Group Action on a Set: Counting Transitive Z6 Sets Up to Isomorphism

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Homework Help Overview

The discussion revolves around counting transitive Z6 sets up to isomorphism. The original poster presents an attempt to identify the number of such sets, noting discrepancies in their findings compared to expected results.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the counts of transitive sets with varying numbers of elements and question the conditions of the problem. There is a focus on isomorphism classes and the implications of different multiplication tables for the same set of elements.

Discussion Status

The discussion is active, with participants questioning the original poster's counts and the nature of isomorphism. There is a recognition of the existence of a transitive G-set with 6 elements, and some participants are examining specific examples of multiplication tables to illustrate their points.

Contextual Notes

Participants note the importance of considering isomorphism classes in their counts and discuss the potential for misunderstanding the problem's conditions. There is mention of a specific reference to a textbook problem, which may influence the interpretation of the task.

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Homework Statement


Up to isomorphism, how many transitive Z6={0,1,2,3,4,5} sets X are there? In other words Z6 acting on X. How many X, up to isomorphism are there?

Homework Equations


A key theorem is Let X be a G-set and let x in X. Then |Gx|=(G:G_{x}).

The Attempt at a Solution


I found 1 set with 1 element. 1 set with 2 elements. 2 sets with 3 elements. 5*5! sets with 6 elements.

However the answer only had 1 set with 1 element. 1 set with 2 elements. 1 set with 3 elements.

Surely there should be at least a set with 6 elements?
 
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There indeed should be a transitive G-set with 6 elements. Unless, of course, you missed a condition of the problem...


As for your counts, remember that you were asked to count isomorphism classes...
 
Hurkyl said:
There indeed should be a transitive G-set with 6 elements. Unless, of course, you missed a condition of the problem...As for your counts, remember that you were asked to count isomorphism classes...

Didn't leave out any of the problem. It is on p196 q17 of Fraleigh's book.

Yes, I see up to isomorphism. Take the set with 3 elements. I am able to produce to different multiplication tables with the same three elements in the set. That must mean two non isomorphic sets.
 
I'm not convinced. What are your two examples?
 
Hurkyl said:
I'm not convinced. What are your two examples?

Attached is the two tables. Once the fixed elements and the multiples of 1 line is determined, the rest of the table is self explanatory by the axiom that a(bx)=(ab)x.
 

Attachments

What do people think?
 
Hurkyl, have you disappeared?
 
I didn't want to download a .doc file, so I was hoping someone else who wasn't bothered would chime in.
 
Strange reason? Are you worried about your memory?

Anyway, here it is without the .doc file

* a b c
0 a b c
1 b c a
2
3 a b c
4
5


* a b c
0 a b c
1 c a b
2
3 a b c
4
5

The above are the two different tables hence two different isomorphisms.
 
  • #10
But those are isomorphic Z6-sets. One isomorphism is

a |--> a
b |--> c
c |--> b
 

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