Let $G$ be a finite simple group, and let $n$ be the smallest integer such that there exists an embedding of $G$ into the permutation group of $n$ elements. Is this embedding unique up to conjugation?
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2$\begingroup$ $Z/2\times Z/2\to S_4$ seems to be a counterexample? $\endgroup$– nsrtCommented Jul 18, 2013 at 15:39
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$\begingroup$ yes, it is truth, sorry, I forgot to add the condition that the group G is simple ... $\endgroup$– ShakiCommented Jul 18, 2013 at 15:46
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1$\begingroup$ @Shaki: That's a rather big condition... and you should have edited the question to include it; I've added it for you now. $\endgroup$– Arturo MagidinCommented Jul 18, 2013 at 16:39
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$\begingroup$ Derek answers up to global conjugation of the image. But it's easier to find a counterexample if you ask about up to isomorphism of $G$-set. Namely, if $G=\mathrm{Alt}_6$, then there are two non-isomorphic faithful $G$-sets of cardinal $6$ (since there are two conjugacy classes of subgroups of index 6, and no non-trivial non-transitive action on 6 elements). $\endgroup$– YCorCommented Mar 14, 2022 at 16:42
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3 Answers
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When you talk about the embedding being unique up to conjugation, I assume that you are asking whether all subgroups of $S_n$ isomorphic to $G$ are conjugate in $S_n$.
The answer is not always, but it's not so easy to find counterexamples. You need a simple group with more than one conjugacy class of subgroups of index $n$, where the classes are not all fused by automorphisms of the group. The groups $G_2(q)$ with $q$ not a power of 3 and $q>4$ satisfy this condition. They have two nonisomorphic maximal subgroups with the structure $q^5.{\rm GL}_2(q)$. (When $q$ is a power of 3, these subgroups are fused by the exceptional graph isomorphism. When $q=4$, there is a subgroup $J_2$ of smaller index.)
So the smallest counterexample appears to be $G_2(5)$ with $n=3906$. You can find it in the ATLAS of Finite Groups.
answered Jul 18, 2013 at 17:22
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More of a supplement to @Derek's excellent answer: the reason you look for counterexamples of the form he gave is that since your group is simple and the embedding is into a minimal $S_n$ your subgroup must be transitive and primitive. The O'Nan-Scott theorem (which you can google) then gives you a list of possible counterexamples.
answered Jul 18, 2013 at 18:56
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Cayley's theorem states that any group is isomorphic to a subgroup of the symmetric group.
This can be seen alternatively from the theory of computability. Any "program" (a group is a set of elements with a group operation) has a representation as a Turing machine, which itself is isomorphic to a subset of a permutation group.
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2$\begingroup$ This does not seem to address the question. $\endgroup$ Commented Jul 18, 2013 at 20:23
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1$\begingroup$ In addition, except for the first sentence, it doesn't make sense. Groups are not programs, nor are Turing machines isomorphic to subsets of permutation groups. (The MO software only allows me to downvote the answer once.) $\endgroup$ Commented Jul 19, 2013 at 4:00