Embedding Sn into An+1 and the Limitations: A Closer Look

[Coca_Cola]

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 A_n+1 on cosets of S_n.

this gives us a homomorphism φ:A_n+1→S_m,

where m = [A_n+1:S_n] = (n+1)!/(n!2) = (n+1)/2.

note that this forces n to be odd.

if n > 4, then the simplicity of A_n+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).

 

[Coca_Cola]

Good answer!

However, no knowledge of cosets is assumed.

 

[Coca_Cola]

If anyone has any tips, it would be very much appreciated.