Every finite group has an abelian normal subgroup whose order is divisible by the orders of all abelian normal subgroups

Statement
Suppose $$G$$ is a finite group. Then, $$G$$ has an fact about::abelian normal subgroup $$N$$ whose order is a multiple of the order of any abelian normal subgroup of $$G$$. Equivalently, the index of $$N$$ divides the index of every abelian normal subgroup of $$G$$.

Any such $$N$$ must also be fact about::maximal among abelian normal subgroups, though not every subgroup that is maximal among abelian normal subgroups necessarily has this largest order.

Note that $$N$$ is not unique, and we may not be able to always find a characteristic subgroup $$N$$ that does the job.

Opposite facts

 * Finite group need not have an abelian subgroup whose order is divisible by the orders of all abelian subgroups

Similar facts

 * Every finite group has an abelian characteristic subgroup whose order is divisible by the orders of all abelian characteristic subgroups

Closely related facts

 * Degree of irreducible representation divides index of abelian normal subgroup