R. Carter prooved that in finite soluble groups $G$ Carter subgroups $C$ exist and that they are conjugated. Furthermore they are exactly the nilpotent projectors: For every normal subgroup $N$ of $G$ the factor group $CN/N$ is maximal nilpotent in $G/N$ . 


My question is whether in non-finite soluble groups Carter subgroups (if they exist) are nilpotent projectors. 


They are maximal nilpotent also in the infinte case and also for non-solvable groups.


For my study I would like to know this for a semidirect product of a nilpotent nomral subgroup with an abelian subgroup.


(This question is on stackexchange, too, unanswered.)



Are Carter subgroups nilpotent projectors?
This is not true for infinite insoluble groups.


Let $K$ be the direct product of countably infinitely many copies $\langle a\_i,b\_i \rangle$ ($i \in {\mathbb N}$) of $S\_3 = \langle a,b \mid a^2=b^3=(ab)^2=1 \rangle$, and let $G$ be the semidirect product of $K$ with a group $\langle t \rangle$ of order $2$ that acts on each copy of $S\_3$ by $a\_i^t=a\_ib\_i$, $b\_i^t=b\_i^{-1}$. (So $t$ is acting in the same way on $\langle a\_i,b\_i \rangle$ as conjugation by $a\_ib\_i^{-1}$.)


Then the subgroup $A = \langle a\_i \mid i \in {\mathbb N} \rangle$ is nilpotent (it's an elementary abelian $2$-group) and self-normalizing in $G$. But $N = \langle b\_i \mid i \in {\mathbb N} \rangle$ (which is an elementary abelian $3$-group) is normal in $G$, $G/N$ is abelian, and $AN/N$ is normal of index $2$ in $G/N$.


Note also that the subgroup $N$ has the elementary abelian complement $\langle t, a\_ib\_i^{-1}\,(i \in {\mathbb N})\, \rangle$ in $G$, so $G$ is a semidirect product of a nilpotent normal subgroup by an abelian group.


