Subnormality is finite-relative-intersection-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., finite-relative-intersection-closed subgroup property)
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Statement

Property-theoretic statement

The property of being a subnormal subgroup is finite-relative-intersection-closed.

Statement with symbols

Suppose H, K are subgroups of G such that H is a subnormal subgroup of G, and K is a subnormal subgroup inside some subgroup L containing both H and K. Then, H \cap K is subnormal in G.

More specifically, if H is h-subnormal in G and K is k-subnormal in L, then H \cap K is (h + k)-subnormal in G.

Related facts

Applications

Facts used

  1. Subnormality satisfies transfer condition: If A \le G is a-subnormal and B \le G, then A \cap B is a-subnormal in B.
  2. Subnormality is transitive: If A \le B \le G are groups, and A is a-subnormal in B and B is b-subnormal in G, then A is (a + b)-subnormal in G.# Transitive and transfer condition implies finite-relative-intersection-closed
  3. Transitive and transfer condition implies finite-relative-intersection-closed

Proof

Given: H, K \le G. H is h-subnormal in G and K is k-subnormal in a subgroup L of G containing both H and K.

To prove: H \cap K is (h + k)-subnormal in G.

Proof:

  1. (Facts used: fact (1); Given data used: H \le L, K is k-subnormal in L): H \cap K is k-subnormal in H: K is k-subnormal in L, and H \le L, so by fact (1), K \cap H is k-subnormal in H.
  2. (Facts used: fact (2); Given data used: H is h-subnormal in G): H \cap K is k-subnormal in H and H is h-subnormal in G, so by fact (2), H \cap K is (h + k)-subnormal in G.