Let G be a group with $o(G)=pq$, $p>q$ are primes.
Prove that $G$ has a subgroup of order $p$ and a subgroup of order $q$.
Remark: I cannot find the topic with this problem.
I have tried different methods but no one of them did not bring result. Please show how to do that. Do not use Sylow's Theorem.
The only way this can fail is if every element other than the identity has order $p$ or has order $q$. I'll assume order $p$; the other case is similar.
Let $x$ be a non-identity element. Its centraliser $C$ has order divisible by $p$, so either $C=G$ or $C=\left<x\right>$. In the first case $\left<x\right>$ is normal in $G$, and as $G/\left<x\right>$ has order $q$, then $y^p$ has order $q$ for any $y\notin\left<x\right>$.
In the second case $x$ has precisely $q$ conjugates. We can assume this for all $x\ne e$. Then the number of non-identity elements is both a multiple of $q$ and equal to $pq-1$.