Composing permutations in cycle notation

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The discussion centers on composing permutations in cycle notation, specifically within the symmetric group S3. Participants clarify that to find the composition of two permutations, such as (2, 3) and (1, 2, 3), one should read from the right, applying the right cycle first. The example demonstrates how to combine the cycles, ultimately resulting in the permutation (1, 3). The conversation highlights the importance of understanding the order of operations in cycle notation for efficient calculations. The participants express satisfaction in grasping the concept and share methods for quick composition.
radou
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I'm currently going through a text about groups, and I'm having problems with composing permutations written in cycle notation, i.e. there are lots of examples and I'm expected to be able to calculate them pretty 'fast', so, is there a way to 'read out' the composition of two permutations direct from the cycle notation (I use standard function composition, which is rather slow), for example, if I have the symmetric group S3, and let's say two of its elements, (2, 3) and (1, 2, 3), how can I figure out what (2, 3) o (1, 2, 3) is? Thanks in advance.
 
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Don't you just read from the right? So, (2 3)(1 2 3) is the permutation \left(\begin{array}{ccc}1&2&3\\3&2&1\end{array}\right), since the right cycle sends 1 to 2, which in turn gets sent to 3 by the left cycle. The right cycle sends 2 to 3, which is then sent to 2 by the left cycle, and finally the right one sends 3 to 1, which remains unchanged by the left.

edit: That might be the way you say you do it. If so, I can't help, as that's the way I was taught!
 
Last edited:
cristo said:
Don't you just read from the right? So, (2 3)(1 2 3) is the permutation \left(\begin{array}{ccc}1&2&3\\3&2&1\end{array}\right), since the right cycle sends 1 to 2, which in turn gets sent to 3 by the left cycle. The right cycle sends 2 to 3, which is then sent to 2 by the left cycle, and finally the right one sends 3 to 1, which remains unchanged by the left.

edit: That might be the way you say you do it. If so, I can't help, as that's the way I was taught!

I *finally* got it. Thanks cristo! :smile:
 
radou said:
I *finally* got it. Thanks cristo! :smile:

You're welcome!
 
Apparently you already have it but I'll stick my oar in anyway!

Here's how I would do that problem:
(1 2 3) means "1 changes to 2, 2 changes to 3, and 3 changes to 1".
(2, 3) means "2 changes to 3 and 3 changes to 2".

Okay, putting them together, 1 changes to 2 and 2 changes to 3. so 1 changes to 3. 2 changes to 3 and 3 changes to 2, so 2 changes to 2. 3 changes to 1 and 1 does not change, 3 changes to 1.
1 changes to 3 and 3 changes to 1. 2 does not change. In cycle notation that is (1 3).
 
HallsofIvy said:
Apparently you already have it but I'll stick my oar in anyway!

Here's how I would do that problem:
(1 2 3) means "1 changes to 2, 2 changes to 3, and 3 changes to 1".
(2, 3) means "2 changes to 3 and 3 changes to 2".

Okay, putting them together, 1 changes to 2 and 2 changes to 3. so 1 changes to 3. 2 changes to 3 and 3 changes to 2, so 2 changes to 2. 3 changes to 1 and 1 does not change, 3 changes to 1.
1 changes to 3 and 3 changes to 1. 2 does not change. In cycle notation that is (1 3).

That's exactly how I got it. I was just trying to find a quick way to look at these, since they occur often in my textbook, at least in the current chapter.
 
That is a quick way to look at it. As quick as the previous one. In fact it is identical to the previous one but written horizontally rather than vertically.
 

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