Join-transitively subnormal of normal implies finite-conjugate-join-closed subnormal
This article describes a computation relating the result of the Composition operator (?) on two known subgroup properties (i.e., Join-transitively subnormal subgroup (?) and Normal subgroup (?)), to another known subgroup property (i.e., Finite-conjugate-join-closed subnormal subgroup (?))
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Statement with symbols
Suppose are subgroups such that is a normal subgroup of and is a join-transitively subnormal subgroup of (in other words, the join of with any subnormal subgroup of is subnormal). Then, is a finite-conjugate-join-closed subnormal subgroup of : a join of finitely many conjugate subgroups of in is again subnormal.
- 2-subnormal implies join-transitively subnormal
- 3-subnormal implies finite-conjugate-join-closed subnormal
- 2-subnormality is conjugate-join-closed
- Join-transitively subnormal of normal satisfies finite-conjugate-join-closed
- Join-transitively subnormal implies finite-automorph-join-closed subnormal
- Finite-automorph-join-closed subnormal of normal implies finite-conjugate-join-closed subnormal
Given: , such that is normal in and is join-transitively subnormal in .
To prove: For any finite set , the subgroup , defined as the join of subgroups , is subnormal in .
- Each is join-transitively subnormal in : Conjugation by defines an automorphism of , since is normal in . Thus, each is the image of under an automorphism of . Since automorphism preserve subgroup properties, each is join-transitively subnormal in .
- is subnormal in : Since is a finite set, is the join of finitely many subgroups , each of which is join-transitively subnormal in . By induction, we see that is also join-transitively subnormal in ; in particular, it is subnormal in .
- is subnormal in : Since is subnormal in and is normal in , we obtain that is subnormal in .
Proof using given facts
The proof follows directly by piecing together facts (1) and (2).