Strict characteristicity is strongly intersectionclosed
This article gives the statement, and possibly proof, of a subgroup property satisfying a subgroup metaproperty
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Contents
Statement
Verbal statement
An arbitrary (possibly empty) intersection of strictly characteristic subgroups of a group is strictly characteristic.
Note: The use of the word strongly is to allow the empty intersection as well. We can also say that normality is intersectionclosed and also identitytrue.
Symbolic statement
Let be an indexing set and be a family of characteristic subgroups of indexed by . Then, the intersection, over all in , of the strictly characteristic subgroups , is also a strictly characteristic subgroup of .
Definitions used
Fully invariant subgroup
A subgroup of a group is said to be strictly characteristic, if given any surjective endomorphism of , we have ≤ .
Strongly intersectionclosed
A subgroup property is termed strongly intersectionclosed if given any family of subgroups having the property, their intersection also has the property. Note that just saying that a subgroup property is intersectionclosed simply means that given any nonempty family of subgroups with the property, the intersection also has the property.
Thus, the property of being strongly intersectionclosed is the conjunction of the properties of being intersectionclosed and identitytrue, viz satisfied by the whole group as a subgroup of itself.
Generalizations
The general result (of which this can be viewed as a special case) is that any invariance property is strongly intersectionclosed.
Here, an invariance property is the property of being invariant with respect to a certain collection of functions on the whole group. For strictly characteristic subgroups, the collection of functions is the surjective endomorphisms.
Proof
Let be a family of strictly characteristic subgroups of indexed by . Suppose is the intersection. We need to show that is strictly characteristic in . In other words, for any in and any surjective endomorphism of , we need to show that is also in .
Now we know that since is in , is also in each . By the normality of , is contained in . Hence, is also contained in the intersection of all the s.
This proof generalizes to the one for an arbitrary invariance property. The idea behind the generalization is that we in no way used any information about the nature and structure of endomorphisms  they were a black box for us.