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The order and signature of a k-cycle

  1. Sep 23, 2009 #1
    Greetings,

    I am faced with a problem in Group Theory. It's not homework. I am trying to study it by myself. The statements are quite obvious, but I want to write the proofs (correctly) with more precision. Could you comment on it or suggest corrections, please?

    1. Let [tex]\sigma \in Sym_n[/tex] be a k-cycle.
    1.1. The order [tex]o( \sigma ) = k[/tex] (intuitively obvious, but I failed to prove it without resorting to prior results. It's likely my proof attempt is wrong, too.)
    1.2. [tex]sgn(\sigma)[/tex] = [tex](-1)^{k-1}[/tex]

    Proof: (1.1.) Let [tex]\sigma = (a_1 \, a_2 \, ... \, a_k)[/tex] be a k-cycle, [tex]a_i \in M \, \forall_i[/tex]. Since [tex]\left< \sigma \right> a_1 \, = \, \bar{a_1}[/tex], the (finite) equivalence class of [tex]a_1[/tex] under the equivalence relation a ~ b [tex]:\Leftrightarrow \, \exists_{m \in \mathbb{Z}} \,\, \sigma^m (a) = b[/tex]; [tex]a,b \in M[/tex] it is known that there exists a least positive integer [tex]k \in \mathbb{N}[/tex] of the property [tex]\sigma^k (a) = a \, \forall_{a \in M}[/tex]. Therefore [tex]o( \sigma)\, = \, k[/tex].

    (1.2.) Let [tex]\sigma = (a_1 \, a_2 \, ... \, a_k)[/tex] be a k-cycle, [tex]a_i \in M \, \forall_i[/tex]. The k-cycle [tex]\sigma = (a_1 \, a_2)(a_2 \, a_3)\,...\,(a_{k-1} \, a_k)[/tex] can be factored into k-1 transpositions. It follows immediately that [tex]sgn(\sigma)[/tex] = [tex](-1)^{k-1}[/tex], since [tex]sgn[/tex] is a homomorphism of groups and transpositions have odd parity.

    In (1.1) I could, of course, give a hand-wavy proof of how [tex]\sigma^k[/tex] passes on its argument internally, eventually resulting in the identity function, but that doesn't sound rigorous enough. I am not even sure whether my proofs work.

    Thanks a lot!

    Cheers,
    Etenim.
     
    Last edited: Sep 23, 2009
  2. jcsd
  3. Sep 27, 2009 #2
    Are those proof attempts so bad that they don't deserve a comment? :( Since I am studying this on my own, some input could be very helpful.
     
    Last edited: Sep 27, 2009
  4. Oct 10, 2009 #3
    I'm sorry, but I can't really make sense out of your proof. Not because I think it is incorrect, but I can't understand your notation. What is, for example [tex]M[/tex]? If I understand your notation correctly, you introduce an equivalence relation to show the existence of a least positive integer k with the required property. However, the existence does not seem to imply that it coincides with the k used to describe the length of the cycle. Or have I misunderstood?


    Anyway, wouldn't some slightly formalized handwaving proof of (1.1) do just well? For example, as we are working with a k-cycle, we let [tex]\sigma = (a_{1}, ... , a_{k})[/tex]. By the definition of a cycle, for all i = 1, 2, ... k, we have [tex]\sigma(a_{i}) = a_{i + 1 (mod(k)}[/tex]. Thus, [tex]\sigma^{l}(a_{i}) = a_{i + l (mod k)}[/tex]. This reduces the problem to the problem of showing that [tex]min\{l \in \mathbb{Z}^{+} | \sigma^{l}(a_{i}) = a_{i}\} = k[/tex], which is equivalent to [tex] min\{l \in \mathbb{Z}^{+} | i + l \equiv i (mod k)\} = k[/tex], which is true. In fact, if there would exist some integer smaller than k with that property, that integer would be congruent to zero mod k, which is not possible.

    Now I might have been to fast thinking this through, but I think that it holds. I hope I could be to some help.
     
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