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

1. Aug 11, 2011

pbxed

1. The problem statement, all variables and given/known data
Consider the group of permutation on the set {123}. Is this group cyclic? Justify your answer

2. Relevant equations

3. 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

2. Aug 11, 2011

jbunniii

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. Aug 11, 2011

pbxed

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.

4. Aug 11, 2011

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. Aug 11, 2011

pbxed

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. Aug 11, 2011

Correct.