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The group ${\rm PSL}(2,7)$ seems to be a $CP$ group according to your definition, but it contains an element $a$ of order $2$ and an element $b$ of order $3$ whose product $ab$ has order $7,$ so you do not have a metric in this group. In fact, as I write, I realise that in a group $G$ where $d$ defines a metric, the product of any two elements of order $2$ can have order at most $3.$ This seems to exclude $G = {\rm PSL}(2,q)$ as a possibility whenever $q >3.$ Also, the Suzuki groups (the only non-Abelian simple groups of order prime to $3$) are excluded, as they always contain a dihedral subgroup of order $10.$

Later edit: note that the structure of a Sylow $2$-subgroup $S$ of a group $G$ such that $d$ defines a metric is extremely restricted. The elements of order at most $4$ in $S$ form a subgroup, and all involutions of $S$ commute with each other. Thus $S$ has no dihedral subgroup of order $8$ and no generalized quaternion subgroup of order $16$ or more.

The group ${\rm PSL}(2,7)$ seems to be a $CP$ group according to your definition, but it contains an element $a$ of order $2$ and an element $b$ of order $3$ whose product $ab$ has order $7,$ so you do not have a metric in this group. In fact, as I write, I realise that in a group $G$ where $d$ defines a metric, the product of any two elements of order $2$ can have order at most $3.$ This seems to exclude $G = {\rm PSL}(2,q)$ as a possibility whenever $q >3.$ Also, the Suzuki groups (the only non-Abelian simple groups of order prime to $3$) are excluded, as they always contain a dihedral subgroup of order $10.$

Later edit: note that the structure of a Sylow $2$-subgroup $S$ of a group $G$ such that $d$ defines a metric is extremely restricted. The elements of order at most $4$ in $S$ form a subgroup, and all involutions of $S$ commute with each other. Thus $S$ has no dihedral subgroup of order $8$ and no generalized quaternion subgroup of order $16$ or more. Even later edit: note also that by an earlier remark, if $t$ and $u$ are involutions of $G,$ then either $t$ and $u$ are conjugate (via an element of order $3$), or else $t$ commutes with every $G$-conjugate of $u.$

The group ${\rm PSL}(2,7)$ seems to be a $CP$ group according to your definition, but it contains an element $a$ of order $2$ and an element $b$ of order $3$ whose product $ab$ has order $7,$ so you do not have a metric in this group. In fact, as I write, I realise that in a group $G$ where $d$ defines a metric, the product of any two elements of order $2$ can have order at most $3.$ This seems to exclude $G = {\rm PSL}(2,q)$ as a possibility whenever $q >3.$ Also, the Suzuki groups (the only non-Abelian simple groups of order prime to $3$) are excluded, as they always contain a dihedral subgroup of order $10.$

The group ${\rm PSL}(2,7)$ seems to be a $CP$ group according to your definition, but it contains an element $a$ of order $2$ and an element $b$ of order $3$ whose product $ab$ has order $7,$ so you do not have a metric in this group. In fact, as I write, I realise that in a group $G$ where $d$ defines a metric, the product of any two elements of order $2$ can have order at most $3.$ This seems to exclude $G = {\rm PSL}(2,q)$ as a possibility whenever $q >3.$ Also, the Suzuki groups (the only non-Abelian simple groups of order prime to $3$) are excluded, as they always contain a dihedral subgroup of order $10.$

Later edit: note that the structure of a Sylow $2$-subgroup $S$ of a group $G$ such that $d$ defines a metric is extremely restricted. The elements of order at most $4$ in $S$ form a subgroup, and all involutions of $S$ commute with each other. Thus $S$ has no dihedral subgroup of order $8$ and no generalized quaternion subgroup of order $16$ or more.

The group ${\rm PSL}(2,7)$ seems to be a $CP$ group according to your definition, but it contains an element $a$ of order $2$ and an element $b$ of order $3$ whose product $ab$ has order $7,$ so you do not have a metric in this group. In fact, as I write, I realise that in a group where $d$ defines a metric, the product of any two elements of order $2$ can have order at most $3.$ This means that such a group $G,$ if it has even order, has cyclic or generalized quaternion Sylow $2$-subgroups. Then $G = O_{2'}(G)C_{G}(t)$ for some involution $t \in G.$ Furthermore, $[O_{2'}(G),t]$ must be a $3$-group. It follows that that $O_{2'}(G)$ is a $3$-group, and that $C_{G}(t)$ is a $2$-group, in light of your other conditions. Hence $O_{3}(G)$ is Abelian and $G$ is a Frobenius group whose kernel is an Abelian $3$-group and whose complement is a cyclic or generalized quaternion $2$-group. In fact, the $3$-group must be elementary Abelian, give the earlier remark.

The group ${\rm PSL}(2,7)$ seems to be a $CP$ group according to your definition, but it contains an element $a$ of order $2$ and an element $b$ of order $3$ whose product $ab$ has order $7,$ so you do not have a metric in this group. In fact, as I write, I realise that in a group $G$ where $d$ defines a metric, the product of any two elements of order $2$ can have order at most $3.$ This seems to exclude $G = {\rm PSL}(2,q)$ as a possibility whenever $q >3.$ Also, the Suzuki groups (the only non-Abelian simple groups of order prime to $3$) are excluded, as they always contain a dihedral subgroup of order $10.$