Groups of permutations and cyclic groups

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Discussion Overview

The discussion revolves around the nature of groups of permutations and their relationship to cyclic groups, focusing on the properties of these groups as functions and the conversion of cyclic groups into products of disjoint cycles. Participants explore both theoretical and practical aspects of these concepts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that a group of permutations can be viewed as a group of one-to-one functions from a set onto itself, sharing properties like associativity and identity.
  • There is a discussion about expressing individual members of a cyclic group as products of disjoint cycles, with some clarification on the direction of evaluation of these cycles.
  • One participant questions the starting point for cycles and how to select the next cycle, suggesting that the smallest unused element is typically chosen.
  • Another participant asserts that the starting element of a cycle does not affect the cycle's identity, as cycles can be represented in different forms but remain equivalent.
  • It is noted that any unused element can be selected for the next cycle, emphasizing the flexibility in choosing elements while maintaining clarity in the process.

Areas of Agreement / Disagreement

Participants generally agree on the properties of permutation groups and the flexibility in choosing elements for cycles, but there are varying opinions on the implications of starting points and the evaluation of cycles.

Contextual Notes

Some assumptions about the evaluation direction of cycles and the selection of starting points are not fully resolved, leading to potential variations in interpretation.

yaganon
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1: Is a group of permutations basically the same as a group of functions? As far as I know, they have the same properties: associativity, identity function, and inverses.

2: I don't understand how you convert cyclic groups into product of disjoint cycles.
A cyclic group (a b c d ... z) := a->b, b->c, c->d, d->e ... y->z, z->a
In the book, it shows that (0 3 6) o (2 7) o (4 8) o (0 4 7 2 6) o (1 8) = (0 8 1 4 2) o (3 6)

How do you get there?

thanks
 
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1. Yes, it's a group of 1-1 functions from the set of elements being permuted (or possibly a superset) onto itself.

2. You don't, you may express individual members of a cyclic group of permutations as the product of disjoint cycles. In your example each cycle goes from left to right but the products are evaluated from right to left. The direction of evaluation could be different in a different book, but the cycles will usually go from left to right I believe.

So for example:

1--(18)--> 8 --(0 4 7 2 6)--> 8--(4 8)--> 4 --(2 7)--> 4 --(0 3 6)--> 4

Which is why 4 follows 1 in the cycle (0 8 1 4 2) on the rhs. Other elements similarly.

If g=(abc...z) then g2 would be (abc...z)(abc...z) evaluated the same way, i.e. (ace...y)(bdf...z) and so on.
 
OK, I kind of get it. So you always start with 0, and the last element ends with 2 because two maps to 0 through the series of functions, completing the cycle. What if there isn't a zero? Also, when you're done with a cycle, how do you start the next one, do you choose the next smallest element that's left? which is why it's (3 6) and not (6 3)?
 
It doesn't matter what you start with. All that happens is your cycles can appear shifted round if you start with something different.

After finding 1->4 I woulve tracked 4 etc. Then my cycle would appear as (14208) instead of (08142) on thr rhs. But it's the same cycle either way.
 
Similarly you can just pick anything you haven't already done for the next cycle. It helps to take them in some sort of order so you don't forget a cycle altogether, but it's up to you.
 
That is to say (63) will do just as well as (36). It's exactly the same mapping.
 

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