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 $\Omega _{j}(P)=P$. However, there may exist smaller $j$ for which $\Omega _{j}(P)=P$.

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

$\{e\}=\Omega _{0}(P)\leq \Omega _{1}(P)\leq \dots \leq \Omega _{r}(P)=P$

The $\Omega$-subgroups may also be studied for a (possibly infinite) p-group.

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 $\Omega _{j}(P)$.

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 $Q\leq P$ is a subgroup, then $\Omega _{j}(Q)\leq \Omega _{j}(P)$.

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 $P$, $\Omega _{j}(\Omega _{j}(P))=P$.