Suppose that $|G| = pq$ where $p$ and $q$ are distinct primes such that $p$ does not divide $q-1$.
Prove that G has a normal Sylow $p$-subgroup
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I know what by Sylow's Theorem, either $n_p=1$ or $n_p=q$.
I am stuck on how to proceed here, can anyone give me a hint?
The 3rd Sylow theorem also tells you that $n_p\equiv 1\bmod p$. Thus $n_p\neq q$, since $q\not\equiv 1\bmod p$. So the only possibility for $n_p$ is it's equal to $1$, which means the $p$-Sylow subgroup is normal, since all $p$-Sylow subgroups are conjugate to each other, by the 2nd Sylow theorem.
This can be proved without an appeal to the Sylow theorems.
WLOG, assume $p \lt q$. By Cauchy's theorem, $G$ has an element of order $q$, let $Q$ be the subgroup generated by this element. Hence $|Q|=q$. Observe that $Q$ has to be normal. For if $Q^*$ is a conjugate of $Q$ and $Q \neq Q^*$, then $Q \cap Q^*=1$, since $q$ is prime. But then $|QQ^*|=\frac{|Q||Q^*|}{|Q \cap Q^*|}=q^2 \gt |G|$, which is absurd.
Since $Q$ is abelian, we have $Q \subseteq C_G(Q) \subseteq N_G(Q)=G$. But $|G|=pq$, so $|G:C_G(Q)|$ equals $1$ or $p$. In the latter case, we have $Q=C_G(Q)$, and we apply the $N/C$ theorem: $G/Q$ embeds homomorphically in $Aut(Q) \cong C_{q-1}$. However, this obstructs $p$ not dividing $q-1$.
We conclude that $G=C_G(Q)$, which is equivalent to $Q \subseteq Z(G)$, the center of $G$.
Finally, again by Cauchy's theorem we can find a subgroup $P$ of order $p$. But then $|PQ|=\frac{|P||Q|}{|P \cap Q|}=pq$, so $G=PQ$, and since $Q$ is central, $P$ is certainly normal in $G$.
Note: in fact we proved that $G$ is cyclic of order $pq$.
I'm assuming that $n_p$ is the index of the normalizer of the $p$-Sylow subgroup. If $n_p=q$,then there are $q$ $p$-Sylow subgroups. But this implies $q\equiv 1\pmod p$.
