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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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$Z/2\times Z/2\to S_4$ seems to be a counterexample? –  nsrt Jul 18 '13 at 15:39
    
yes, it is truth, sorry, I forgot to add the condition that the group G is simple ... –  Shaki Jul 18 '13 at 15:46
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@Shaki: That's a rather big condition... and you should have edited the question to include it; I've added it for you now. –  Arturo Magidin Jul 18 '13 at 16:39

3 Answers 3

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.

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thank you very much! –  Shaki Jul 18 '13 at 19:16

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.

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thank you for the reference! –  Shaki Jul 18 '13 at 19:16

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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This does not seem to address the question. –  András Bátkai Jul 18 '13 at 20:23
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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.) –  Andreas Blass Jul 19 '13 at 4:00

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