Transitive and transfer condition implies finite-relative-intersection-closed

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Statement

Suppose p is a Transitive subgroup property (?) satisfying the Transfer condition (?). Then, p is a Finite-relative-intersection-closed subgroup property (?).

Definitions used

Transitive subgroup property

Further information: Transitive subgroup property

A subgroup property p is termed transitive if whenever L \le H \le G are groups such that L has property p in H, and H has property p in G, we have that L has property p in G.

Transfer condition

Further information: Transfer condition

A subgroup property p is said to satisfy the transfer condition if whenever H, K \le G are subgroups such that H satisfies property p in G, we have that H \cap K satisfies property p in K.

Finite-relative-intersection-closed subgroup property

Further information: Finite-relative-intersection-closed subgroup property

A subgroup property p is termed finite-relative-intersection-closed if whenever H, K \le G are subgroups such that H satisfies property p in G and K satisfies property p in some subgroup containing both H and K, then H \cap K satisfies property p in G.

Examples

Proof

Given: A group G with subgroups H, K such that H satisfies property p in G and K satisfies property p in some subgroup containing both H and K. p is both transitive and satisfies the transfer condition.

To prove: H \cap K satisfies property p in G.

Proof:

  1. (Given data used: K satisfies p in L, p satisfies transfer condition): H \cap K satisfies property p in H: K satisfies property p in L, and H is a subgroup of L. Since p satisfies the transfer condition, we have that K \cap H satisfies property p in H.
  2. (Given data used: p is transitive, H satisfies p in G): By the previous step, we have H \cap K satisfies property p in H, and H satisfies property p in G. Since p is transitive, we obtain that H \cap K satisfies p in G.