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What is the smallest degree symmetric group with an element of order 50?

I just had this question on an exam and reasoned out that it should be $S_{27}$ since we could have cycles $(\underline2)(\underline{25})$. So $\operatorname{lcm}(2, 25) = 50$. (Where $(\underline m)$ is an $m$-cycle).

I'm wondering if there is a more systematic approach to this general kind of question? Other than just reasoning it out like this. (Assuming my answer is correct.)

Thanks!

Davide Giraudo
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EgoKilla
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1 Answers1

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This is A008475 on OEIS:

http://oeis.org/A008475

The OEIS links to a paper which discusses this problem (bottom of p. 205):

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.124.8582&rep=rep1&type=pdf

The least degree symmetric group with an element of order $m$ is $S_n$, where $n = \sum_{i} p_i^{r_i}$ and $m = \prod_{i} p_i^{r_i}$ is the prime factorization of $m$ ($p_i$ are distinct primes).

Perry Elliott-Iverson
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  • This is really cool, thank you! – EgoKilla Feb 27 '14 at 16:09
  • Neat. Also, the individual terms in $m$'s prime factorization give the lengths of the corresponding disjoint cycles for such an element. For example, the prime factorization of $60=2^235$ so the smallest symmetric group containing an element of order $60$ is $S_{2^2+3+5}=S_{12}$. So there is element in $S_{12}$ with order $60$ and a cycle pattern of $3\cdot4\cdot5$. Is this pattern unique? That is, in general, for a permutation $\sigma$ of order $m$ and the symmetric group $S$ with the smallest degree that contains $\sigma$, is there only one possible cycle pattern for $\sigma$ in $S$? – Display name Nov 08 '23 at 01:34