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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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?

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  • $\begingroup$ 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
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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 $G$.

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  • $\begingroup$ 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

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