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Group Theory, permutations formula

  1. Apr 12, 2008 #1
    When proving that [tex]A_n[/tex] with [tex]n \geq 5[/tex] is simple, we require the following lemma:

    If N is a normal subgroup of [tex]A_n[/tex] with [tex]n \geq 5[/tex] and N contains a 3-cycle, then [tex]N = A_n[/tex].

    The proof is actually given for us in the lecture notes, however he utilizes a formula that I'm not sure how he derived:

    Let [tex]f \in S_n[/tex] be a permutation and let [tex]x = (i_1 i_2 ... i_k)[/tex] be a k-cycle. Then:

    [tex] fxf^{-1} = f(i_1 i_2 ... i_k)f^{-1} = ( f(i_1) f(i_2) ... f(i_k) ) [/tex]

    where the right side is a cycle. I'm blind to how he gets from left to right.. and we were left with the ubiquitous "it is easy to check that..."

    Any help to get started on seeing how this is true would be appreciated. Thanks,
    Last edited: Apr 12, 2008
  2. jcsd
  3. Apr 12, 2008 #2


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    Hi Coto! :smile:

    For any number j:

    fxf^{-1}(f(i_j)) = fx(i_j) = f(i_j+1);

    and that is the definition of ( f(i_1) f(i_2) ... f(i_k) ), isn't it? :smile:
  4. Apr 12, 2008 #3
    Thanks tiny-tim. I must be having a brain-dead day, I'm still having troubles seeing this.

    Specifically I get lost in the reasoning from moving left to right. Could you elaborate between each equality, specifically why you've decided to add (f(i_j)) to fxf^{-1}, and why fx(i_j) = f(i_j+1) ?

    The definition of the cycle on the right would be all f(i_j) move to f(i_j+1) (j < k)? But I'm having troubles making that connection with your help given above.
  5. Apr 12, 2008 #4


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    Hi Coto! :smile:

    ok … I think what you're missing is the basic definition of f{-1}, which of course is that f{-1}f = I, the identity … so in particular f{-1}f (i_j) = (i_j);

    so fxf^{-1}(f(i_j)) = fx [ f^{-1}f (i_j)) ] = fx [ (i_j) ] = f [ x(i_j) ] = f(i_j+1). :smile:
  6. Apr 12, 2008 #5
    Thanks again for your time tiny-tim.

    Actually I understood where the derivation was coming from in terms of the inverse.. but I seem to be stuck more on why fxf^{-1}(f(i_j)) appears.

    1) We are looking for fxf^{-1} so why do we look at fxf^{-1}(f(i_j))? Why do we append the cycle (f(i_j)) to it?

    2) we have x(i_j) = (i_1 ... i_k)(i_j) = (i_j+1) ?? I guess I'm lost on this step too. I'm not sure why this is true. If we're considering (i_j) as a "cycle" of one element, then wouldn't x(i_j) = x?

    I suppose these may seem basic.. my mind is having troubles wrapping around this.

    Thanks again for the help. Sorry for being a bit dense right now.. I'm just not quite seeing it for some reason.
  7. Apr 12, 2008 #6


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    Because we want to prove that fxf^{-1} is the cycle that takes f(i_j) to f(i_j+1). :smile:

    AND (f(i_j)) is not a cycle … it's an element, just as i_j is.
    Ah … all is clear now … you're misreading the cycle notation. And you're confusing cycles with elements. AND there's no such thing as a "cycle" of one element … how would you define it?! :wink:

    If C = (a b c d …) is the notation for a cycle C, then that means that Ca = b, Cb = c, …

    So if C = (i_1 ... i_k), then that means C(i_1) = i_2, C(i_2) = i_3, … and generally C(i_j) = i_j+1.

    x is a permutation, just like f, not an element … x operates on the same elements as f does. :smile:
  8. Apr 12, 2008 #7
    Gotcha. I think that actually was my problem. I didn't realize this notation at all. It makes sense now what is going on.

    By mentioning (i_j).. it was sometimes used when writing a permutation as a product of disjoint cycles, i.e. (1345)(2) would be the cycle as 1>3>4>5>1 and the invariant 2>2. Superfluous given that we known when using cycle notation that all other elements remain the same, but it seems that it was the source of my confusion when interpreting x(i_j).

    Combine that with not understanding the meaning behind Ca = b and you have a confused student! ;)
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