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Descendant not implies subnormal
From Groupprops
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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Contents |
Statement
A descendant subgroup of a group need not be subnormal.
Related facts
- Normality is not transitive
- There exist subgroups of arbitrarily large subnormal depth
- Ascendant not implies subnormal
Definitions used
Descendant subgroup
Further information: Descendant subgroup
Subnormal subgroup
Further information: Subnormal subgroup
Proof
Example of the group of 2-adic integers
Let K be the group of 2-adic integers under addition. This is the inverse limit of the chain:
.
Let G be the semidirect product of K with a group H of order two, acting via the inverse map.
- H is a descendant subgroup of G: Consider a descending chain Kn defined as follows. K0 = K, and Kn is the kernel of the quotient map to
. Define Gn as the semidirect product of Kn with H. Then, the Gn 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 Gns is equal to H, and thus, H is a descendant subgroup of G.
- H is not a subnormal subgroup of G: If H were a k-subnormal subgroup of G, then the image of H in G / Kn would be a k-subnormal subgroup of G / Kn for every n. On the other hand, we know that the image of H in G / Kn has subnormal depth exactly n in G / Kn, which is a contradiction for n > k. (For more on this, refer the example of the dihedral group in: there exist subgroups of arbitrarily large subnormal depth).
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