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.

  • $\begingroup$ Not an answer, but a quick check on GAP says that it's true for $n \leq 8$ as well. $\endgroup$ – Carl-Fredrik Nyberg Brodda May 27 at 5:40
  • 4
    $\begingroup$ This is actually true in a stronger form for all finite simple groups, see this nice survey. $\endgroup$ – Andreas Caranti May 27 at 7:31
  • $\begingroup$ Really helps a lot! Thank you.@AndreasCaranti $\endgroup$ – Hongyi Huang May 27 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.