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Let $P$ and $Q$ be distinct abelian $p$-Sylow subgroups of a finite group $G$.

Is their intersection $P\cap Q=\{1\}?$

This is true for order $p$ $P$ and $Q$.

I've been trying for a while now but I can't get this totally elementary-looking result.

But someone told me it is true (not sure if realiable tho).

Maybe a Hint?

I don't want to waste much time on this. Thanks.

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    $\begingroup$ It is not even true for cyclic $P$ and $Q$, so I am not sure why you think it is obvious. $\endgroup$ – Derek Holt Nov 17 '17 at 21:23
  • $\begingroup$ I was thinking order $p$ lol; you're right. Maybe intersection of all the $p$-Sylows? Not sure. Do you have easy counter-examples? $\endgroup$ – Shoutre Nov 17 '17 at 21:32
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    $\begingroup$ There is a group of order $12$ with $3$ cyclic Sylow $2$-sugroups that intersect in a subgroup of order $2$. $\endgroup$ – Derek Holt Nov 17 '17 at 21:52
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Not true in general: look at the Sylow $2$-subgroups of $S_3 \times S_3$ and you will find a pair having non-trivial intersection.

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As mentioned above this is not true in general. It IS true however for different primes p and q. That is, if $P$ is a p-group and $Q$ is a q group then $$P \cap Q = \{e\}$$

This is because an element in this intersection must have an order that divides both $p$ and $q$, which are obviously co-prime. I'm wondering if that's what he meant.

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