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Embedding Sn into An+1

  1. Nov 23, 2011 #1

    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.
  2. jcsd
  3. Nov 23, 2011 #2


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    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).
  4. Nov 23, 2011 #3
    Good answer!

    However, no knowledge of cosets is assumed.
  5. Nov 28, 2011 #4
    If anyone has any tips, it would be very much appreciated.
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