. $\endgroup$ – mesel Nov 16 '13 at 17:17
  • $\begingroup$ But it can be 1. My above comment was kind of a joke of course, I guess you are looking for a deeper criterion. $\endgroup$ – azimut Nov 16 '13 at 18:30
  • $\begingroup$ Actually,it can not be 1 by my assumption :)thanks anyway. $\endgroup$ – mesel Nov 17 '13 at 18:41

I can prove the following, which provides a criterion to start with.

Theorem Let $G$ be a finite group. Then the following are equivalent.

$(a)$ For all $S,T \in Syl_p(G)$ with $S \neq T$, $S \cap T=1$.
$(b)$ For each non-trivial $p$-subgroup $P$ of $G$, $N_G(P)$ has a unique Sylow $p$-subgroup.
$(c)$ For each non-trivial $p$-subgroup $P$ of $G$, $P$ is contained in a unique Sylow $p$-subgroup of $G$.

| cite | improve this answer | |
  • $\begingroup$ Nicky, I am sorry not give any response. I guess you found conditions for a Sylow subgroup to be T.I. subgroup. If you supply proofs, I would read. $\endgroup$ – mesel Jan 28 '18 at 15:49

I have made up such a condition.

We know that the action of $G$ on $Syl_p(G)$ by conjugation is transitive. If this action is double transitive then the intersection of any two Sylow p-subgroups is conjugate and they must have same order.

Proof: Let $P,Q,R,S$ be elements of $Syl_p(G)$ such that $P\neq Q$ and $R\neq S$. By double transivity, $\exists x$ in $G$ such that $P^x=R$ and $Q^x=S$, thus $(P\cap Q)^x=P^x\cap Q^x =R\cap S$.

But I do not know when $G$ is double transive on $Syl_p(G)$. Should I ask it as a new question?

| cite | improve this answer | |
  • $\begingroup$ Did you ever find anything about when $G$ acts 2-transitively on its Sylow $p$-subgroups? There seems to be a fair amount of variety, but I guess it would be possible to give a classification of $G/R$ where $R$ is the intersection of normalizers of the Sylow $p$-subgroups. $\endgroup$ – Jack Schmidt Feb 18 '14 at 19:30
  • $\begingroup$ Not something remarkable.But there are some groups which acts 2-transitively on its Sylow$p-$subgroup.I really expect interesting result about $G/R$. $\endgroup$ – mesel Feb 18 '14 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.