Invariance implies strongly intersection-closed

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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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Verbal statement

Any subgroup property that arises as an invariance property with respect to the function restriction formalism is strongly intersection-closed, viz it is both intersection-closed and identity-true.

Symbolic statement

Let p be a function property (that is, a property of functions from a group to itself). let I be a (possibly empty) indexing set. Let H_i is a family of subgroups of G indexed by I. Assume that for every function f on G satisfying p, f(H_i)H_i (viz H_i satisfies the invariance property for p).

Then, if H denotes the intersection of all H_is, H also satisfies the invariance property for p. In other words, whenever f is a function on G satisfying p, f(H)H.


Definitions used

Invariance property


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 G be a group and I an indexing set. Let H_i be a family of subgroups of G indexed by I. Suppose, for every function f on G with property p, it is true that each H_i is invariant under f. Let H denote the intersection of all H_is. Then, we need to show that H is also invariance under every f satisfying p.

To show this, let f be a function satisfying p and x be in H. Then, for each H_i, f(H_i)H_i, hence f(x) is contained in H_i. Since this is true for every i, f(x) is also contained in the intersection of all the H_is and hence f(x) is in H. Hence proved.

Particular cases of proof

The general proof technique can also be seen explicitly in some cases: