What is maximal possible order of an element in $S_{10}$? Why? Give an example of such an element. How many elements will there be in $S_{10}$ of that order?
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In the future you can format latex using the notation guide. – Oct 13 '14 at 03:27
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1There is only one conjugacy class of elements of order $30$, and the cyclic group generated is self-centralizing, so the total number of elements of that order is $10!/30$. – Derek Holt Oct 13 '14 at 09:05
3 Answers
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This is an example of Landau's function.
There are asymptotic estimates (info attributed to André Nicolas).
Edward Jiang
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A few helpful ideas to get you started here. If you think through these, and define the terms appearing below, you should find yourself at the doorstep of the answer to your question.
- The order of an $n$-cycle is $n$.
- Every permutation can be written as a product of disjoint cycles.
- Disjoint cycles commute, so the the order of a permutation is the least common multiple of the lengths of the cycles in its cycle decomposition.
Now, try to write down an element of $S_{10}$ such that the least common multiple of the lengths of the cycles in its decomposition is maximal.
Jared
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Landau's function $g(n)$ gives the maximum order of any element in $S_n$, the symmetric group of order $n!$.
From the OEIS, we have $$g(10) = 30$$
Zubin Mukerjee
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