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let $G$ be a group of order divisible by $p,q$. for any $p$-sylow subgroup $P$ and $q$-sylow subgroup $Q$. I know the intersection $P\cap Q = \{e\}$ because the orders are different. what about two $p$-sylow subgroups, can I say their intersection is only $e$?

thanks. benny.

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    It is not enough for the orders of two subgroups to be "different" in order to conclude that their intersection is trivial. You need to use the fact that their orders are *relatively prime*.2012-02-21

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Firstly, having different orders does not guarantee trivial intersection. For instance in a cyclic group of order $4$, we subgroup of order $2$ and a subgroup of order $4$ will intersect in a subgroup of $2$. What guarantees trivial intersection is coprime orders.

Will two Sylow $p$ subgroups intersect trivially always?

Definitely not. There are plenty of examples, where intersection of two Sylow $p$-Subgroups is not identity.

Firstly, two Sylow $p$-Subgroups intesect trivially if the order of the Sylow is prime. That is, the maximal exponent of $p$ in $|G|$ is $1$. The proof of this fact follows from Lagrange's Theorem. However, I give no guarantee about the converse.

Another related question on when two Sylow Subgroups have trivial intersection is here.

However, you can still say something about the index of intersection of Sylow Subgroups. For instance, look here. Related is a fact called Poincare's Theorem which is stated in many forms. One of them is:

Let $G$ be a finite simple group. Let $H$ be a proper subgroup of $G$. Then the order of $G$ divides $[G:H]!$