We have this theorem
Theorem. Let $x$ be any nontrivial element of the symmetric group $S_n$. If $n\ne 4$, then there exists an element $y\in S_n$ such that $S_n = \langle x,y\rangle$.
My question: is the similar statement of $A_n$ valid? That is, if we pick an arbitrary nontrivial element $x$ of $A_n$, can we always find a corresponding $y\in A_n$ such that $\langle x,y\rangle = A_n$?
I checked $A_4$, the statement is also valid for $n=4$. I think it is true but don't know how to prove it. It seems not to be a corollary of the above theorem.
One proof of the above theorem uses Jordan's theorem:
If $G$ is a primitive subgroup of $S_n$ which contains a $p$-cycle with $p\le n-3$ be a prime, then $G = A_n$ or $G = S_n$.
Is this useful for the proof of $A_n$?
Thank you in advance.