Write (13257)(23)(47512) as a product of disjoint cycles

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Homework Help Overview

The problem involves expressing the product of three permutations, (13257)(23)(47512), as a product of disjoint cycles. The context is within the study of permutations and their properties in abstract algebra.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster considers evaluating the product directly to find the disjoint cycles but questions if there is a more efficient method. Some participants note the complexity introduced by the number 2 appearing in all three cycles, suggesting that a straightforward approach may not be applicable.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. There is a mention of a related question about calculating the order of a different permutation, which introduces additional considerations about the properties of disjoint cycles.

Contextual Notes

Participants are navigating the implications of the permutations involved, particularly focusing on the shared elements across cycles and the potential complexity this introduces to the problem.

Mr Davis 97
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Homework Statement


Write (13257)(23)(47512) as a product of disjoint cycles. Each bracket is a permutation of seven elements written in cycle notation.

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The Attempt at a Solution


This isn't too hard of a problem. One easy way would be to evaluate the entire product, and then write that product in cycle notation. However, is there an easier, faster way of doing this, just be looking at the expression (13257)(23)(47512) directly?
 
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Not that I know of. And ##2## occurs in all three cycles, so any possible "rule" is likely more complicated than simply multiply them.
 
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Also, one quick related question. I need to calculate the order of (125)(34). Is there a quick way to do this, or do I have to literally keep composing the permutation with itself until I get the identity permutation?
 
If they are disjoint, they commute. So ##(ab)^n=a^nb^n## and the order is the least common multiple of both orders. And a cycle of length ##n## is of order ##n##.
 
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