Purely definable subgroup

This article defines a subgroup property: a property that can be evaluated to true/false given a group and a subgroup thereof, invariant under subgroup equivalence. View a complete list of subgroup properties[SHOW MORE]

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VIEW FACTS THAT:use, or start with, a subgroup satisfying the property|prove, or show that, a subgroup satisfies the property

This subgroup property is the version, in a pure group, of the following subgroup property for logicians: definable subgroup | See other such examples

This is a variation of characteristicity|Find other variations of characteristicity | Read a survey article on varying characteristicity

Definition

A subgroup of a group is said to be purely definable if it is definable as a subset in the first-order theory of the pure group. By pure group, we mean the set equipped only with the structure of the group operations and with no additional first-order data supplied.

By definable subset, we mean that there is a first-order formula with one free variable such that the set of elements satisfying the formula is precisely that subset.

Metaproperties

Metaproperty name	Satisfied?	Proof	Statement with symbols
transitive subgroup property	Yes	pure definability is transitive	If $H\leq K\leq G$ are groups and $K$ is purely definable in $G$ and $H$ is purely definable in $K$, then $H$ is purely definable in $G$. In fact, we can actually compose a formula defining $H$ within $K$ with a formula defining $K$ within $G$.
trim subgroup property	Yes	(obvious)	The trivial subgroup and whole group are purely definable.
strongly finite-intersection-closed subgroup property	Yes	pure definability is strongly finite-intersection-closed	Suppose $G$ is a group and $H,K$ are purely definable subgroups of $G$. Then the intersection $H\cap K$ is also a purely definable subgroup of $G$.
quotient-transitive subgroup property	Yes	pure definability is quotient-transitive	Suppose $G$ is a group and $H\leq K\leq G$ are such that $H$ is purely definable in $G$ and the quotient group $K/H$ is purely definable in $G/H$. Then, $K$ is purely definable in $G$.

Relation with other properties

Stronger properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Intermediate notions
characteristic subgroup of finite group	characteristic subgroup and the whole group is a finite group			|FULL LIST, MORE INFO
verbal subgroup of finite type	image of a word map			|FULL LIST, MORE INFO
finite verbal subgroup		(via verbal subgroup of finite type)	(via verbal subgroup of finite type)	|FULL LIST, MORE INFO
marginal subgroup of finite type				|FULL LIST, MORE INFO

Weaker properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Intermediate notions
purely definably generated subgroup	generated by a purely definable subset		derived subgroup not is purely definable, but it is purely definably generated	|FULL LIST, MORE INFO
elementarily characteristic subgroup	no other elementarily equivalently embedded subgroup			|FULL LIST, MORE INFO
second-order purely definable subgroup	definable in the second-order theory of the group			|FULL LIST, MORE INFO
characteristic subgroup	invariant under all automorphisms			|FULL LIST, MORE INFO
normal subgroup	invariant under all inner automorphisms			Characteristic subgroup|FULL LIST, MORE INFO