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

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Homework Statement


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


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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 im 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
jbunniii
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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 im 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).
 
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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
 
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jbunniii
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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?
 
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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?
 
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jbunniii
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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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