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As far as I know, the only groups of order $n$ that aren't subgroups of $S_{n-1}$ are cyclic groups with prime power order and the Klein four-group. Is there any nonabelian group of order $n$ that isn't a subgroup of $S_{n-1}$?

Shaun
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mathlander
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    https://math.stackexchange.com/questions/1597347/finding-the-minimal-n-such-that-a-given-finite-group-g-is-a-subgroup-of-s-n – Jean-Claude Arbaut Nov 29 '22 at 21:42
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    Yes. In particular for $n=8$, the quaternion group $Q_8$ is not isomorphic to a subgroup of $S_7$ - see [this post](https://math.stackexchange.com/questions/481957/q-8-is-isomorphic-to-a-subgroup-of-s-8-but-not-isomorphic-to-a-subgroup-of?rq=1). – Dietrich Burde Nov 29 '22 at 21:51

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