I want to prove that $A/B$ is cyclic when $B$ is non-trivial normal subgroup of $A$, and $|A|$ = $pq$ such that $p$ and $q$ are primes.
I had the idea of using Lagrange's Theorem which states that 'the order of any subgroup of group $A$ is a divisor of the order of $A$', along with the idea that $A$ must contain an element '$a$' of order $q$; then prove that $A/B$ is cyclic since the '$a$' of order $q$ could be the left side generator of the set of (left) cosets of $B$ in $A$ (thus using the trivial normal subgroup $B$ to form a cyclic quotient group $A/B$)...
but I am coming up empty when trying to prove this directly...
If $B$ is a non-trivial subgroup of $A$, then $|B| \in \{p,q,pq\}$. In any case, $|A/B| = |A|/|B| \in \{q,p,1\}$, and recall that, any group of prime order is cyclic.