FPM quiz question on permutations

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Here is another question and (partial) answer from my FPM Piazza forum last autumn, this time related to the "challenge question" from my First quiz on permutations.

First, here is a screenshot of the relevant quiz question. (You can click on the image to view it full size.)

PermutationChallenge

Challenge question from first FPM quiz on permutations

Student's question: Why in this question does \(12\) being divisible by \(3\) and \(4\) mean that \(\rho^{12}\) is the identity permutation?

[Joel notes: The identity permutation on a set is that very special permutation which doesn't move anything. So it is the map defined on your set by the rule \(x \mapsto x\).]

My response: I think some of you may still want to think about this, so I don't want to give the whole thing away at once. But I'll give you some things to think about.

Have a look at \(\rho = (1\,7\,2) (3\,6\,5\,4)\), and notice that this is a product of disjoint cycles, and disjoint cycles commute. Because of this,
     \(\rho^2=(1\,7\,2) (3\,6\,5\,4)(1 \,7\, 2)(3\, 6 \,5 \,4) = (1 \,7 \,2)^2 (3\, 6 \,5\, 4)^2\)
and then an easy induction shows that, for all natural numbers \(n\),
     \(\rho^n=(1 \,7\, 2)^n (3 \,6 \,5 \,4)^n\,.\)
Actually, I never defined \(\rho^1\), but you will be unsurprised to know that this is just defined to be \(\rho\). (Also, \(\rho^0\) is, by convention, the identity permutation.)
So your task now is to think about the permutations \((1 \,7\, 2)^n\) and \((3 \,6\, 5 \,4)^n\).
For which \(n\) is \((1 \,7\, 2)^n\) equal to the identity permutation?
For which \(n\) is \((3\, 6\, 5\, 4)^n\) equal to the identity permutation?
Best wishes,
Dr Feinstein

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