Let $G$ group of order $pq$ where $p,q$ primes.Show that if $G$ contains normal groups $N$ and $K$ with $|N|=p$ and $|K|=q$ then is cyclic

Question

Let $G$ group of order $pq$ where $p,q$ primes.Show that if $G$ contains normal groups $N$ and $K$ with $|N|=p$ and $|K|=q$ then is cyclic

Any ideas or hints for showing this?

Answer 1

$N$ and $K$ are certainly cyclic. Take generators $n$ and $k$, respectively. Now we know that $nk\in nK=Kn$, so $nk=k^rn$ for some $0\le r\le q-1$. But also $nk\in Nk=kN$ so $nk=kn^s$ for some $0\le s\le p-1$. Then $$k^rn=kn^s$$ that is $$k^{r-1}=n^{s-1}$$ Since $n$ and $k$ were generators, this implies that both sides of the latter equation are the identity and thus, $r=s=1$. So, $G$ is abelian.

We must add the condition $p\neq q$ to guarantee that $G$ is cyclic.

Answer 2

Since $N$ and $K$ are normal, $NK$ is normal. Also $p,q$ are prime, so $N\cap K=1$. Since $NK\subset G$, by $$ |NK|=\frac{|N||K|}{|N\cap K|}=pq=|G| $$ we have $NK=G$. So $G\cong N\times K$. Since $|N|$ and $|K|$ are prime, $N$ and $K$ are cyclic, and thus $G$ is cyclic.

Answer 3

Firstly, if $p=q$ then $G$ is abelian but not necessarily cyclic (think of $\Bbb Z_p\times\Bbb Z_p$). So I guess you are implicitly assuming $p\ne q$. In this case, if $|G|=pq$ then $G$ is centerless or abelian. In fact, assume $|Z(G)|=p$; then, all the noncentral elements have centralizer of order $p$, and hence the class equation yields: $$pq=p+kq$$ for some positive integer $k$: contradiction, because $q\nmid p$. Likewise $|Z(G)|\ne q$. But if $G$ is centerless, then the class equation yields: $$pq=1+kq+lp$$ where both $k,l$ are positive integers (as $p,q\nmid 1$): contradiction, because then there's no way to build up a normal subgroup of order $p$ or $q$ as union of conjugacy classes, one of which must be the singleton (containing the identity). Therefore $G$ is abelian and, given some $x,y\in G$ of order $p$ and $q$ respectively (Cauchy), the element $xy$ has order $pq$, namely $G$ is cyclic.

(Incidentally, this shows that the claim holds also with the weaker condition that $G$ has either a normal subgroup of order $p$ or some of order $q$.)