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Normal groups of permutation

  1. May 26, 2012 #1
    good day! i need to prove that the alternating group An is a normal subgroup of symmetric group, Sn, and i just want to know if my proving is correct.

    we know that normal subgroup is subgroup where the right and left cosets coincides. but i got this equivalent definition of normal group from fraleigh's book, which states that for all gεG and hεH, a subgroup H of G is normal iff gHg-1=H.

    now here's my proof using the definition i got,

    I. I need to show that for all τεSn, τAnτ-1 is subset of An.

    Let λετAnτ-1, then λ=τστ-1, for all σεAn. but since multiplication of transpositions are commutative and therefore,

    λ=τστ-1=σττ-1=σ, thus, λεAn, and therefore τAnτ-1 is a subset of An.

    II. I need to prove that An is a subset of τAnτ-1.

    Let σεAn and τεSn, since σ is an even transposition, τσ must be an odd transposition since no permutation is a product of both even or odd transposition. Also, since multiplication of transposition is commutative I now have,


    thus, An is a subset of τAnτ-1.

    since I've shown that the two sets are subsets of each other, I therefore conclude that τAnτ-1=An and An is normal.

    thanks and God bless!
  2. jcsd
  3. May 26, 2012 #2
    looks hard to read. your grader isn't looking forward to digging through that either.

    maybe there's a slicker approach.

    isn't tA the same size as At? and there's only two cosets, since A is half of S. so either tA=At, or they are disjoint. but e is in A, so tA intersects At.

    i used facts about size of cosets being same, A being half of S, and cosets are always disjoint or equal, so if you don't have these facts or close to them, then maybe you're on the right track.
    Last edited: May 26, 2012
  4. May 26, 2012 #3
    Huh, since when??? And why can you limit your proof to transpositions??

    Maybe another approach will help. Can you prove that An is the kernel of a suitable homomorphism??
  5. May 26, 2012 #4


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    As micromass says, you cannot assume [itex]\sigma[/itex] and [itex]\tau[/itex] commute.
    I also note that you have nowhere used any property of An other than the fact that it is a subgroup. You need to start with some property of An that subgroups in general don't have. algebrat's approach of using the fact that An is half of Sn does work, though I would word it differently. Left cosets (tH) are not necessarily disjoint from or identical to right cosets (Hu) but they are disjoint from (or identical to) each other. Since all cosets have the same size, either tA = A or tA = S-A. Likewise At = A or At = S-A. Therefore either tA = At or tA = S-At. Since t is an element of both, tA = At.
    The 'disjoint or equal' property of left cosets is easily proved. Suppose xH and yH have a common element, xh = ym. Let xk ε xH. So ymh-1k = xk. mh-1k ε H, therefore xk ε yH.
    That they all have the same size is also easy.

    Btw, this wording is a bit sloppy:
    Better is
    a subgroup H of G is normal in G iff (for all gεG gHg-1=H).
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