Let $G$ be a finite group, $|G|=p^n m$. Artin is using the following nice trick to show existence of a Sylow $p$-subgroup of $G$. Let $S$ be the set of all subsets of $G$ of cardinality $p^n$. Then, $G$ acts on $S$ and $|S|= \binom{p^n m}{p^n}$ is not divisible by $p$. Then there is an orbit of order not divisible by $p$, say orbit $GU$. So $$ |Stab(U)| |GU|=|G|=p^n m. $$ Then he claims $|Stab(U)|=p^n$ and so $Stab(U)$ is a Sylow $p$-subgroup.
I don't understand why we could not have $m=m_1 m_2$, $m_1 >1$ and $m_2 >1$ and $|Stab(U)|=p^n m_1$?