# 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:

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

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, $\Omega_k(P) = H$.