# Embedding Sn into An+1

NOT HOMEWORK.

I know how to embed Sn into An+2, just with the extra transposition, etc...

But how to show it can not be embedded into An+1. We don't have the extra transposition.

Using Lagrange's Theorem, we can say when n+1 is odd, then n! does not divide (n+1)!/2 therefore a subgroup of order n! can not exist in An+1.

However, we must also not use Lagrange's Theorem.

Any help?

Thank you.

Deveno
suppose we had such an embedding.

consider the action of An+1 on cosets of Sn.

this gives us a homomorphism φ:An+1→Sm,

where m = [An+1:Sn] = (n+1)!/(n!2) = (n+1)/2.

note that this forces n to be odd.

if n > 4, then the simplicity of An+1 forces (n+1)!/2 to be a divisor of m!

writing k = (n+1)/2, we have that:

(2k)! ≤ 2(k!), which is untrue if k ≥ 2, thus when n ≥ 3, and a fortiori when n > 4.

so it suffices to prove that S3 cannot be embedded into A4.

(note the theorem is FALSE for n = 1, as S1 is trivial).