Invariance implies strongly intersection-closed

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 $$p$$, $$f(H_i)$$ &sube; $$H_i$$ (viz $$H_i$$ satisfies the invariance property for $$p$$).

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

Examples

 * 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.

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 $$H_i$$s. 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 $$H_i$$, $$f(H_i)$$ &sube; $$H_i$$, 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 $$H_i$$s 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:


 * Normality is strongly intersection-closed
 * Characteristicity is strongly intersection-closed
 * Strict characteristicity is intersection-closed
 * Full characteristicity is intersection-closed