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 $, denoted by $P_G(H)$.

A group $G$ is said to satisfy the permutizer condition in $G$ 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 $G$.

Now can say every group that satisfy in maximal permutizer condition then satisfy then 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 ?