# Finite group with two prime factors

1. Nov 19, 2012

### moont14263

1. The problem statement, all variables and given/known data
I am trying to prove the following:

Let $G$ be a finite group and let $\{p,q\}$ be the set of primes dividing the order of $G$. Show that $PQ=QP$ for any $P$ Sylow $p$-subgroup of $G$ and $Q$ Sylow $q$-subgroup of $G$. Deduce that $G=PQ$.

2. Relevant equations
The set $PQ=\{xy: x \in P \text{ and } y \in Q\}$

3. The attempt at a solution
I know that $PQ=QP$ means that I must prove $PQ$ is a subgroup of $G$. Let $r$, $s$ be elements of $PQ$. Then $r=x_{1}y_{1}$ and $s=x_{2}y_{2}$.

Now $rs^{-1}=x_{1}y_{1}(x_{2}y_{2})^{-1}=x_{1}y_{1}y_{2}^{-1}x_{2}^{-1}$ but I am not sure whether this is an element of $PQ$ or not.

The second part by the following:
$|G|=|P||Q|=\frac{|P||Q|}{|P \cap Q|}=|PQ|$ as $P$,$Q$ are Sylow for distinct primes of $G$ and $P \cap Q=\{1\}$. This imply that $G=PQ$ as $PQ$ is a subgroup of $G$ with the same order of $G$.

2. Nov 20, 2012

### jbunniii

What happens if you reverse the roles of P and Q?

3. Nov 20, 2012

### moont14263

In case $PQ$ is a subgroup of $G$, then $PQ=QP$.