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Let $G$ be a group of order $p^2q$ ($p$ and $q$ are distinct primes) and $P_1$ and $P_2$ be distinct $p$-Sylow subgroups of $G$. Then if $Z(G) = \{e\}$, $P_1\cap P_2 = \{e\}$

How do I go about proving the above? Using Lagrange's theorem, I have that $|P_1\cap P_2|\in\{1,p\}$ (excluding $p^2$ as the subgroups are distinct), so it only remains to exclude $p$. But if $|P_1\cap P_2| = p$, then the intersection is a cyclic subgroup of $G$. How does this lead to a contradiction? I cannot see how.

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Here is a sketch of a proof, with details missing (for you to work out): note that $P_1$ and $P_2$ are abelian and that, together, they generate $G$, so their intersection must be central.

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  • $\begingroup$ Is there any other way of doing this problem? $\endgroup$ – wrb98 Sep 7 '17 at 23:26
  • $\begingroup$ This is probably the easiest way. Did you manage to fill in the details? $\endgroup$ – verret Sep 8 '17 at 0:07
  • $\begingroup$ I understand why they must be abelian, but why do they together generate G? $\endgroup$ – wrb98 Sep 8 '17 at 0:08
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    $\begingroup$ Think about the possibilities for the size of the group they generate. $\endgroup$ – verret Sep 8 '17 at 0:29
  • $\begingroup$ They can't generate $G$ :s $\endgroup$ – José Alejandro Aburto Araneda Sep 8 '17 at 1:57

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