# 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)}
View subgroup property satisfactions for subgroup-defining functions $|$ View subgroup property dissatisfactions for subgroup-defining functions

## 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.

## Proof

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