Intermediate isomorph-conjugacy is normalizing join-closed
This article gives the statement, and possibly proof, of a subgroup property (i.e., intermediately isomorph-conjugate subgroup) satisfying a subgroup metaproperty (i.e., normalizing join-closed subgroup property)
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- Isomorph-freeness is strongly join-closed
- Intermediate automorph-conjugacy is normalizing join-closed
- Pronormality is normalizing join-closed
CONVENTION WARNING: This article/section uses the left-action convention. The left and right action conventions are equally powerful and statements/reasoning here can be converted to the alternative convention (the main reason being that every group is naturally isomorphic to its opposite group via the inverse map). For more on the action conventions and switching between them, refer to switching between the left and right action conventions.
This proof follows the left-action convention, though it can be stated equally well using the right-action convention (a similar proof for pronormality uses the right-action convention -- the proofs use essentially the same idea). We denote by the operation of conjugation by .
Given: A group , intermediately isomorph-conjugate subgroups such that normalizes . is an isomorphism from to some subgroup of .
To prove: There exists such that .
- (Given data used: is intermediately isomorph-conjugate in ): and are isomorphic subgroups, so, since is intermediately isomorph-conjugate, there exists a such that . Note that .
- (No given data used): Define as a map from to . Now, , because it involves composing with conjugation by an element inside . Thus, .
- (Given data used: normalizes ): We have . Now, by construction. Thus, . Thus, .
- (Given data used: is intermediately isomorph-conjugate in ): Since is intermediately isomorph-conjugate in , the conclusion of step (3) tells us that there exists such that . In particular, , so . Thus, , with . By the conclusion of step (2), this yields .
- Now, consider the product We claim that this works (No problem data used): Clearly, since both and are elements of , so is . Further, . In particular, . Thus, and are conjugate in the subgroup they generate.