I'm reading Beardon's, Algebra and Geometry. Here:

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I'm having a little trouble understanding the marked part. In the expression $\rho_j(x)=\rho(x)$, who is $\rho_j(x)$? Is it one cycle that is equal to the entire permutation? And in $\rho_i(x)=x$, is it one (trivial?) cycle that fixes each integer in $\rho_i$? I'm a little confused at:

  • If they are actually what I think they are;

  • Assuming they are, why are they used? I guess that the author wants to use $\rho_i$ to represent the identity of a particular cycle and $\rho_j$ to represent the entire permutation as one element of the factorization. Is that it?


For example $\rho$ is a permutation acting on a set of numbers. For example it maps the tuple $[1,2,\ldots 10]$ to $[ 6, 5, 7, 3, 8, 10, 2, 4, 9, 1 ]$. Then $\rho$ partitions the set $\{1,\ldots, 10\}$ into orbits $\{1, \rho(1), \rho^2(1)\} = \{1,6, 10\}$ etc. The author then associates each orbit with cycles $\rho_1, \rho_2,$ , etc. And now for the underlined part. For any element $x$ in the set there is a cycle $\rho_j$ that moves the point $x$ in the same way $\rho$ does whence $\rho_j(x) = \rho(x)$ while the other cycles $\rho_i$ with $i \neq j$ leave $x$ invarinant whence $\rho_i(x) = x$.



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