Abelian permutable complement to core-free subgroup is self-centralizing

Statement

Suppose $G$ is a group, $M$ is a Core-free subgroup (?) and $N$ is an Abelian subgroup of $G$ such that $NM=G$. Then, $N=C_{G}(N)$.

Related facts

• Plinth theorem: This is a sort of generalization of the result to the non-Abelian case. In the general version (where the complement is not necessarily Abelian), while the complement need not be self-centralizing, it is isomorphic to its centralizer, and equals its double centralizer.
• Abelian normal subgroup and core-free subgroup generate whole group implies they intersect trivially
• Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements

Applications

• Primitive implies Fitting-free or elementary Abelian Fitting subgroup

Proof

Given: A group $G$, a core-free subgroup $M$, an Abelian subgroup $N$ of $G$ such that $NM=G$ and $N\cap M$ is trivial.

To prove: $N=C_{G}(N)$.

Proof: We first prove that $C_{M}(N)$ is trivial. Suppose $g\in C_{M}(N)$, in other words $g\in M$ commutes with every element of $N$. Then, $g\in nMn^{-1}$ for every $n\in N$. Since $NM=G$, any $h\in G$ can be written as $nm$ for $n\in N,m\in M$, so $nMn^{-1}=hMh^{-1}$. Thus, $g\in hMh^{-1}$ for every $h\in G$, and so $g$ is in the normal core of $M$.

Thus, no element of $M$ centralizes $N$. But we know that since $N$ is Abelian, $N\leq C_{G}(N)$, so $C_{G}(N)=N(M\cap C_{G}(N))=NC_{M}(N)$. Hence, $C_{G}(N)=N$.