Omega subgroups of group of prime power order

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Suppose P is a finite p-group, i.e. a group of prime power order where the prime is p. Then, we define:

\Omega_j(P) := \langle x \in P \mid x^{p^j} = e \rangle

In other words, it is the subgroup generated by all elements whose order divides p^j.

If the exponent of P is p^r, then \Omega_r(P) = P. However, there may exist smaller j for which \Omega_j(P) = P.

The \Omega-subgroups form an ascending chain of subgroups:

\{ e \} = \Omega_0(P) \le \Omega_1(P) \le \dots \le \Omega_r(P) = P

The \Omega-subgroups may also be studied for a (possibly infinite) p-group. Since every element in a p-group, by definition, has order a power of p, the union of the \Omega_j(P), for all finite j, is the whole group P. It may still happen that \Omega_j(P) = P for some finite j.

Subgroup properties satisfied

All the \Omega_j are clearly characteristic subgroups, and in fact, they're all fully characteristic subgroups: any endomorphism of P sends each \Omega_j(P) to within \Omega_j(P). Even more strongly, all the \Omega_js are homomorph-containing subgroups, and for P a finite p-group, they are thus also isomorph-free subgroups.

Further information: Omega subgroups are homomorph-containing

Subgroup-defining function properties


This subgroup-defining function is monotone, viz the image of any subgroup under this function is contained in the image of the whole group

If Q \le P is a subgroup, then \Omega_j(Q) \le \Omega_j(P).


This subgroup-defining function is idempotent. In other words, applying this twice to a given group has the same effect as applying it once

Applying \Omega_j twice is equivalent to applying it once. In other words, for any P, \Omega_j(\Omega_j(P)) = \Omega_j(P).