Show that there exists $a$, $b$ in $G$ such that $|a| = p$ and $|b| = q$ Show that there exists $a$, $b$ in $G$ such that $|a| = p$ and $|b| = q$, where $G$ is a non-abelian group with $|G| = pq$ where $2 < p, q$ are distinct primes.
Is there a way to do this without Sylow or Cauchy Theorem? Using the conjugacy class equation $|G| = |Z(G)|+∑[G:N_G(x_j)]$?

Pick any element $g$ different from 1 and consider the cyclic subgroup $C$ it generates. In view of Lagrange's theorem this has order $p$ or $q,$ so the problem is solved for $a$ or for $b.$ Without loss of generality assume $|C|=p$ so we are done for $a.$
There are $q$ distinct left cosets of $C$ in $G.$
Pick any element $h$ outside $C$ and consider the cyclic subgroup $D$ it generates.
Either every element of $D$ is in a different left coset of $C$, or there are two different elements of $D$ in the same left coset of $C$.
In the first case the order of $D$ is $q$ and we are done for $b$ as well.
In the second case we have a nontrivial power of $h$ belonging to $C$ so $C\subset D.$ By Lagrange's theorem, the order of $D$ is a multiple of $p$ and a divisor of $pq$. But it cannot be equal to $p$ since $h\notin C,$ therefore $D=G$ and we can choose $b=h^q.$
