The order and signature of a k-cycle

[Etenim]

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

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

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

In (1.1) I could, of course, give a hand-wavy proof of how \sigma^k 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.

 

[Etenim]

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.

 

[Jösus]

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 M? 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 \sigma = (a_{1}, ... , a_{k}). By the definition of a cycle, for all i = 1, 2, ... k, we have \sigma(a_{i}) = a_{i + 1 (mod(k)}. Thus, \sigma^{l}(a_{i}) = a_{i + l (mod k)}. This reduces the problem to the problem of showing that min\{l \in \mathbb{Z}^{+} | \sigma^{l}(a_{i}) = a_{i}\} = k, which is equivalent to min\{l \in \mathbb{Z}^{+} | i + l \equiv i (mod k)\} = k, 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.