This means that the order of $(4,5)$ is $2$. Hence the final answer is $6$.Īddendum: I just wanted to add a bit about orders of these elements. Now it is not to hard to see that the order of $\sigma$ is exactly the least common multiple of $2$ and $3$ (since we need both $(4,5)^m = (1)$ and $(2,3,7)^m = (1)$ and the smallest $m$ where this happens is exactly the least common multiple). So the order of $\sigma$ is exactly the smallest natural number $n$ such that $(4,5)^n = (1)$ and $(2,3,7)^n = (1)$ (think about this fact for a moment).īut what is the order of a each of $(4,5)$ and $(2,3,7)$? Since the cycles $(4,5)$ and $(2,3,7)$ are disjoint you have the element that sends every number to itself). The order, by definition, is the the smallest natural number $n$ such that $\sigma^n = (1)$ (i.e. This article incorporates material from cycle on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.So take $\sigma = (4,5)(2,3,7)$. The name RSA is used for multiple things: A specific trapdoor one-way permutation, several public-key encryption schemes build on this permutation, several public key signature schemes build on this permutation, and a company which markets these algorithms (and other security-related stuff). (1991), The Symmetric Group / Representations, Combinatorial Algorithms & Symmetric Functions, Wadsworth & Brooks/Cole, ISBN 978-0-7 (2006), A First Course in Abstract Algebra with Applications (3rd ed.), Prentice-Hall, ISBN 978-0-13-186267-8
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