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

Statement
Suppose $$G$$ is a group, $$M$$ is a fact about::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, $$N \le C_G(N)$$, so $$C_G(N) = N(M \cap C_G(N)) = NC_M(N)$$. Hence, $$C_G(N) = N$$.