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 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:

$H\leq \Omega _{k}(P)=\langle x\mid x^{p^{k}}=e\rangle$.

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 $p^{l}$, $l\leq k$. Then, $\langle x\rangle \cong \langle y^{p^{k-l}}\rangle$. Thus, $\langle x\rangle$ is isomorphic to a subgroup of $H$, so $\langle x\rangle \leq H$. Thus, $\Omega _{k}(P)=H$.