Normal not implies finite-pi-potentially characteristic in finite

Statement
It is possible to have a finite group $$G$$ and a normal subgroup $$H$$ of $$G$$ such that there is no finite group $$K$$ containing $$G$$ and satisfying both the following conditions:


 * Every prime divisor of the order of $$K$$ is also a prime divisor of the order of $$G$$.
 * $$H$$ is a characteristic subgroup of $$K$$.

Facts used

 * 1) uses::Every finite p-group is a subgroup of a finite p-group that is not characteristic in any finite p-group properly containing it

Proof
By fact (1), we can definitely find a finite $$p$$-group $$H$$ with the property that $$H$$ is not a characteristic subgroup in any finite $$p$$-group properly containing it.

Let $$G$$ be the direct product of $$H$$ and a cyclic group of order $$p$$. $$H$$ is thus a normal subgroup of $$G$$. However, for any finite $$p$$-group containing $$G$$, $$H$$ is a proper subgroup of it and therefore not a characteristic subgroup of it. Hence, $$H$$ cannot be made to be characteristic in any finite $$p$$-group $$K$$ that contains $$G$$.