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For any $n$, does there exist a group $G$ of order $n$ so that $G$ is not isomorphic to any subgroup of $S_j$ for $j<n$?

Kushal Bhuyan
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    Cayley's theorem is for "every group". – Kushal Bhuyan Jun 12 '16 at 11:49
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    Whenever $n$ is a prime number the (up to isomorphism) only group of order $n$ has this property. But what exactly is the question? For which $n$ does there exist such a group, or? AFAICT $n=4$ also works. – Jyrki Lahtonen Jun 12 '16 at 11:52
  • @lhf that is the title and also the description. What is unclear? –  Jun 12 '16 at 11:58
  • @JyrkiLahtonen That is exactly the question. –  Jun 12 '16 at 11:59
  • For any (= some) or for every $n$? – lhf Jun 12 '16 at 12:02
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    See also http://math.stackexchange.com/questions/191446/efficient-version-of-cayleys-theorem-in-group-theory. – lhf Jun 12 '16 at 12:05
  • @lhf Thanks! The link answer my question, should this be marked as duplicate or....? –  Jun 12 '16 at 12:17
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    @user336-iactuallychosethis: Yes, we appreciate when the OP himself closes his question as a duplicate (it is faster, too, because no extra votes are required then). – Alex M. Jun 12 '16 at 14:22

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