Omega subgroups are variety-containing in regular p-group

Definition

Suppose $P$ is a Regular p-group (?). Then, for any natural number $k$, the subgroup:

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

is a Variety-containing subgroup (?) of $P$ (i.e., a 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 Subhomomorph-containing subgroup (?) and a 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. 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.