# Can $A_n$ be generated by two elements $x,y$ with $x\ne 1$ arbitrarily chosen?

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.

• Not an answer, but a quick check on GAP says that it's true for $n \leq 8$ as well. – Carl-Fredrik Nyberg Brodda May 27 at 5:40
• This is actually true in a stronger form for all finite simple groups, see this nice survey. – Andreas Caranti May 27 at 7:31
• Really helps a lot! Thank you.@AndreasCaranti – Hongyi Huang May 27 at 8:49