$G$ a finite group with two non-trivial normal subgroups. $|G| = pq$. Why is $G$ cyclic?

Question

How can I prove that if $G$ is a finite group , and the order of $G$ is $pq$ while $p$ and $q$ are primes, and in addition , $G$ with two normalic subgroups , so --> $G$ is cycle? Ideas? Hwo can i show that have just one subgroup for any divisior of pq?

Answer 1

By Sylow theory the two subgroups are then the only non-trivial subgroups of $G$.

Hint: Show that the must exist an element that does not belong to either of those subgroups. What is the subgroup it generates?

Answer 2

Let $P, Q$ the subgroups with order $p, q$. As @Jyrki indicated earlier these are the only nontrivial subgroups of G. Let $a \in P, b \in Q$ have orders $p, q$. Consider the element $ab$. Since $p, q$ are relatively prime $a$ is not a power of $b$ and vice versa. Then $ab$ is not in $P, Q$. Since $ab$ is an element of $G$, the order of $ab$ must divide the order of $G$. Then the order of $ab$ is in $\{1, p, q, pq\}$. As $a, b$ are in different groups they are not inverses so it is not $1$. $ab$ is not in $P$ or $Q$ so $ab$ does not have order $p$ or $q$ and therefore must have order $pq$. Then $G$ is generated by $ab$. Then $G$ is cyclic.