Omega subgroups are isomorph-free

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) uses::Omega subgroups are homomorph-containing
 * 2) uses::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.