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Does the permutation group $S_8$ contain elements of order $14$?

My answer: If $\sigma =\alpha \beta $where $\alpha$ and $ \beta$ are disjoint cycles, then $|\sigma| =lcm(|\alpha|, |\beta|)$. Therefore the only possible disjoint cycle decompositions for a permutation $\sigma \in S_8$ with $|\sigma| =14$ is $(7,2)$. Since $7+2\neq 8$ so there is no element of order 14 in $S_8$.

Is my answer right? If no, what's the right answer?

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You've only considered the possibility that $\sigma$ is product of two disjoint cycles. So no, your answer is not right. – Chris Eagle Feb 26 '13 at 19:28
also the issue is not that $7+2\neq 8$, but that $7+2>8$. – Dustan Levenstein Feb 26 '13 at 19:29
The proof needs to consider more than 2 cycles explicitly, but is otherwise fine. – vonbrand Feb 26 '13 at 22:48


1) Every element in $\,S_n\,$ can be expressed as a product of disjoint cycles

2) The order of a product of disjoint cycles is the least common multiple of their lengths

So...can you see now why there is no element of order $\,14\,$ in $\,S_8\,$? There though are elements of order $\,15\,,\,6\,,\,10\ldots$

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Assume that $g$ is an element of $S_8$ of order $14$. Write $g$ as a product of disjoint cycles and let $a_1,\ldots,a_n$ be the cycle lengths (fixed points are counted as cycles of length 1). Then you have the equation $$a_1 + a_2 + \ldots + a_n = 8$$ and since the order is the least common multiple of the cycle lengths $$\operatorname{lcm}(a_1,a_2,\ldots,a_n) = 14.$$ So you have to decide if there is a solution in positive integers $a_i$ to this system of equations.

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