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The permutizer of a subgroup $H$ of $G$ is defined to be the subgroup generated by all cyclic subgroups of $G$ that permute with $H$, i.e. $\langle x \in G | \langle x \rangle H = H \langle x \rangle \rangle $. The permutizer is denoted by $P_G(H)$.

A group $G$ is said to satisfy the permutizer condition if $P_G(H)$ strictly contains $H$ for any subgroup $H$ of $G$.

A group $G$ is said to satisfy the maximal permutizer condition if $P_G(M) = G$ for any maximal subgroup $M$ of $G$.

Question: Can one say that every group satisfying the maximal permutizer condition must satisfy the permutizer condition ?

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  • $\begingroup$ The converse is also true? $\endgroup$
    – Simin Eftekhari
    Commented Apr 24, 2015 at 9:03
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    $\begingroup$ @GeoffRobinson I don't think that's right. $S_3$ satisfies the maximal permutizer condition. $\endgroup$
    – Derek Holt
    Commented Apr 24, 2015 at 9:19
  • $\begingroup$ @DerekHolt : Yes, you are right I was mistaken. Thanks for pointing it out. $\endgroup$
    – Geoff Robinson
    Commented Apr 24, 2015 at 11:35
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    $\begingroup$ @Simin: In your definition of "permutizer condition", I guess you should require $H$ to be a proper subgroup of $G$. This is already understood in the notion of "maximal subgroup". (By the way, is there a reason for including the tag 'algebraic-groups'?) $\endgroup$
    – Jim Humphreys
    Commented Apr 24, 2015 at 12:54
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    $\begingroup$ I suspect that all groups satisfying the maximal permutizer condition must be solvable. I think I could prove that if I knew that no nonabelian simple group satisfied the condition. If a simple group satisfies the maximal permutizer condition, then every maximal subgroup is complemented by a cyclic group, and that seems unlikely. In fact, a subgroup of even index in a simple group cannot be complemented, but unfortunately, a simple group can fail to have a maximal subgroup of even index. A_7 is a counterexample. $\endgroup$
    – Marty Isaacs
    Commented May 26, 2015 at 23:43

1 Answer 1

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An example of a group satisfying the maximal permutizer condition but not the permutizer condition is given in Example 1, page 213 of

O. H. Kegel, On Huppert’s characterization of finite supersoluble groups, in ‘‘Proc. Internat. Conf. Theory of Groups, Canberra, 1965,’’ pp 20-215, Gordon and Breach, New York, 1967.

The group $G$ has the structure (in ATLAS notation) $2^4.(2^2 \times 2^2):(S_3 \times S_3)$. The minimal normal subgroup $2^4$ is equal to $\Phi(G)$, and $G/\Phi(G) \cong S_4 \times S_4$ satisfies the maximal permutizer condition, and hence so does $G$. But in an earlier paper,

James C. Beidleman, Derek J. S. Robinson On Finite Groups Satisfying the Permutizer Condition, Journal of Algebra 191, 68-703,

it is proved that, in a group satisfying the permutizer condition, the chief factors have either prime order or order $4$. Since $G$ has a chief factor of order $16$, it does not satisfy this condition.

Concerning Marty Isaacs'xcomment, I couldn't find an record of anyone having proved that the maximla permutizer conditino implies solvability.

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  • $\begingroup$ I don't have access to this article, can you help me? $\endgroup$
    – Soroush
    Commented Nov 27, 2015 at 7:23
  • $\begingroup$ @Soroush Help you with what? If you ask a specific question then I will try and answer it! $\endgroup$
    – Derek Holt
    Commented Nov 27, 2015 at 8:54
  • $\begingroup$ i don't understand $ G $ has the structure and $ G/\Phi(G) \cong S_{4} \times S_{4} $ . $\endgroup$
    – Soroush
    Commented Nov 27, 2015 at 9:22
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    $\begingroup$ Sorry, I don't have access to the paper by Kegel either! I read a description of the group somewhere and constructed it on the computer. It is constructed as a nonsplit extension of an elementary abelian group $N$ of order $2^4$ by $S_4 \times S_4$, and it turns out that $N=\Phi(G)$, $\endgroup$
    – Derek Holt
    Commented Nov 27, 2015 at 10:16
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    $\begingroup$ In GAP, the group is $\mathtt{TransitiveGroup}(24,9958)$. $\endgroup$
    – Derek Holt
    Commented Nov 27, 2015 at 10:46

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