Variety-containing implies omega subgroup in group of prime power order

Definition
Suppose $$P$$ is a group of prime power order, i.e., a finite $$p$$-group for some prime number $$p$$. Suppose $$H$$ is a fact about::variety-containing subgroup of $$P$$: a subgroup of $$P$$ such that any subgroup of $$P$$ isomorphic to a subgroup of $$H$$ is itself contained in $$H$$. Then, $$H$$ is one of the omega subgroups of $$P$$. More specifically, if the exponent of $$H$$ is $$p^k$$, then:

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

Note that by the equivalence of definitions of variety-containing subgroup of finite group, assuming that $$H$$ is a variety-containing subgroup of $$P$$ is equivalent to assuming that it is a subhomomorph-containing subgroup or that it is a subisomorph-containing subgroup.

Proof
Given: A finite $$p$$-group $$P$$, a variety-containing subgroup $$H$$ of $$P$$.

To prove: $$H = \Omega_k(P)$$ for some natural number $$k$$.

Proof: Let $$p^k$$ be the exponent of $$H$$. Then, we clearly have:

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

Next, we show that $$H = \Omega_k(P)$$. Suppose $$x \in P$$ is such that $$x^{p^k} = e$$. Then, since $$p^k$$ is the exponent of $$H$$, there exists $$y \in H$$ such that the order of $$y$$ is $$p^k$$. Suppose the order of $$x$$ is $$p^l$$, $$l \le k$$. Then, $$\langle x \rangle \cong \langle y^{p^{k-l}} \rangle$$. Thus, $$\langle x \rangle$$ is isomorphic to a subgroup of $$H$$, so $$\langle x \rangle \le H$$. Thus, $$\Omega_k(P) = H$$.