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!
Composite group homomorphism between alternating groups
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abstract-algebra
group-theory
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0For $n>4$, $A_n$ 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
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0Yes, but he is not allowed to use the simplicity of $A_n$. Perhaps that is what he is trying to prove? – 2012-12-03