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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.

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# The order and signature of a k-cycle

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