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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.)

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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.

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  • $\begingroup$ I believe that if the definition of Carter subgroup is the nilpotent self-normalizing subgroup, then an infinite solvable group may fail to contain a Carter subgroup. But I'm not so sure, may I ask is that right? $\endgroup$ Apr 29 '19 at 8:40

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