Omega subgroups of group of prime power order

Definition

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 $\Omega_j$s 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

Monotonicity

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 $\Omega_j(Q) \le \Omega_j(P)$.

Idempotence

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 $P$, $\Omega_j(\Omega_j(P)) = \Omega_j(P)$.