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What can we say about the intersection of the $p$-sylow and $q$-sylow subgroups of group $G\;$?

It's not necessary that $p=q$.

Is there general statements about the intersections of sylows subgroups ?

You can give me restricted conditions, this is also welcome !

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    $\begingroup$ Hint: any element in that intersection has order a power of $\,p\,$ and also a power of $\,q\,$ ... $\endgroup$
    – DonAntonio
    Jun 14 '13 at 23:41
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    $\begingroup$ Yes that certainly covers the case $p \neq q.$ $\endgroup$ Jun 14 '13 at 23:43
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    $\begingroup$ @DonAntonio so the intersection is trivial ! $\endgroup$ Jun 14 '13 at 23:48
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    $\begingroup$ Withou any further data I don't think there's much more to be said, unless we begin with very particular cases, as if the Sylow $\,p$-subgroups are of order $\,p\,$ or some restriction/condition on the order of the normalizer of these groups... $\endgroup$
    – DonAntonio
    Jun 14 '13 at 23:48
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    $\begingroup$ Yes @MathsLover, exactly. $\endgroup$
    – DonAntonio
    Jun 14 '13 at 23:49
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Let $R = P\cap Q$, where $\;P\leq G$ a p-sylow subgroup and $Q\leq G$ is a q-sylow subgroup, and $p\neq q$.

Then for any $r \in R$, $r$ has order equal a power of $p$ and equal to a power of $q$.

What must $R$ "look like"?

The same can not necessarily be said about the case $p = q$, for $\;p, q$ prime.

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  • $\begingroup$ $R$ must be the trivial subgroup . $\endgroup$ Jun 15 '13 at 0:11
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    $\begingroup$ Exactly. Precisely because $p, q$ are distinct primes. $\endgroup$
    – amWhy
    Jun 15 '13 at 0:14
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    $\begingroup$ yes because , if $p^a = q^b$ for some $a,b \in Z$ then $p^{a-1} = q^b / p $ so p divides q ,contradiction since q is prime. so $p^a \neq q^b$ for any $a,b \in Z$ $\endgroup$ Jun 15 '13 at 0:17
  • $\begingroup$ @amWhy: Very nice! +1 $\endgroup$
    – Amzoti
    Jun 16 '13 at 0:37
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The intersection of two different Sylow $p$-subgroups for the same prime $p$ is less predictable. If a finite group $G$ has a normal $p$-subgroup $U,$ then $U$ is contained in any Sylow $p$-subgroup of $G.$ It is quite difficult to give precise conditions which will guarantee the existence of two Sylow $p$-subgroups $P_{1}$ and $P_{2}$ such that $P_{1} \cap P_{2} = U.$ A theorem of J. Brodkey asserts that there are such subgroups if $G/U$ has Abelian Sylow $p$-subgroups. ut, for example there is a finite group $G$ with $|G| = 144$ such that $G$ has no non-identity normal $2$-subgroup, but there do not exist Sylow $2$-subgroups $P_{1}$ and $P_{2}$ of $G$ with $P_{1} \cap P_{2} = 1.$ ( because $|G| < |P_{1}||P_{2}|$). The group $G$ in this example has a semi-dihedral Sylow $2$-subgroup.

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We know that the intersection of two Sylow $p$-subgroups for the same prime $p$ is a prime power. Without more data, there is really nothing more to say. In fact, for any integer $n \geq 1$ we can find a finite group $G$ with the following properties:

  1. The Sylow $p$-subgroups of $G$ have order $p^n$.
  2. For any $0 \leq i \leq n$, there exists Sylow $p$-subgroups $P$ and $Q$ such that $P \cap Q$ has order $p^i$.

To construct such $G$, first let $q$ be a prime such that $q \equiv 1 \mod{p}$ and $q > p$ and $q > n$. The existence of such $q$ follows from Dirichlet's theorem.

Let $T$ be the nonabelian group of order $pq$. Define $G = T \times T \times \ldots \times T$, where $T$ appears in the direct product exactly $n$ times. Now $T$ has exactly $q$ Sylow $p$-subgroups, so let $P_1, P_2, \ldots, P_n, P$ be distinct Sylow $p$-subgroups of $T$. Let

\begin{align*} Q_0 &= P \times P \times P \times \ldots \times P \\ Q_1 &= P_1 \times P \times P \times \ldots \times P \\ Q_2 &= P_1 \times P_2 \times P \times \ldots \times P \\ \ldots \\ Q_{n-1} &= P_1 \times P_2 \times P_3 \times \ldots \times P_{n-1} \times P \\ Q_n &= P_1 \times P_2 \times P_3 \times \ldots \times P_{n-1} \times P_n \end{align*}

Now each $Q_i$ is a Sylow $p$-subgroup of $G$ and $|Q_i \cap Q_n| = p^i$ for all $0 \leq i \leq n$.

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    $\begingroup$ Very nice observations,m.k. (+1) $\endgroup$
    – Mikasa
    Jun 15 '13 at 12:50
  • $\begingroup$ @m. k. , i have some qestions , 1- how did you prove that $n_p \neq 1 $ ? because it has two possibilities namely , $1$ and $q$ ? 2- how did you know that such group $T$ must exist ? i know that if $p$ doesn't divides $q-1$ then $T$ is abelian , but this is not the case here , so , is there a theorem or something like that says that if $p$ divides $q-1$ then there is a non-abelian group of order $pq$ ? $\endgroup$ Jun 15 '13 at 14:04
  • $\begingroup$ for my first question , i got the answer ! if $n_p =1 $ then this unique sylow subgroup is normal and then $T$ is cyclic hence abelian , contradiction ! so $n_p =q $ . $\endgroup$ Jun 15 '13 at 14:10
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    $\begingroup$ Yes, whenever $p < q$ are primes and $q \equiv 1 \mod{p}$, there exists a nonabelian group of order $pq$. It can be constructed as a semidirect product, see for example this question. $\endgroup$ Jun 15 '13 at 14:16
  • $\begingroup$ @m.k. , great ! my last question , why $|Q_i \cap Q_n|=p^i$? $\endgroup$ Jun 15 '13 at 20:32

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