The order of a permutation cycle

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The order of a k-cycle (a(1), a(2), ..., a(k)) is determined by the length of the cycle, which is k. According to the theorem on the order of permutations, the order is the least common multiple of the lengths of the cycles. Since there is only one cycle in this case, the order is simply k. The specific elements a(1), a(2), ..., a(k) do not affect this outcome. Therefore, the order of the k-cycle is k.
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Homework Statement



what is the order of k-cycle (a(1),a(2),...,a(k))


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The Attempt at a Solution



According to the theorem of the order of a permutation: the order of a permutation set written in disjoint cycle form is the least common multiple of the lengths of the cycles.(Ruffini-1799)

in this case , the length of the k-cycle is k, for all the common multiples of a(1), a(2) ... and a(k) would be k. So the order of cycle k is k right?
 
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The length of the cycle is k so the order is k. It doesn't matter what a(1) etc are. If I understand your notation.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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