Omega subgroups not are variety-containing

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., omega subgroups of group of prime power order) does not always satisfy a particular subgroup property (i.e., variety-containing subgroup)
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Statement

We can have a group of prime power order (i.e., a finite $p$-group) $P$ such that the first omega subgroup $\Omega _{1}(P)$, defined as:

$\Omega _{1}(P):=\langle x\mid x^{p}=e\rangle$,

is not a variety-containing subgroup (i.e., Variety-containing subgroup of group of prime power order (?)) of $P$: there exists a subgroup $H$ of $P$ isomorphic to a subgroup of $\Omega _{1}(P)$ but that is not contained in $\Omega _{1}(P)$.

In particular, because of the equivalence of definitions of variety-containing subgroup of finite group, $\Omega _{1}(P)$ need not be a Subisomorph-containing subgroup (?) and it need not be a Variety-containing subgroup (?).

Related facts

• Omega subgroups are homomorph-containing

Proof

Further information: wreath product of groups of order p

Suppose $A$ is a wreath product of groups of order p, i.e., $A$ is a group of order $p^{p+1}$ obtained as the semidirect product of an elementary abelian group of order $p^{p}$ by a cyclic group of order $p$ acting as automorphisms. $A$ is isomorphic to the $p$-Sylow subgroup of the symmetric group of degree $p^{2}$. In particular, $A$ has a cyclic subgroup of order $p^{2}$.

Suppose $B$ is a cyclic group of order $p^{2}$.

Define:

$P:=A\times B$.

Then, $\Omega _{1}(P)=A\times C$ where $C$ is the subgroup of order $p$ in $B$. Consider the subgroup $H=\{e\}\times B$. $H$ is isomorphic to a subgroup of order $p^{2}$ in $A\times \{e\}$, which in turn is contained in $\Omega _{1}(P)$, but $H$ itself is not contained in $\Omega _{1}(P)$.