Full invariance does not satisfy image condition

From Groupprops
Jump to: navigation, search
This article gives the statement, and possibly proof, of a subgroup property (i.e., fully invariant subgroup) not satisfying a subgroup metaproperty (i.e., image condition).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about fully invariant subgroup|Get more facts about image condition|


Suppose G is a group, K is a fully invariant subgroup of G, and \varphi:G \to H is a surjective homomorphism. Then, \varphi(K) need not be fully invariant in H.


Example of an Abelian group of prime-cube order

(This example uses additive notation).

Suppose G is the direct product of a cyclic group A of order p and a cyclic group of order B of order p^2. Define:

\{ x \in G \mid px = 0 \}.

N := \{ y \in G \mid \exists x, px = y \}.

  • Observe that K is fully invariant in G (more generally, all omega subgroups are fully invariant). However, \varphi(K) is a subgroup of order p in \varphi(G) which is elementary abelian of order p^2 -- hence \varphi(K) is not fully invariant in \varphi(G).

Example of a non-abelian group of prime-cube order

Further information: Prime-cube order group:p2byp, Subgroup structure of prime-cube order group:p2byp

Let p be an odd prime. Suppose A is a cyclic group of order p^2 and B is a cyclic group of order p, with B acting on A via multiplication by p+1. Then, the semidirect product of A by B is a non-Abelian group of order p^3. Call this group P. Define \Omega_1(P) (see omega subgroups of a group of prime power order) as the subgroup generated by all elements of order p in P. By the fact that Omega-1 of odd-order class two p-group has prime exponent, \Omega_1(P) is a subgroup of prime exponent. This forces it to be a subgroup of order p^2 generated by the elements of B and the multiples of p in A. All the omega subgroups are fully characteristic, so \Omega_1(P) is fully characteristic.

The center of P, namely Z(P), simply comprises the multiples of p in A. Thus, in the quotient map P \to P/Z(P), the image of \Omega_1(P) is cyclic of order p, while the whole group is elementary Abelian of order p^2. Thus:

  • \Omega_1(P) is fully characteristic in P.
  • The image of \Omega_1(P) in P/Z(P) is not fully characteristic in P/Z(P).