If $P_1 , P_2 $ are two sylow $p$-subgroups of the group $G$ prove that:

$ P_1 \bigcap $ $P_2$ = $ { 1 } $

I tried to prove it by induction as follows: proved it when $P_1 , P_2$ have the order p for some prime p then supposed it is true when the sylow p-subgroup has the order $p^n$ and supposed that there is some element in the intersection , made $H$ = the subgroup generated by this element " say , x "

I proved that H is normal subgroup of $P_1$ , $P_2$ , and made the factor group, $P_1$ mod $H $ = $Q_1$ and $P_2$ mod $H$ = $Q_2$

So by the induction, if $h$ $\in$ the intersection of $Q_1 , Q_2$ then $Q_1$= $Q_2$

But, I couldn't determine the element which is in this intersection--I don't know if this element $h$ must be exist or not -

I don't know what is the next step now; I need some hints to prove this statement:

I found that the text - dummit and foote - use the fact that the intersection of two sylow p-subroup is the identity element, but it didn't prove this fact so I look for a proof.

  • $\begingroup$ See here: math.stackexchange.com/questions/111730/… $\endgroup$ – Cortizol Feb 25 '13 at 20:56
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    $\begingroup$ The claim is false (not only because the condition $P_1\ne P_2$ is missing). The $2$-Sylow subgroups of $S_4$ have order $8$ and there are $16$ elements of power-of-two order. If there are $k$ Sylow groups and these have pairwise trivial intersection, they cover $7k+1\ne 16$ elements - contradiction. $\endgroup$ – Hagen von Eitzen Feb 25 '13 at 20:57
  • $\begingroup$ Who set you this problem? You need more information, since the result as stated is false in general, as others have already said. $\endgroup$ – Geoff Robinson Feb 25 '13 at 20:59
  • $\begingroup$ I am also wondering how you showed that $H$ must be a normal subgroup. Would you like to provide with some details? Thanks. $\endgroup$ – awllower Feb 26 '13 at 5:02

This is not true in general. For example the group $S_6$ has the following two Sylow 2-subgroups. Let $P_1$ be generated by $(12),(13)(24)$ and $(56)$ (the first two permutations generate a Sylow 2-subgroup of $S_4$). Let $P_2$ be generated by $(12),(35)(46)$ and $(56)$ (here we have a Sylow 2-subgroup of $S_4$ in a diguise that $S_4$ is now acting on the set $\{3,4,5,6\}$). Obviously $P_1$ and $P_2$ intersect non-trivially as they share two generators. Yet they are not the same subgroup as the orbits of $P_1$ are $\{1,2,3,4\}$ and $\{5,6\}$ whereas the orbits of $P_2$ are $\{1,2\}$ and $\{3,4,5,6\}$.

  • $\begingroup$ There are smaller examples, and looks like this will be merged with another question soon. Switching to CW. All the interested people are advised to study the linked question and Hagen's comment under the OP. $\endgroup$ – Jyrki Lahtonen Feb 25 '13 at 21:02
  • $\begingroup$ thanks for you correction :) $\endgroup$ – Fawzy Hegab Feb 26 '13 at 0:39

Although, as pointed out above, the statement is not true in general, you might wonder for which groups it is true indeed. Your question can be reformulated as which p-groups appear as non-normal T.I. Sylow p-subgroups? Here T.I. stands for "trivial intersection". A trivial intersection set is one that intersects each of its conjugates fully or trivially.

These have been extensively studied, since groups that have T.I. sets exhibit some interesting representation theoretic behavior (see e.g. Chapter 7 of Isaacs's famous book Character Theory of Finite Groups). See further Jack Schmidt's post of September 2011 and the discussion following this.

Another somewhat more specialized angle at this is asking the question which p-groups can be realized as a Frobenius complement: $G$ is a Frobenius group if and only if $G$ has a proper, non-identity subgroup $H$ ("the Frobenius complement") such that $H \cap H^g = 1$ for every $g \in G − H$. It can be proved that if $H$ is a Sylow p-subgroup of $G$, it must be cyclic or generalized quaternion.


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