Sylow and TI implies SCDIN
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Sylow TI-subgroup) must also satisfy the second subgroup property (i.e., SCDIN-subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about Sylow TI-subgroup|Get more facts about SCDIN-subgroup
Suppose is a finite group, is a prime, and is a -Sylow subgroup of , such that is a TI-subgroup (?) of : any conjugate of is either equal to or intersects trivially. Then is a SCDIN-subgroup of : any two elements of that are conjugate by in are conjugate by some element in , and moreover, and have the same element-wise action on .
In fact, something stronger is true: itself is in .
- Sylow and TI implies CDIN
- Sylow implies WNSCDIN: Combines Sylow implies pronormal and pronormal implies WNSCDIN
- Sylow implies MWNSCDIN: Combines Sylow implies pronormal and pronormal implies MWNSCDIN
- Abelian and Sylow implies SCDIN: Combines Sylow implies pronormal and abelian and pronormal implies SCDIN
CONVENTION WARNING: This article/section uses the right-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.
Given: A finite group , a prime , a -Sylow subgroup of such that the intersection of with any distinct conjugate of itself is trivial.
To prove: If are conjugate by , then and are conjugate in . In fact, .
Proof: If or is empty or has only one element, the same is true for the other, and in this case we are done. So, we can assume that neither nor is contained in the trivial group.
By fact (1), there exist -Sylow subgroups such that is a tame intersection, and elements such that and for .
However, since is TI, its intersection with any other Sylow subgroup (note also fact (2)) is trivial. Since , we are forced to have that all the are nontrivial, and hence, equal to . This forces for all , hence . Hence, , which is the product of the s, is in , and and are conjugate via an element in .