RIGHT ANSWER BY DUMMIT: Here
Suppose $|G|=pq$ for primes $p$ and $q$ with $p<q$. Let $P \in Syl_{p}(G)$ and let $Q \in Syl_{q}(G)$. We are going to show that $Q$ is normal in $G$ and if $P$ also normal in $G$, then $G$ is cyclic.
Now the three conditions: $n_q=1+kq$ for some $k\geq 0$, $n_q$ divides $p$ and $p<q$, together force $k=0$. Since $n_q=1$, $Q\trianglelefteq G$.
Since $n_p$ divides the prime $q$, the only possibilities are $n_p=1$ or $q$. In particular, if $p\nmid (q-1)$, then $n_p$ cannot equal $q$, so $P\trianglelefteq G$.
Let $P=\langle x \rangle$ and $Q=\langle y \rangle$. If $P\trianglelefteq G$, then since $G/C_{G}(P)$ is isomorphic to a subgroup of $Aut(\mathbb{Z}_p)$ and the latter group has order $p-1$, Lagrange's Theorem together with the observation that neither $p$ or $q$ can divide $p-1$ implies that $G=C_{G}(P)$. In this case $x \in P \leq Z(G)$ so $x$ and $y$ commute. ($G\cong \mathbb{Z}_{pq}$).
If $p| (q-1)$, we shall see in other chapter that there is a unique non-abelian group of order $pq$ (in which, necessarily, $n_p=q$). We can prove the existence of this group now. Let $Q$ be a Sylow $q-\operatorname{subgroup}$ of the symmetric group of degree $q$ in $S_q$. By Exercise 34 in section 3, ${\color{red}{|N_{S_q}(Q)|=q(q-1)}}$. By assumption, $p|(q-1)$ so by Cauchy's Theorem $N_{S_q}(Q)$ has a subgroup, $P$, of order $p$. By Corollary 15 in Section 3.2, $PQ$ is a group of order $pq$. Since ${\color{red}{C_{S_q}(Q)=(Q)}}$, $PQ$ is a non-abelian group.
MY QUESTIONS ARE IN COLOR ${\color{red}{\text{RED}}}$.
Can someone provide a step by step answer for the following questions, please?
- ${\color{red}{|N_{S_q}(Q)|=q(q-1)}}$ I've done the exercise 34 suggested. However I still not understanding this equality, how to calculate this?
- ${\color{red}{C_{S_q}(Q)=(Q)}}$ Although I've tried to understand the corollary mentioned I didn't understand this equality in this case!
Dummit has evidently broken it down for you step by step. Each of your questions in red is answered with an exercise or example.
To summarize, if $p\not|q-1$, then the group is the product of its Sylow subgroups, and is cyclic, isomorphic to $\Bbb Z_{pq}$.
If $p|q-1$, there exists a unique nonabelian group of order $pq$. Note that $C_{S_q}(Q)=Q$ implies that $G$ is nonabelian. Uniqueness is postponed for the moment.