Talk:Equality of Cycles

Possibly better statement
Let $S_n$ denote the symmetric group on $n$ letters, realised as the permutations of $\left\{1,\ldots,n\right\}$.

Let $\rho = \begin{bmatrix} a_0 & \cdots & a_{k-1} \end{bmatrix}$, $\sigma = \begin{bmatrix} b_0  & \cdots & b_{k-1} \end{bmatrix} \in S_n$ be $k$-cycles of $S_n$.

Then there exists $j \in \{0,\ldots,k-1\}$ such that $a_0 = b_j$.

Moreover $\sigma = \rho$ iff we have for all $i = 0,\ldots,k-1$:
 * $a_i = b_{\operatorname{rem}(j+i)}$

where $\operatorname{rem}(j+i)$ is the remainder of the quotient of $j+i$ by $n$. --Linus44 (talk) 21:37, 25 October 2012 (UTC)


 * That is a lot clearer to me. But, take care to put up a page Definition:Remainder defining that concept. Good initiative though. --Lord_Farin (talk) 21:44, 25 October 2012 (UTC)


 * Too complicated, and relies on a series of statements which themselves require proofs.


 * Perhaps we need to refer to the "canonical definition" for a cycle, i.e. with its lowest element first. The concept is simple - we need a way to encapsulate that concept somewhere else, so as to say that two cycles are equal if, when expressed as ordered tuples, their canonical definitions are equal. --prime mover (talk) 21:44, 25 October 2012 (UTC)


 * Could make sense. First, IMO, it is required that the definition of $k$-cycle be separated from the mapping it represents; otherwise, we could just as well refer to the underlying mappings in place of a "canonical definition" (which is not but a way of assigning a cycle to a mapping in a bijective fashion). --Lord_Farin (talk) 21:50, 25 October 2012 (UTC)


 * The definition of a cycle in Definition:Cyclic Permutation is a good one. From this the 'canonical' representation of a cycle is obtained by taking the minimal possible $i$. Possibly the ambiguity of the representation could be mentioned on this page, and then the canonical representation linked. --Linus44 (talk) 22:00, 25 October 2012 (UTC)