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

In summary, the group of permutations on the set {123} is not cyclic, as there is no single permutation that can generate all permutations within the group. However, the subgroup of even permutations is cyclic.
  • #1
pbxed
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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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  • #2
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).
 
  • #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
 
  • #4
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?
 
  • #5
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?
 
  • #6
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.
 

What is a cyclic group of permutations?

A cyclic group of permutations is a mathematical concept in abstract algebra that refers to a group of permutations that can be generated by a single element, also called a generator. In other words, all the elements in a cyclic group can be obtained by repeatedly applying a single permutation to itself.

What is a permutation?

A permutation is a rearrangement of a set of elements. In the context of a cyclic group, a permutation refers to a specific ordering or arrangement of the elements in the set. In the case of the set {123}, a permutation could be 132, which rearranges the elements in the set to be in the order 1, 3, 2.

How can you tell if a group of permutations is cyclic?

To determine if a group of permutations is cyclic, you can check if there is a single element that can generate all the other elements in the group. If such an element exists, then the group is cyclic. In other words, if you can obtain all the elements in the group by repeatedly applying a single permutation to itself, then the group is cyclic.

Is the group of permutations on the set {123} cyclic?

Yes, the group of permutations on the set {123} is cyclic. This is because the element 123 can generate all the other elements in the group by repeatedly applying it to itself. For example, 123^2 = 132, 123^3 = 213, and so on.

What is the justification for the group of permutations on the set {123} being cyclic?

The justification for the group of permutations on the set {123} being cyclic is that it follows the definition of a cyclic group, which states that there must be a single element that can generate all the other elements in the group. In this case, the element 123 satisfies this condition, making the group cyclic.

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