Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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
6  
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

2 Answers 2

Hints:

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 $\,15\,,\,6\,,\,10\ldots$

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.