Is the group of permutations on the set {123} Cyclic? Justification required

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Homework Help Overview

The discussion revolves around the group of permutations on the set {123} and whether this group is cyclic. Participants are exploring the properties of permutations and their compositions, particularly focusing on the implications of the Cayley table and the nature of cyclic groups.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of composing the permutation (123) multiple times and question whether this is sufficient to demonstrate that the group is cyclic. There are also inquiries about handling larger groups and finding quicker methods to determine cyclicity.

Discussion Status

Some participants have provided guidance on the nature of the group and the number of permutations involved. There is an ongoing exploration of the subgroup of even permutations and its cyclicity, with some participants expressing uncertainty about their conclusions.

Contextual Notes

Participants note that there are six permutations of the set {1,2,3}, which challenges the initial assumption about the number of elements in the group. The discussion also touches on the properties of groups with prime order.

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Homework Statement


Consider the group of permutation on the set {123}. Is this group cyclic? Justify your answer


Homework Equations





The Attempt at a Solution



I wrote out the cayley table for this group, and noticed that if we take (123)^3 = e . Seeing as we can get back to the original orientation of the permutation by composition of (123) three times and that any permutation can be written as a product of transpositions is this enough to show that the group is cyclic? I think it is but I am not totally convinced.

Also what happens if we are given a larger group and have to show that it is cyclic or not? Say {1234567}. Is there a quicker way than writing out all permutations manually trying to find some (g^n) = e (where n is a member of the integers) ?

Thanks
 
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pbxed said:
I wrote out the cayley table for this group, and noticed that if we take (123)^3 = e . Seeing as we can get back to the original orientation of the permutation by composition of (123) three times and that any permutation can be written as a product of transpositions is this enough to show that the group is cyclic? I think it is but I am not totally convinced.

No, this isn't correct. If (1 2 3), (1 2 3)^2, and (1 2 3)^3 were the only elements of the group, it would be right, but that is not the case. There are six permutations of {1,2,3}, not three. In general, there are n! permutations of {1,2,...n}.

The specific permutations you didn't account for are (1 2), (2 3), and (1 3).
 
Oh okay. I think I was just outleveling myself for a minute. The next part of the question ask if the subgroup of even permutations is cyclic. I guess what I have shown is the proof that it is cyclic.

Thanks for your help jbunniii
 
pbxed said:
Oh okay. I think I was just outleveling myself for a minute. The next part of the question ask if the subgroup of even permutations is cyclic. I guess what I have shown is the proof that it is cyclic.

Thanks for your help jbunniii

Yes, that's correct - the subgroup of even permutations is cyclic. (It has order 3, which is prime, and any group with prime order is cyclic.)

So what about the original question: is the group of all permutations of {1,2,3} cyclic?
 
Its not cyclic because no permutation can be a generator for the group. That is, no one permutation when composed with itself however many times can generate all the permutations within the group.

Is that correct?
 
pbxed said:
Its not cyclic because no permutation can be a generator for the group. That is, no one permutation when composed with itself however many times can generate all the permutations within the group.

Is that correct?

Correct.
 

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