# If $P_1 , P_2$ are two $p$-sylow subgroups, prove that $P_1 \bigcap$ $P_2$ = ${ 1 }$

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.

• – Cortizol Feb 25 '13 at 20:56
• 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. – Hagen von Eitzen Feb 25 '13 at 20:57
• Who set you this problem? You need more information, since the result as stated is false in general, as others have already said. – Geoff Robinson Feb 25 '13 at 20:59
• I am also wondering how you showed that $H$ must be a normal subgroup. Would you like to provide with some details? Thanks. – 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 $\{3,4,5,6\}$.
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.