Does every group that satisfies the maximal permutizer condition then satisfy the permutizer condition? 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 ?
 A: 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. 
