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This is a follow up question of this

(Edit: The following question is for primitive group $G$ which lies in the case (i) and does not lie in case (ii) of Theorem 5.6C of the textbook "Finite permutation group" (page no. 167))

It is mentioned in Theorem 5.6C(i) that, the socle $H$ is permutation isomorphic to $A_m^d$ for some $d >0 $. Also, the Theorem 5.6B (on the same page) says that, any primitive group has order at most $exp\{c' \sqrt{n}(\log n)^2 \}$.

Question In the case of Theorem 5.6C(i), does the index of $H$ in $G$ have a bound in terms of $n$ (degree of $G$)? (I mean bound like $|G/H|\leq n^c$ or $n^{poly(\log n)}$ )

Note: I am trying to understand the O'Nan-Scott Theorem and its application. There I came up with the above doubt. Any help will be appreciated.

Derek Holt
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Jins
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    You need to make your assumptions clearer. – Derek Holt May 09 '22 at 14:31
  • @DerekHolt, I apologize. I have edited the question. Thank you! – Jins May 10 '22 at 07:21
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    But Theorem 5.6C does not say that, it says that at least one of conditions (i) and (ii) hold. It does not say that Condition (i) holds. Do you mean you want to assume that Condition (i) holds? If so, then why not say so? – Derek Holt May 10 '22 at 08:26
  • @DerekHolt, Yes, I am assuming that only Condition (i) holds for $G$. Because, if for some $G$, Condition (ii) also hold then it solve my doubt in this case, right ? – Jins May 10 '22 at 09:28
  • I am really just asking you to edit the question so that it makes sense and it is clear exactly what you are asking. – Derek Holt May 10 '22 at 11:05
  • I have made it more clear. I am sorry about that. – Jins May 10 '22 at 13:34

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We have $n \ge m^d$ and $|G:H| \le 2^dd!$, so I guess that gives a bound of $n^{O(\log \log(n))}$.

Derek Holt
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  • I am not able to prove $|G : H| \leq 2^d d!$. Assuming it together with $n \geq m^d$ we can show the bound you mentioned. Is there any reference or hint to prove $|G : H| \leq 2^d d!$ ? – Jins May 10 '22 at 16:09
  • That;s the order of the outer automorphism group of $A_m^d$ (for $m > 6$). – Derek Holt May 10 '22 at 16:13
  • I apologize, but still I didn't get it. It will be good if you give some detailed explanation @Derek holt. – Jins May 10 '22 at 18:13
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    Automorphisms of $A_m^d$ permute the direct factors and there are $d!$ such permutations. The kernel of this permutation action consists of automorphisms that fix all direct factors. Since $|{\rm Aut}((A_n):A_n| = |S_n:A_n| = 2$ for $n \ge 6$, the order of the outer automorphism group induced by the kernel of the permutation action is $2^d$, so $|{\rm Aut}(A_m^d):A_m^d| = 2^dd!$. – Derek Holt May 10 '22 at 19:06
  • Thank you! It is clear now. – Jins May 11 '22 at 05:48