Let $G$ be a group with $\#G=p \cdot q$ for $p,q \in \mathbb{P}$.
W.l.o.g. $p>q$. According to Sylow theorems we have one Sylow p-subgroup, let's call it $P$. Let $Q$ be a Sylow q-subgroup.
How can I prove that $\#P=p$ and $\#Q=q$?
Generally speaking, should not $\#P=p^n$ and $\#Q=q^m$ for $n,m \in \mathbb{N}$ be also possible?
Thx
By definition, if $p$ is a prime dividing $|G|$, then a Sylow $p$-subgroup of $G$ is of order $p^n$ where $|G|=p^nk$ such that $(p,k)=1$. In other words, the order of a Sylow $p$-subgroup is the highest power of $p$ in the order of $G$.
So since $|G|=pq$ where $p,q$ are distinct primes, a Sylow $p$-subgroup of $G$ must be of order $p$.