Subnormality is normalizing join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) satisfying a subgroup metaproperty (i.e., normalizing join-closed subgroup property)
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Suppose H,K \le G are subnormal subgroups, with the property that K \le N_G(H): in other words, K normalizes H. Then the join of subgroups \langle H, K \rangle is also subnormal. Moreover, the subnormal depth of \langle H, K \rangle is bounded from above by the products of subnormal depths of H and K.

Related facts

Facts used

  1. Join of normal and subnormal implies subnormal of same depth: If L is normal in G and K is k-subnormal in G, then KL is subnormal in G with subnormal depth at most k.
  2. Normality is upper join-closed: If a subgroup is normal in two intermediate subgroups, it is normal in their join.
  3. Subnormality satisfies intermediate subgroup condition: More specifically, if A \le B \le G are groups such that A is k-subnormal in G, then A is also k-subnormal in B.
  4. Subnormal subgroup has a unique fastest descending subnormal series, where the series members are obtained by taking successive normal closures.


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Given: A group G, subnormal subgroups H, K \le G such that K \le N_G(H), i.e., K normalizes H. H has subnormal depth h and K has subnormal depth k.

To prove: HK = \langle H, K \rangle is a subnormal subgroup, with subnormal depth at most hk.


Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the descending chain G_i defined by G_0 = G, and G_{i+1} is the normal closure of H in G_i. This is the fastest descending subnormal series for H, and thus, G_h = H. Fact (4) H is h-subnormal in G
2 K normalizes G_i for all i. In particular, for any i, \langle G_i, K \rangle = G_iK. K normalizes H Step (1) Any subgroup of G defined deterministically in terms of H must be invariant under any automorphism that leaves H invariant.
3 For each i, G_{i+1} is normal in G_iK. Fact (2) Steps (1), (2) By construction, G_{i+1} is normal in G_i, and as observed in Step (2), K normalizes G_{i+1}, so G_{i+1} is normal in G_iK (fact (2)).
4 For each i, K is k-subnormal in G_iK Fact (3) K is k-subnormal in G Given-fact combination direct
5 For each i, G_{i+1}K is k-subnormal in G_iK Fact (1) Steps (3), (4) Step-fact combination direct
6 HK is hk-subnormal in G Steps (1), (5) We have a chain:
HK = G_hK \le G_{h-1}K \le \dots \le G_1K \le G_0K = G
where each member is k-subnormal in its successor. This tells us that HK is hk-subnormal in G.


Textbook references

  • A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, More info, Page 387, Section 13.1 (Joins and intersections of subnormal subgroups)
  • Subnormal subgroups of groups by John C. Lennox and Stewart E. Stonehewer, Oxford Mathematical Monographs, ISBN 019853552X, Page 3, Section 1.2 (First results on joins), Theorem 1.2.1, More info