# Omega subgroups not are prehomomorph-contained

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., omega subgroups of group of prime power order) does not always satisfy a particular subgroup property (i.e., prehomomorph-contained subgroup)
View subgroup property satisfactions for subgroup-defining functions $|$ View subgroup property dissatisfactions for subgroup-defining functions

## Statement

It is possible to have a group of prime power order $P$ (i.e., a finite $p$-group $P$ for some prime number $p$), a subgroup $K$ of $P$, and a surjective homomorphism $\varphi:K \to \Omega_1(P)$, such that $\Omega_1(P)$ is not contained in $K$.

Analogous examples can be constructed for $\Omega_k(P)$ for any $k \ge 1$.

## Proof

### Example for $k = 1$

Further information: quaternion group, direct product of Q8 and Z2

Let $Q$ be the quaternion group of order eight and $C$ be the cyclic group of order two. Define:

$P := Q \times C$

Let $K = Q \times \{ e \}$.

Then, $K/Z(K) \cong Q/Z(Q)$ which is a Klein four-group. Also, $\Omega_1(P) = \Omega_1(Q) \times \Omega_1(C) = Z(Q) \times C$ which is again a Klein four-group. Hence, $K/Z(K) \cong \Omega_1(P)$, so there is a surjective homomorphism <mah>\varphi:K \to \Omega_1(P)[/itex]. However, $\Omega_1(P)$ is not contained in $K$.