# embeddings of finite group into permutation groups

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
@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

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