Invariance implies strongly intersection-closed
This article gives the statement and possibly, proof, of an implication relation between two subgroup metaproperties. That is, it states that every subgroup satisfying the first subgroup metaproperty (i.e., Invariance property (?)) must also satisfy the second subgroup metaproperty (i.e., Strongly intersection-closed subgroup property (?))
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Let be a function property (that is, a property of functions from a group to itself). let be a (possibly empty) indexing set. Let is a family of subgroups of indexed by . Assume that for every function on satisfying , ⊆ (viz satisfies the invariance property for ).
Then, if denotes the intersection of all s, also satisfies the invariance property for . In other words, whenever is a function on satisfying , ⊆ .
- An intersection of normal subgroups is normal. Here, the subgroup property of being normal is the property of being invariant under inner automorphisms, and is hence the invariance property for the function property of being an inner automorphism.
- An intersection of characteristic subgroups is characteristic. Here the subgroup property of being characteristic is the invariance property with respect to the function property of being an automorphism.
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Strongly intersection-closed subgroup property
A subgroup property is termed strongly intersection-closed if given any family of subgroups having the property, their intersection also has the property. Note that just saying that a subgroup property is intersection-closed simply means that given any nonempty family of subgroups with the property, the intersection also has the property.
Thus, the property of being strongly intersection-closed is the conjunction of the properties of being intersection-closed and identity-true, viz satisfied by the whole group as a subgroup of itself.
Proof with symbols
Let be a group and an indexing set. Let be a family of subgroups of indexed by . Suppose, for every function on with property , it is true that each is invariant under . Let denote the intersection of all s. Then, we need to show that is also invariance under every satisfying .
To show this, let be a function satisfying and be in . Then, for each , ⊆ , hence is contained in . Since this is true for every , is also contained in the intersection of all the s and hence is in . Hence proved.
Particular cases of proof
The general proof technique can also be seen explicitly in some cases: