I was working on a problem asking to show that a group of order $132=2^2*3*11$ is not simple. The solution basically involves counting up elements and deciding how many are in a Sylow $2$-subgroup, $3$-subgroup, or $11$-subgroup. In the solution, we use the fact that the intersection between all of these subgroups is trivial. Is this always the case? Under what circumstances is the intersection of two Sylow $p$-subgroups trivial?
[Math] When is the intersection between two Sylow subgroups trivial
abstract-algebragroup-theorysylow-theory
Best Answer
Let $p$ be a fixed prime dividing the order of $G$. Suppose $H_1,H_2,\dots,H_n$ are the $p$-Sylow subgroups of $G$. Then $H=H_1\cap H_2\cap\dots\cap H_n$ is normal in $G$, because $p$-Sylow subgroups are conjugate of each other. If $H$ is not trivial, then $G$ is not simple.
So, in order to show $G$ is not simple, you can assume that, for each $p$ dividing $|G|$, the $p$-Sylow subgroups intersect trivially. Note that, in general, the intersection of all the $p$-Sylow subgroups is not trivial: just consider a $p$-group or, more generally, a group having a unique $p$-Sylow subgroup.
Of course the intersection of a $p$-Sylow and a $q$-Sylow subgroups is trivial when $p$ and $q$ are distinct.