Omega subgroups not are prehomomorph-contained

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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)</math>. However, \Omega_1(P) is not contained in K.