Structure of a group, $G$, of order $pq$ where $p, q$ are prime. [duplicate]

There is a proposition in Beachy and Blair's Abstract Algebra that I don't entirely follow. The proposition is the following:

Let $G$ be a group of order $pq$, where $p > q$ are primes.

a) If $q$ is not a divisor of $p-1$, the $G$ is cyclic.

b) If $q$ is a divisor of $p-1$, then either $G$ is cyclic or else $G$ is generated by two elements $a$ and $b$ satisfying the following equations: $$a^p = e, \\ b^q = e,\\ ba = a^nb $$ where $n \not \equiv 1 \ (mod \ p)$ but $n^q \equiv 1 \ (mod \ p)$.

Can one of you prove how this is true? I understand a similar proof for when $q=2$, but this one is more complicated.

Solution 1:

Let $H$ be a subgroup of $G$ with order $p$ and $K$ be a subgroup of order $q$,

It is easy to see that $H$ must be normal in $G(*)$

Then $K$ act on $H$ by conjugation so we have $\phi:K\to Aut(H)\cong \mathbb Z_{p-1}$.

a) If $q\nmid p-1$ then $\phi$ is trivial homomorphism which implies $G=HK\cong H\times K$ which conclude that $G$ is cylic.

b) if $q\mid p-1$ then $G$ need not to be cyclic but clearly $G=\left<a,b\right>$ where $H=\left<a\right>$ and $K=\left<b\right>$ as $G=HK$.

And $bab^{-1}\in H$ since $H$ is normal and since it is cyclic then $bab^{-1}=a^k$ for some $k$. I think you can get from this point.