Show that the normalizer of a Sylow $p$-subgroup contains a nonabelian group of order $pq$ I am stuck with the following question: let $ P $ be a Sylow $p$-subgroup of the alternating group $ A_p $ where $ p $ is an odd prime. Show that for an odd prime $ q \mid p-1 $, $ N_{A_p}(P) $ contains a nonabelian group of order $pq$.
I'm not sure how to even get started. I know the various orders of $ P $, $ N_{A_p}(P)$ and also the index of $ N_{A_p}(P) $ in $ A_p $ (the number of Sylow $p$-subgroups), but how do I show that it has a group of order $ pq $?
 A: When $q\mid (p-1)$, there are only two groups of order $pq$: the cyclic group and a nonabelian group generated by the relation:
$$x^p = 1, y^q = 1, yxy^{-1} = x^i$$
where $i$ is a nontrivial solution to $i^q \equiv 1 \pmod{p}$.
Any sylow $p$-subgroup of $A_p$ is cyclic. Hence it suffices to prove the following: 


Let $i$ be a nontrivial solution to $i^q \equiv 1 \pmod{p}$, and let $x=(1\quad 2\quad \cdots \quad p)$. There exists a permutation $y\in A_p$ of order $q$ such that $yx = x^iy.$ 

We will explicitly construct such an $y$. We will intrepret the indices as numbers modolo $p$. Note that $\langle i \rangle$ has order $q$ in the multiplicative group modulo $p$, denotes its cosets as $\langle i \rangle, a_1\langle i \rangle, \cdots, a_r\langle i \rangle$. We form the permutation $$y=(i\quad i^2\quad \cdots \quad i^q)(a_1i\quad a_1i^2\quad \cdots \quad a_1i^q)\cdots(a_ri\quad a_ri^2\quad \cdots \quad a_ri^q)$$
The only number does not appear in the above product is $0$ modulo $p$.
Then $y$ satisfies the condition $yx = x^i y$. This is easy to see, once we note that the action of $x$ on a number $u\in \mathbb{Z}/p\mathbb{Z}$ is to increase it by $1$, while the action of $y$ on $u$ is to multiply it by $i$. Hence $$yx(u) = y(u+1) = i(u+1)$$ and $$x^iy(u) = x^i(iu) = iu+i$$ Therefore they are indeed equal. The proof is completed.
