# 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

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?

#### Solutions Collecting From Web of "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"

Since $N$ and $K$ are normal, $NK$ is normal. Also $p,q$ are prime, so $N\cap K=1$. Moreover $N$ and $K$ are commute. 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.

$N$ 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.