1
$\begingroup$

Let $N$ a non-trivial normal subgroup of $A_n$ and $H = N \cap A_{n-1}$. I would like to show that $A_{n-1} \hookrightarrow A_n \to A_n/N$ is surjective, where $A_n \to A_n/N$ is the canonical homomorphism, and that the kernel of this composite map is precisely $H$. Any help will be greatly appreciated!

  • 0
    For $n>4$, $A_n$ has no non-trivial normal subgroups (other than itself), so then $A_n/N = A_n/A_n$. It would be trivially true for the second step mapping. So you just need to check cases for $n<5$.2012-12-03
  • 0
    Yes, but he is not allowed to use the simplicity of $A_n$. Perhaps that is what he is trying to prove?2012-12-03

2 Answers 2