Full invariance does not satisfy image condition
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).
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Suppose is a group, is a fully invariant subgroup of , and is a surjective homomorphism. Then, need not be fully invariant in .
Example of an Abelian group of prime-cube order
(This example uses additive notation).
Suppose is the direct product of a cyclic group of order and a cyclic group of order of order . Define:
- (see omega subgroups of a group of prime power order), i.e., is the subgroup comprising all the elements:
- is the quotient map by the normal subgroup (see agemo subgroups of a group of prime power order), i.e., is the quotient map by the subgroup:
- Observe that is fully invariant in (more generally, all omega subgroups are fully invariant). However, is a subgroup of order in which is elementary abelian of order -- hence is not fully invariant in .
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 be an odd prime. Suppose is a cyclic group of order and is a cyclic group of order , with acting on via multiplication by . Then, the semidirect product of by is a non-Abelian group of order . Call this group . Define (see omega subgroups of a group of prime power order) as the subgroup generated by all elements of order in . By the fact that Omega-1 of odd-order class two p-group has prime exponent, is a subgroup of prime exponent. This forces it to be a subgroup of order generated by the elements of and the multiples of in . All the omega subgroups are fully characteristic, so is fully characteristic.
The center of , namely , simply comprises the multiples of in . Thus, in the quotient map , the image of is cyclic of order , while the whole group is elementary Abelian of order . Thus:
- is fully characteristic in .
- The image of in is not fully characteristic in .