If $p|q-1$ with $p,q$ prime, then there exists a non-abelian group of order $pq$. I have some trouble in the proof of this theorem in Dummit & Foote (Page-142):

Problem. For any two prime p and q with $~p|q-1$, there exists a non-abelian group of order $pq$.

Solution. Let $Q$ be the Sylow $q$ subgroup of the symmetric group  $S_q$. Then we know, $|N_{S_q}(Q)|=q(q-1)$. By assumption, $~p|q-1$ so by Cauchy's theorem $N_{S_q}(Q)$ has a subgroup, $P$, of order $p$. Since $P \le N_{S_q}(Q)$, $PQ$ is a subgroup of $G$ of order $pq$. Now since $C_{S_q}(Q)=Q$, $PQ$ is a non- abelian group.
Now, I have two question here:

1.If $Q$ be the Sylow $q$ subgroup of the symmetric group  $S_q$ then $|N_{S_q}(Q)|=q(q-1)$.
2.How does the group $PQ$ become non-abelian?

First I try to solve (1):
I know, $(q-1)!=[S_q:Q]=[S_q:N_{S_q}(Q)][N_{S_q}(Q):Q]$. But I can't proceed further with this.....So I think in other direction. Any such $q$ Sylow subgroup, $Q$ is of order $q$, So must be generated by a $q$-cycle.Let $Q=<\sigma>$, $\sigma$ is a $q$-cycle. Let $\tau \in S_q$ such that $\tau \sigma \tau^{-1} \in Q=<\sigma>$. Then we have to calculate the total number of such $\tau \in S_q$ (Which is equal to $|N_{S_q}(Q)|$). Now any two Sylow $q$-subgroups is conjugate in $S_q$. Let $\sigma =(1 ~2~ 3\dots q)$ Then we have to find the total number of $\tau \in S_q$ such that $\tau \sigma \tau^{-1} =(\tau(1)~\tau(2)~ \dots \tau(q))$ is a power of $\sigma$. Here I again can't conclude further.....Please help..
Next for (2):
I know, since $Q$ is abelian so $Q \le C_{S_q}(Q) \le S_q$ also it can be checked that $C_{S_q}(Q) \ne S_q$ so $Q =C_{S_q}(Q)$ (since, $Q$ has prime index). But I can't understand how does $PQ$ become non-abelian?
Please help... Thank you .
 A: If $p \mid q-1$, then there is a non-trivial homomorphism from $C_p$ to $\operatorname{Aut}(C_q) = C_{q-1}$, which gives a non-trivial semidirect product, and they're all non-abelian.

What is it with DF giving a hard proof...


*

*We know that the Sylow $q$-subgroup of $S_q$ has order $q$, and the subgroup generated by $\sigma = (1~2~3 \cdots q)$ also has order $q$, so we can identify that as our Sylow $q$-subgroup. We know that $\tau \sigma \tau^{-1} = (\tau(1)~\tau(2)~\tau(3) \cdots \tau(q))$, and if we want that to equal $\sigma^n$ for some $n$ then $\tau$ is basically determined up to the order of the elements (i.e. $(123)$ vs $(231)$), so there is $q-1$ choices for $n$ (can't be the identity!) and for each choice there is $q$ ways to order the elements. To see what I mean, let $q=3$. If we want $\tau \sigma \tau^{-1} = \sigma^2$, i.e. $(\tau(1)~\tau(2)~\tau(3)) = (1~3~2)$, then either $\tau(1)=1; \tau(2)=3; \tau(3)=2$ or $\tau(1)=3; \tau(2)=2; \tau(3)=1$ or $\tau(1)=2; \tau(2)=1; \tau(3)=3$

*If $PQ$ were abelian, then we would have $C_{PQ}(Q) = PQ$. Recall that $PQ$ is a subgroup of $S_q$.
