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 $|G| = |Z(G)|+∑[G:N_G(x_j)]$?
 A: 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 $b=h^q.$
A: Yes, the class equation is the right tool, but first you have to rule out the options $|Z(G)|=p,q$. In fact, suppose $|Z(G)|=p$; then there is some $x\in G\setminus Z(G)$, and hence  $Z(G)<C_G(x)<G$; but then (Lagrange) $p\mid |C_G(x)|$ and $|C_G(x)|\mid pq$; namely, $|C_G(x)|=kp$ for some integer $k>1$, and $kp\mid pq$; so, $k\mid q$ and hence ($k>1$) $k=q$. Therefore $|C_G(x)|=pq$, namely $C_G(x)=G$: contradiction. Exactly the same argument leads to $|Z(G)|\ne q$. Therefore, your non-abelian $G$ must have trivial center, and every non-trivial element must have centralizer of order $p$ or $q$. It is not possible that all such elements have centralizer of order $p$, though, because then the class equation would yield $pq=1+kq$ (contradiction, as $q\nmid 1$). Likewise, it is not possible that all such elements have centralizer of order $q$, because then the class equation would yield $pq=1+lp$ (contradiction, as $p\nmid 1$). Therefore, there must be centralizers of order $p$ and centralizers of order $q$. But a group of prime order is generated by any of its nontrivial elements, so there are elements of order $p$ and elements of order $q$. (Moreover, the assumption $p,q>2$ didn't enter the argument, and hence the claim holds also in case one of the two primes is $2$.)
