Omega subgroups of group of prime power order

Definition
Suppose $$P$$ is a finite $$p$$-group, i.e. a group of prime power order where the prime is $$p$$. Then, we define:

$$\Omega_j(P) := \langle x \in P \mid x^{p^j} = e \rangle$$

In other words, it is the subgroup generated by all elements whose order divides $$p^j$$.

If the exponent of $$P$$ is $$p^r$$, then $$\Omega_j(P) = P$$. However, there may exist smaller $$j$$ for which $$\Omega_j(P) = P$$.

The $$\Omega$$-subgroups form an ascending chain of subgroups:

$$\{ e \} = \Omega_0(P) \le \Omega_1(P) \le \dots \le \Omega_r(P) = P$$

The $$\Omega$$-subgroups may also be studied for a (possibly infinite) p-group. Since every element in a p-group, by definition, has order a power of $$p$$, the union of the $$\Omega_j(P)$$, for all finite $$j$$, is the whole group $$P$$. It may still happen that $$\Omega_j(P) = P$$ for some finite $$j$$.

Subgroup properties satisfied
All the $$\Omega_j$$ are clearly characteristic subgroups, and in fact, they're all fully characteristic subgroups: any endomorphism of $$P$$ sends each $$\Omega_j(P)$$ to within $$\Omega_j(P)$$. Even more strongly, all the $$\Omega_j$$s are homomorph-containing subgroups, and for $$P$$ a finite $$p$$-group, they are thus also isomorph-free subgroups.

Subgroup-defining function properties
If $$Q \le P$$ is a subgroup, then $$\Omega_j(Q) \le \Omega_j(P)$$.

Applying $$\Omega_j$$ twice is equivalent to applying it once. In other words, for any $$P$$, $$\Omega_j(\Omega_j(P)) = P$$.