Omega subgroups are isomorph-free

This article gives the statement, and possibly proof, of the fact that for any group, the subgroup obtained by applying a given subgroup-defining function (i.e., omega subgroups of a group of prime power order) always satisfies a particular subgroup property (i.e., isomorph-free subgroup)}
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Statement

Statement with symbols

Let $P$ be a group of prime power order. Then, for any natural number $j$, the subgroup:

$\Omega _{j}(P)=\langle g\in P\mid g^{p^{j}}=e\rangle$

is an isomorph-free subgroup.

Facts used

1. Omega subgroups are homomorph-containing
2. Homomorph-containing implies isomorph-free for co-Hopfian subgroup

Proof

The proof follows directly from facts (1) and (2), along with the fact that any finite group is co-Hopfian.