Descendant not implies subnormal
This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., descendant subgroup) need not satisfy the second subgroup property (i.e., subnormal subgroup)
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- Normality is not transitive
- There exist subgroups of arbitrarily large subnormal depth
- There exist subgroups of arbitrarily large descendant depth
- Ascendant not implies subnormal
Further information: Descendant subgroup
Further information: Subnormal subgroup
Example of the group of 2-adic integers
Let be the group of 2-adic integers under addition. This is the inverse limit of the chain:
Let be the semidirect product of with a group of order two, acting via the inverse map.
- is a descendant subgroup of : Consider a descending chain defined as follows. , and is the kernel of the quotient map to . Define as the semidirect product of with . Then, the form a descending chain of subgroups, each having index two in its predecessor, so each is normal in its predecessor. The intersection of all the s is equal to , and thus, is a descendant subgroup of .
- is not a subnormal subgroup of : If were a -subnormal subgroup of , then the image of in would be a -subnormal subgroup of for every . On the other hand, we know that the image of in has subnormal depth exactly in , which is a contradiction for . (For more on this, refer the example of the dihedral group in: there exist subgroups of arbitrarily large subnormal depth).
Example of the infinite dihedral group
Suppose is the infinite dihedral group with cyclic maximal subgroup (isomorphic to the group of integers) and a complementary subgroup of order two. Then, is a descendant subgroup, as it is the intersection of the descending chain:
Note that each member has index two in its predecessor, hence is normal in its predecessor.
This example is very similar in nature to the 2-adic example.