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:

${\mathrm{\Omega }}_j(P):=⟨x\in P\mid x^{p^j}=e⟩$

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

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

$\{e\}={\mathrm{\Omega }}_0(P)\le {\mathrm{\Omega }}_1(P)\le \dots \le {\mathrm{\Omega }}_r(P)=P$

The $\mathrm{\Omega }$-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 ${\mathrm{\Omega }}_j(P)$, for all finite $j$, is the whole group $P$. It may still happen that ${\mathrm{\Omega }}_j(P)=P$ for some finite $j$.

Subgroup properties satisfied

All the ${\mathrm{\Omega }}_j$ are clearly characteristic subgroups, and in fact, they're all fully characteristic subgroups: any endomorphism of $P$ sends each ${\mathrm{\Omega }}_j(P)$ to within ${\mathrm{\Omega }}_j(P)$. Even more strongly, all the ${\mathrm{\Omega }}_j$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 ${\mathrm{\Omega }}_j(Q)\le {\mathrm{\Omega }}_j(P)$.

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 ${\mathrm{\Omega }}_j$ twice is equivalent to applying it once. In other words, for any $P$, ${\mathrm{\Omega }}_j({\mathrm{\Omega }}_j(P))={\mathrm{\Omega }}_j(P)$.