Proof of G is a cyclic group with |G|=|pq|

Question

I have a question about the proof of following lemma:

Let $G$ be a group such that $Ord(G)=pq$ for some primes $p<q$ and $p$ does not divide $q-1$. Then G is cyclic.

In the proof they are saying, that when $n$ is the number of all Sylow $q$-subgroups and $m$ is the number of all Sylow $p$-subgroups, then $n$ divides $pq$ and $q$ divides $n-1$. Also $m$ divides $pq$ and $p$ divides $m-1$. Since $p<q$ we conclude that $n=1$. Also since $q$ does not divde $q-1$, $m=1$.

Well, I understand that $n$ and $m$ divides $pq$, but why $q$ divides $n-1$ and $p$ divides $m-1$? I don't get it. The rest is also unclear for me. Hope anyone can help.

Answer 1

The 3rd Sylow theorem states that

If a group $G$ has order $p^ns$, where $s$ is not divisible by $p$, the number $n_p$ of $p$-Sylow subgroups of $G$ satisfies:

1. $n_p\equiv 1\mod p$ ;
2. $n_p \mid s$.

So here $m$ divides $q$ and $n$ divides $p$, hence both divide $pq$. But this last point is unimportant. What matters is this:

$n$ is one of $1, 1+q, 1+2q,\dots$, and it must be a divisor of $p$. In the list only $1$ can divide $p$ since $p<qw1+q$. Similarly for $m$.

The consequence of this is that the Sylow-subgroups of $G$ are normal subgroups.

Answer 2

By Cauchy exists an element $x$ of order $q,$ then $H=\langle x \rangle$ is a normal subgroup of $G$ because $[G:H]=p$ which is the smallest prime who divide $|G|.$ Too, exists an element of order $p$ and let $K$ its generated subgroup on $G.$ Since $(p,q)=1,$ we have $|G|=|HK|,$ then $G$ is a semi-direct product between $H$ and $K,$ but if $\phi:K\to \operatorname{Aut}(H),$ since $H\cong \mathbb{Z}_q$ and $K\cong \mathbb{Z}_p$ and $|\operatorname{Aut}(\mathbb{Z}_q)|= |(\mathbb{Z}_q)^*|=q-1$ and by hypothesis $p$ does not divide $q−1$, $\phi$ has to be the trivial morphism so $G\cong \mathbb{Z}_p\times \mathbb{Z}_q\cong \mathbb{Z}_{pq} ,$ so is cyclic.