Omega subgroups are variety-containing in regular p-group

Definition
Suppose $$P$$ is a fact about::regular p-group. Then, for any natural number $$k$$, the subgroup:

$$\Omega_k(P) := \langle x \mid x^{p^k} = e \rangle$$

is a fact about::variety-containing subgroup of $$P$$ (i.e., a fact about::variety-containing subgroup of group of prime power order): it contains any subgroup of $$P$$ that is in the subvariety of the variety of groups generated by $$\Omega_k(P)$$.

In particular, $$\Omega_k(P)$$ is a fact about::subhomomorph-containing subgroup and a fact about::subisomorph-containing subgroup of $$P$$.

Related facts

 * Subisomorph-containing implies omega subgroup in group of prime power order
 * Omega subgroups not are subisomorph-containing: In particular, the analogous statements break down for groups that are not regular.

Facts used

 * 1) uses::Omega subgroup equals set of elements of the exponent dividing the prime power in regular p-group

Proof
By fact (1), $$\Omega_k(P)$$ is precisely the subset of $$P$$ comprising the elements of $$P$$ whose order divides $$p^k$$.

Let $$\mathcal{V}$$ be the variety of all groups in which the order of every element divides $$p^k$$. Note that this is a variety and it contains $$\Omega_k(P)$$. Thus, the subvariety generated by $$\Omega_k(P)$$ is contained in $$\mathcal{V}$$. Further, any element of $$\mathcal{V}$$ that is a subgroup of $$P$$ is contained in $$\Omega_k(P)$$. Thus, $$\Omega_k(P)$$ contains all subgroups of $$P$$ in the subvariety it generates.