On generation of $A_n$ by elements of prime order

Question

There is a question regarding generation of finite simple groups with elements of prime order. Recently, Guralnick, Shareshian, Woodroofe and Teräväinen made advances in this direction. We have, for example, the following result:

Theorem. Assume that $n>7$. Let $p$ be a prime satisfying $n/2 < p < n-2$ and let $y \in A_n$ have order $p$. If $n$ is odd let $x \in A_n$ be an $n$-cycle. If $n$ is even let $x \in A_n$ be the product of two disjoint $\frac{n}{2}$-cycles. Then $x$ and $y$ generate $A_n$ invariably.

If we choose $n$ to be a prime, I think using the Prime Number Theorem we can get infinitely many pairs of primes $(n,p)$ satisfying the hypothesis of the Theorem so that we have infinitely many alternating groups generated by elements of prime order.

Now, if we have an infinite set of primes not necessarily having all primes this is not true. I mean, I think we can construct an infinite set of primes having no pairs satisfying the hypothesis of the Theorem.

Question. Given arbitrarily an infinite set of primes $\mathcal{P}$, can we choose some elements $x,y$ of orders $p,q \in \mathcal{P}$ such that $x$ and $y$ generates $A_n$ for some $n$?

Answer 1

For any two odd primes $p\le q$ there exist $x,y\in A_q$ of order $p,q$ respectively such that $A_q=\langle x,y\rangle$. First assume $p<q$ and take $x=(12\cdots p),y=(12\cdots q)$. Then $yxy^{-1}x^{-1}$ is a 3-cycle and by Jordan's theorem $\langle x,y\rangle$ is all of $A_q$. For $p=q$ take $x=(12\cdots p),\,y=(2134\cdots p)$ and note that $xy^{-1}$ is a 3-cycle.