I'm trying to get a better handle on characteristic subgroups, and many nice examples are given with some sort of "natural" definition. For example, it's clear that the center, torsion subgroup, and commutator subgroup of a given group are all characteristic, just because of the way they are defined. How can we formalize this "naturality"? The latter two have functors associated to them, but I'm not entirely certain if that's the reason for the subgroups being characteristic (and there isn't a similar functor for the center of a group).

EDIT: Let me explain why I'm asking this question. It's useful to know how various characteristic subgroups interact with direct products, quotients, and other group constructions. Henry's construction below is perhaps the right formalism, but it isn't clear to me what extra that buys us. On the other hand, knowing that a certain subgroup arises as the image of a functor means that we ought to be able to use categorical considerations to determine some properties of this subgroup. So here are a few related questions:

  1. Is it possible to use definable subsets to see how characteristic subgroups act with respect to direct product, quotients, or other group constructions?
  2. What properties does the Comm functor on Grp have? What about the Tors functor on Ab? Furthermore, which properties does a functor have to have in order to define a characteristic subgroup? (Other than the obvious property that F(G) is a subgroup of G for all G!)
  • 1
    $\begingroup$ I don't understand what you're getting at with 1. I don't see any reason to think that all characteristic subgroups can be described using definable subsets; and characteristic subgroups don't behave very well with respect to direct products, free products, quotients etc. $\endgroup$
    – HJRW
    Nov 27, 2009 at 21:56
  • $\begingroup$ I certainly didn't mean to imply that all characteristic subgroups could be described using definable subsets -- but when we know that one is, what extra does this tell us about the group? Similarly, I don't expect all characteristic subgroups to behave nicely with respect to, say, direct products, but the center does, and so does the commutator subgroup. Is there a big-picture reason why these subgroups respect the direct product? $\endgroup$ Nov 28, 2009 at 2:28

4 Answers 4


The commutator subgroup of a group is given by a functor on the category whose objects are groups and whose morphisms are all homomorphisms. We can say the similar statement for the torsion subgroup of an abelian group, and this is why the subgroups are fully invariant and not just characteristic. The center is not fully invariant (see e.g., the dihedral group of order two times four), and it is only given by a functor on the core of Grp, i.e., the category whose objects are groups and whose morphisms are isomorphisms.

If you have an endofunctor on the category of (abelian) groups, whose morphisms are all homomorphisms, such that objects are taken to subobjects, then those subobjects are fully invariant (hence characteristic) subgroups. If you have the endofunctor that is only defined on the core of Grp (or Ab), such that objects are taken to subobjects in Grp (not in the core), then the subobjects are characteristic subgroups. Proof: look at what the functor does to an endomorphism of the source.

  • $\begingroup$ Thanks -- that's quite helpful! I'm still a category theory newbie, and it hadn't occurred to me to think of the center as a functor on the core of Grp. But I'm glad that my suspicion that the naturality stemmed from functorality was correct. $\endgroup$ Nov 28, 2009 at 2:35
  • 1
    $\begingroup$ @GabeCunningham This answer not quite true about the center. The center is given as a functor on the category of groups and surjective group homomorphism (not just isomorphisms). $\endgroup$ Apr 4, 2016 at 0:00

This isn't particularly deep, but one way of getting at these subgroups is via definable subsets. A first-order predicate defines a subset of a group. For instance, the predicate p_g(x)={there exists y, yxy^{-1}=g} defines the conjugacy class of the element g. The predicate p_g used a coefficient, g, but a subset defined by a coefficient-free predicate will clearly be invariant under any group automorphism.

Now, the centre of a group is easily seen to be a definable subset, on the nose.

The other two examples probably aren't actually definable, in general, but they are unions of definable subsets. I believe it follows from an as-yet-unpublished theorem of Bestvina and Feighn that the commutator subgroup of a free group is not definable. But the commutator subgroup is of course the infinite union of subsets defined by predicates of the form

q_n(x)={there exist a_1,....,a_2n , x=[a_1,a_2]...[a_{2n-1},a_2n]}.

Alternatively, you can just think of it as the subgroup generated by the definable set that consists of all commutators - as this set is invariant, the subgroup is invariant too.

As for the torsion subgroup (in the abelian case), it's the union of subsets defined by predicates of the form


over all n. I don't know an example, but there's no particular reason to think that it can be defined by a single predicate. (Note that you're only allowed to quantify over group elements, so a predicate like {there exists n, x^n=1} isn't allowed when discussing definable sets. Although it's a perfectly good proof that the torsion subgroup is characteristic!)

Regarding verbal subgroups, they are precisely the subgroups generated by definable subsets with predicates of a particular form.


I have plenty to say, so to simplify matters, I'll use the term "subgroup-defining function" for an isomorphism-invariant mapping that associates a unique subgroup to each group. [NOTE: It is not really a "function" because the "domain" and "range" are not sets.]

Any subgroup-defining function yields a characteristic subgroup, because the isomorphism-invariance implies invariance under automorphisms. However, it is not usually the case that a subgroup-defining function gives an endofunctor (in the sense that Scott describes), i.e., it is not necessary that it obey covariance with arbitrary group homomorphisms.

There are a few special examples of endofunctors, notably the commutator subgroup, members of the derived series, members of the lower central series, and other verbal definitions (that give verbal subgroups). In particular, if a subgroup-defining function is an endofunctor, it should, as Scott mentions, output a fully invariant subgroup, which is a weaker condition than being verbal but still stronger than being characteristic.

In fact, most subgroup-defining functions of interest are not endofunctors, and this might be related to the fact that category-theoretic, or functorial, thinking, does not come up much in the study of groups. Subgroup-defining functions such as the center, member of the upper central series, Frattini subgroup, Fitting subgroup, socle, are not endofunctors. Moreover, the subgroup they yield are not necessarily fully invariant. Interestingly, many of these subgroups are still strictly characteristic subgroups (also called distinguished subgroups) -- they are sent to within themselves by surjective endomorphisms of the group. However, this is not true for all of them.

The next question might be: is there some sense in which the subgroups obtained through easy-to-write subgroup-defining functions are more special than arbitrary characteristic subgroups? There are two related ideas that we can borrow from logic to define notions of "uniqueness" stronger than characteristicity:

  1. There should be no other subgroup of the group such that the theories of the two group-subgroup pairs are "equivalent" in whatever logic we are working with. In first-order logic, there should be no other subgroup whose embedding in the whole group is elementarily equivalent to that of the given subgroup.

  2. The subgroup can be defined (as a subset of the group) in the pure theory of the group, using whatever logic we are working with. In first-order logic, this is what we'd call a "purely definable subgroup" (or "definable subgroup" -- the purely is to emphasize that no additional structure has been tacked on to the group). As mentioned in Henry's reply, the center is purely definable (more generally, all members of the finite part of the upper central series are) but the commutator subgroup is not purely definable for free groups.

  3. Even more generally, we may want that there is a pure definition that works for the subgroup-defining function, as opposed to just requiring that the output of the subgroup-defining function for each input group is definable in that group.

For each logic, (2) is stronger than (1). For instance, a purely definable subgroup cannot be elementarily equivalent to any other subgroup. Moreover, the more powerful the logic, the weaker the notions (1) and (2) in that logic.

[NOTE: That (2) in general is stronger than (1) can be seen from the analogous situation for real numbers: any two real numbers can be distinguished by a first-order predicate in the theory of real numbers, so every real number is "unique" in the sense of (1). However, there are only countably many definable real numbers, so almost all real numbers are not uniquely definable in the sense of (2).]


One nice family of subgroups is the one of those which are 'verbal'; this does not include the center or the torsion subgroup though. On the other hand, you can get the torsion subgroup as the set of elements which satisfy (one of a) set of equations, and this generalizes to other types of equations.


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