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Possible Duplicate:
relation between p-sylow subgroups

If I have two p-sylow groups $P_{1}, P_{2}$ can I then be sure that

$x\in P_{1} \Rightarrow x \notin P_{2}$?

That is to say that all the elements in $P_{1}$ are distinct from the elements in $P_{2}$?

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    $\begingroup$ Well, identity element? $\endgroup$ – user21436 Mar 20 '12 at 11:04
  • $\begingroup$ Exact Duplicate, Look at my answer and other linked questions...: math.stackexchange.com/q/111730/21436 $\endgroup$ – user21436 Mar 20 '12 at 11:07
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No. An easy counterexample would be the direct product $G=P\times N$, where $P$ is a $p$-group, and $N$ is a group with several Sylow $p$-subgroups. Then all the Sylow $p$-subgroups of $G$ are of the form $P\times P'$, where $P'$ is a Sylow $p$-subgroup of $N$. All these intersect non-trivially at $P$.

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  • $\begingroup$ Nice Answer. In the question, I link there, I have been hand wavy about an example. But, good example. +1! $\endgroup$ – user21436 Mar 20 '12 at 11:08
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The intersection of all the Sylow $p$-subgroups is called the $p$-core of $G$ and is the biggest normal $p$-subgroup of $G$. In other words, if you have a normal $p$-subgroup in $G$, then this subgroup is contained in all the Sylows. Easy example: dihedral group of order 20. The centre is of order 2 and is contained in every Sylow 2-subgroup.

This is not the only way that two Sylows can intersect. The intersection of two Sylows can be non-trivial even if the $p$-core is trivial. For example in $A_{10}$, the Sylow 5-subgroups are each generated by 2 disjoint 5-cycles. The groups $\langle (1,2,3,4,5), (6,7,8,9,10)\rangle$ and $\langle (1,2,3,4,5), (6,7,8,10,9)\rangle$ intersect in a cyclic subgroup of order 5. But the 5-core is trivial, since $A_{10}$ is simple.

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  • $\begingroup$ Nice Answer, +1. :-) $\endgroup$ – user21436 Mar 20 '12 at 14:13
  • $\begingroup$ +1 Good to see an example, where the $p$-core is trivial. $\endgroup$ – Jyrki Lahtonen Mar 20 '12 at 20:07

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