The Intersection of a Sylow p-Subgroup and a Sylow q-Subgroup

Proposition 1: Let $G$ be a finite group. If $H$ is a Sylow $p$-subgroup of $G$, and $K$ is a Sylow $q$-subgroup of $G$ where $p$ and $q$ are primes with $p \neq q$ then $H \cap K = \{ e \}$.

• Proof: Let $G$ be a finite group of order $n$ and let $n = p^aq^bm$ where $p$ and $q$ are primes, $p \neq q$, and $p \nmid m$, $q \nmid m$. Let $H$ be a Sylow $p$-subgroup and let $K$ be a Sylow $q$-subgroup. Then $|H| = p^a$ and $|K| = q^b$.

• Now by definition, $H \cap K$ is a subgroup of $H$. By Lagrange's Theorem, the possible orders of $H \cap K$ are $1$, $p$, $p^2$, …, $p^a$.

• Similarly, $H \cap K$ is a subgroup of $K$. Again by Lagrange's Theorem, the possible orders of $H \cap K$ are $1$, $q$, $q^2$, …, $q^b$.

• Since $p \neq q$ we see that $p^s \neq p^t$ for all $1 \leq s \leq a$ and for all $1 \leq t \leq b$. Thus we must have that $|H \cap K| = 1$.

• So $H \cap K = \{ e \}$. $\blacksquare$