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The following is the statement of O'Nan-Scott Theorem.

Theorem: Let $G$ be a finite primitive group of degree $n$, and let $H$ be the socle of G. Then either

(a) $H$ is a regular elementary abelian $p$-group for some prime $p$, $n=p^m:=|H|$, and $G$ is isomorphic to a subgroup of the affine group ${\rm AGL}_m(P)$; or

(b) H is isomorphic to a direct power $T^m$ of a nonabelian simple group $T$ and one of the following holds:

(i) m = 1 and G is isomorphic to a subgroup of Aut(T);

(ii) $m\geq2$ and $G$ is a group of "diagonal type" with $n = |T|^{m-1}$.

(iii) $m\geq2$ and for some proper divisor $d$ of $m$ and some primitive group $U$ with a socle isomorphic to $T^d$, $G$ is isomorphic to a subgroup of the wreath product $U\,\wr\, {\rm Sym}(m/d)$ with the product action, and $n=l^{m/d}$, where $l$ is the degree of $U$.

(iv) $m\geq6$, $H$ is regular, and $n=|T|^m$.

Question: In the Case (b), the socle of $G$ which is $H$ is isomorphic to a direct power $T^m$ of a nonabelian simple group $T$. In the case (b)(iii), does any bounds on the order of $H$ in terms of $n$ is known ? (e.g. in part(b)(ii), $|T|^{m-1}=n)$.

Note: Trivial bound is $|H| \leq |G| \leq n^{(\sqrt n)\log n}$. (Thus I am asking about some nontrivial one like $|H|\leq n^c$ or $n^{poly(\log n)}$ ).

Edit: See page 126 of the book "Finite permutation group", where it is mentioned that (last line of the first paragraph), the Groups of types (b)(ii) and (b)(iii), are generally distinguished as having a small orders of socle with respect to their degree.

Derek Holt
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Jins
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    Use $\le$ for $\le$. – Shaun Apr 28 '22 at 19:49
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    I think you want $n = |T|^{m-1}$ rather than $n=|T|^m$ in (b)(ii). In the group $S_m \wr C_2$ in (b)(iii), we have $|H| = (m!)^2$ and $n = m^2$, so there is no bound of the type you are suggesting. – Derek Holt Apr 28 '22 at 21:44
  • @DerekHolt, Here, $H$ is the socle of $G$. It is direct product of non-abelian simple groups. Do you mean that, in the above example $T \cong A_m$? Since $T$ can not be $S_m$. – Jins Apr 29 '22 at 04:33
  • Sorry, yes, I got your notation confused. $T = A_m$, $H = A_m^2$ so $|H| = (m!)^2/4$, , $U = S_m$. – Derek Holt Apr 29 '22 at 07:50
  • Okay right. What if we discard the case when $T$ is $A_m$? I am asking this because I have read somewhere that, Groups of types (b)(ii) and (b)(iii), are generally distinguished as having a small orders of socle with respect to their degree. Added a link for the reference (in the question). – Jins Apr 29 '22 at 10:04
  • Sorry once again for the confusion of notation, the $m$ is my comment is different from the $m$ in your question. I am unsure exactly what you are asking now. Do you want to exclude the case when $T$ is isomorphic to $A_k$ for some $k$? – Derek Holt Apr 29 '22 at 10:35
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    My question is: can we bound (non-trivial bound which mentioned in the question) the order of socle (i.e. $|T^m|)$ in terms of the degree of $G$ (which is $n$) ? If it is not true in general, then for what non abelian simple group $T$ it is true ? Like does it holds if we exclude the case when $T$ is not $A_k$ for some $k$? – Jins Apr 29 '22 at 15:17

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Theorem 5.6B on Page 167 of Dixon and Mortimer's book states that for any primitive group of degree $n$ other than $A_n$ and $S_n$, we have $|G| \le \exp(c \sqrt{n}(\log n)^2)$ for some constant $c$.

The examples $G=S_k \wr C_2$ with product action, where $k = \sqrt{n}$, that I mentioned in my comment have $|G:H| = 8$ and order greater than $\exp(d \sqrt{n}(\log n)^2)$ for some constant $d$. So we cannot improve on this bound for $|G|$ as a bound for $|H|$.

But if we assume that $T$ is not isomorphic to $A_k$ for any $k$ then, using the classification of finite simple groups, we can use Theorem 5.6C (ii) to get the better bound of $\exp( b (\log n)^2)$ for $|G|$ and $|H|$, for some constant $b$.

The example $U \wr C_2$ with product action, with $U = {\rm GL}_k(2)$ in its natural primitive action of degree $2^k-1$ shows that this is best possible (apart from estimating the constant $b$).

Derek Holt
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