Analysis of Thompson's critical subgroup theorem

Thompson's critical subgroup theorem is an extremely important result in the theory of finite $$p$$-groups: it allows us to, starting with any finite $$p$$-group, obtain a characteristic subgroup that is substantially less complicated in structure, but at the same time, is reasonably big in the whole $$p$$-group (namely, it is self-centralizing and also coprime automorphism-faithful).

The critical subgroup theorem is somewhat different from other theorems that guarantee the existence of characteristic subgroups, because there is no natural choice of a subgroup-defining function that yields the critical subgroup. Rather, we get a collection of possible critical subgroups. In this article, we discuss various kinds of nilpotent groups, and what can be said about critical subgroups in these.

Definition of critical subgroup
A subgroup $$H$$ of a group $$G$$ is critical if it is characteristic in $$G$$ and satisfies the following:


 * 1) $$\Phi(H) \le Z(H)$$: In other words, $$H/Z(H)$$ is elementary Abelian
 * 2) $$[G,H] \le Z(H)$$
 * 3) $$C_G(H) \le Z(H)$$: In other words, $$H$$ is a self-centralizing subgroup

Thompson's constructive procedure
Thompson's constructive procedure has two cases (for a starting group $$G$$):


 * 1) If there exists a maximal among Abelian normal subgroups that is also characteristic, then that is a critical subgroup. Hence, we get an Abelian critical subgroup.
 * 2) Consider any subgroup maximal among Abelian characteristic subgroups $$H$$. Then, intersect its centralizer with the inverse image of $$\Omega_1(Z(G/H))$$ of the quotient group. This is a critical subgroup, with the center equal to $$H$$.

Constraints on critical subgroups
The self-centralizing property implies that, in general, any critical subgroup must contain the center of the group. Moreover, for a non-Abelian group, any critical subgroup must contain the center as a proper subgroup.

Another thing we can do is classify critical subgroups by their center. Given a group $$G$$, we want to find all the critical subgroups $$H \le G$$ such that $$Z(H) = K$$, where $$K$$ is an Abelian characteristic subgroup of $$G$$ containing $$Z(G)$$.

Then, the set of subgroups $$H$$ is precisely the set of subgroups satisfying the following five constraints (the final constraint is automatically satisfied when $$K$$ is maximal among Abelian characteristic subgroups):


 * $$H$$ is characteristic in $$G$$
 * $$H$$ is contained in $$C_G(K)$$
 * $$H$$ contains $$K$$
 * The image $$H/K$$ is contained in $$\Omega_1(Z(G/K))$$
 * $$Z(H)$$ is contained in $$K$$

In the particular case where $$K = C_G(K)$$, of course, the only possibility is $$H = K$$.

Limitations of Thompson's procedure
Thompson's procedure cannot construct all possible critical subgroups. There are two limitations:


 * 1) Limitations on center: It can only construct critical subgroups whose center is maximal among Abelian characteristic subgroups.
 * 2) Limitations on choice of group given its center: It constructs the largest possible critical subgroup for that particular choice of center.

More precisely, a critical subgroup that can be yielded by Thompson's procedure is termed a constructibly critical subgroup. A subgroup $$H$$ of $$G$$ is termed constructibly critical if it satisfies the following two conditions:


 * 1) The center $$K := Z(H)$$ is maximal among Abelian characteristic subgroups of $$G$$
 * 2) $$H$$ is the intersection of $$C_G(K)$$ with the inverse image in $$G$$ of the subgroup $$\Omega_1(Z(G/K))$$ of $$G/K$$.

Every Abelian critical subgroup is constructibly critical, but as we shall see here, there are non-Abelian critical subgroups that are not constructibly critical.

For Abelian groups
For an Abelian group, there is only one possible critical subgroup: the whole group itself. That's because any critical subgroup is self-centralizing, and hence must contain the center of the whole group. But an Abelian group is its own center.

For Frattini-in-center groups
If $$G$$ is a $$p$$-group with the property that $$\Phi(G) \le Z(G)$$, i.e. $$G$$ is a Frattini-in-center group, then $$G$$ is a critical subgroup of itself. Let's check the three conditions:


 * 1) $$\Phi(G) \le Z(G)$$: This is by assumption
 * 2) $$[G,G] \le Z(G)$$: That's because $$[G,G] \le \Phi(G)$$ for any finite $$p$$-group $$G$$
 * 3) $$C_G(G) \le Z(G)$$: In fact, they're equal by definition

In fact, a group is a critical subgroup of itself if and only if it is a Frattini-in-center group. In particular, any special group is a critical subgroup of itself, and in particular, any extraspecial group is a critical subgroup of itself.

This doesn't immediately, however, rule out the possibility of other critical subgroups. Let's try to figure out constraints on a critical subgroup $$H$$ of $$G$$.

First, $$H$$ must contain $$Z(G)$$. Let's see what we can say about conditions (1) and (2) for $$H$$ to be a characteristic subgroup, given that $$Z(G) \le H$$:


 * 1) Always satisfied: Since $$Z(G) \le H$$, $$Z(G) \le Z(H)$$. So $$H/Z(H)$$ is a subquotient of $$G/Z(G)$$, hence it is elementary Abelian. Thus, $$\Phi(H) \le Z(H)$$.
 * 2) Always satisfied: Since $$Z(G) \le H$$, we conclude that $$Z(G) \le Z(H)$$. Also, we know that $$[G,H] \le [G,G] \le \Phi(G) \le Z(G)$$, so $$[G,H] \le Z(H)$$. Thus, any subgroup $$H$$ containing $$Z(G)$$ satisfies the property $$[G,H] \le Z(H)$$.
 * 3) Not necessarily satisfied: It is not necessarily true that any subgroup $$H$$ containing $$Z(G)$$ is self-centralizing.

The upshot is that for Frattini-in-center groups, a critical subgroup is the same as a characteristic self-centralizing subgroup. The other two conditions are automatically guaranteed.

The example of the dihedral group
Consider the dihedral group with eight elements. This is an extraspecial group. It has two ccritical subgroups: the whole group, and the unique cyclic subgroup of order four.

Recall that Thompson's procedure can yield a particular critical subgroup only if the center of that critical subgroup is maximal among Abelian characteristic subgroups. In the dihedral group, there is only one subgroup maximal among Abelian characteristic subgroups: the cyclic group of order four. This is also a self-centralizing subgroup, so the only critical subgroup with this as center is the group itself. Thus, Thompson's procedure can only yield the cyclic group of order four, and fails to yield the whole group.

This group thus gives examples of the following:


 * 1) A non-Abelian group with an Abelian critical subgroup
 * 2) A group with two possible critical subgroups, one of which is contained in the other
 * 3) A group with a critical subgroup that cannot be obtained by an application of the constructive procedure of Thompson's critical subgroup theorem

The example of the quaternion group
Consider the quaternion group. This is an extraspecial group, and it has a unique critical subgroup: the whole group. This group arises from the constructive procedure for Thompson's critical subgroup theorem. First, none of the subgroups that are maximal among Abelian normal subgroups is characteristic. Taking any of them, and using the procedure of Thompson's critical subgroup theorem, yields the whole group.

The non-Abelian groups of prime-cubed order and prime exponent
There are two non-Abelian groups of order $$p^3$$, for an odd prime $$p$$. Of these, precisely one has exponent $$p$$. This can be described in many ways, such as:


 * 1) It is the semidirect product of an elementary Abelian group of order $$p^2$$, via an automorphism of order $$p$$.
 * 2) It is the group $$U(3,p)$$ of upper triangular unipotent matrices over the field of $$p$$ elements.

It turns out that the only characteristic subgroups of this group are the trivial subgroup, the whole group, and the center. Thus, there's only one possibility for a critical subgroup: the whole group.

Let's see how we get the whole group using the constructive procedure of Thompson's critical subgroup theorem:


 * 1) First, we look at all the maximal among Abelian normal subgroups. There are various subgroups of order $$p^2$$. However, none of these is characteristic, so we have no hope of getting a critical subgroup this way.
 * 2) Next, we look at all the maximal among Abelian characteristic subgroups. There's precisely one such subgroup: the center. Applying the procedure to this subgroup yields the whole group as the critical subgroup (both  $$C_G(Z(G))$$ and the inverse image of $$\Omega_1(Z(G/Z(G)))$$ are the whole group $$G$$).

The other non-Abelian group of prime-cubed order
The other non-Abelian group of prime-cubed order can be described as the semidirect product of the cyclic group of order $$p^2$$ and a cyclic group of order $$p$$, acting as automorphisms. This group has four characteristic subgroups: the trivial subgroup, the whole group, the center (which is of order $$p$$) and an elementary Abelian subgroup of order $$p^2$$, generated by the multiples of $$p$$ in the subgroup of order $$p^2$$, and the automorphism element of order $$p$$.

Thus, there are two critical subgroups:


 * 1) The whole group
 * 2) The characteristic subgroup of order $$p^2$$

Just as was the case with the dihedral group, it is only the latter critical subgroup that emerges from the procedure of Thompson's critical subgroup theorem. In fact, it is the only maximal among Abelian characteristic subgroups, and is also a self-centralizing subgroup.

Odd-order group with cyclic maximal subgroup
For an odd prime $$p$$, there exists a unique non-Abelian $$p$$-group with a cyclic maximal subgroup. This group has nilpotence class two, and is described as the semidirect product of a cyclic group with a group of order $$p$$ acting as automorphisms.

This is a group of nilpotence class two: the center and commutator subgroup are both of order $$p$$. However, it is not a Frattini-in-center group: the Frattini subgroup has index $$p^2$$ in the whole group, and is therefore strictly bigger than the center for groups of order $$p^4$$ or higher. Hence, it is not a critical subgroup of itself.

It turns out that $$G$$ has a unique subgroup that is maximal among Abelian characteristic subgroups -- in fact, it is also a maximal subgroup, and is isomorphic to $$\mathbb{Z}/p^{n-2}\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$$. This is also a self-centralizing subgroup. Thus, by the observations made earlier, this is the only possible critical subgroup that Thompson's procedure can yield.

Let us now study more closely what the possible critical subgroups are.

Groups with an Abelian Frattini subgroup
If $$P$$ is a $$p$$-group such that $$\Phi(P)$$ (the Frattini subgroup of $$P$$) is an Abelian group, then the group $$C_P(\Phi(P))$$ is a critical subgroup (though it may not necessarily be Abelian).

One example is the maximal class group of order $$p^{p+1}$$ described as follows:


 * 1) It is the wreath product of the cyclic group of order $$p$$ with itself, under the regular action. Hence, it is a group of order $$p^{p+1}$$, with an elementary Abelian normal subgroup $$N$$ of order $$p^p$$, and a cyclic group acting on it by cyclically permuting the coordinates.
 * 2) It is the $$p$$-Sylow subgroup on the symmetric group of order $$p^2$$

Here, the Frattini subgroup, which also equals the commutator subgroup, is an Abelian characteristic subgroup of order $$p^{p-1}$$: it is the subgroup comprising those elements inside $$N$$ whose coordinates add up to zero. Clearly, $$N$$ centralizes this subgroup. We now have two cases:


 * The case $$p = 2$$: In this case, the group is the dihedral group of order eight, and the centralizer of the Frattini subgroup is the whole group, which is critical as a subgroup of itself.
 * The case $$p$$ is an odd prime: In this case, the centralizer of the Frattini subgroup is the subgroup $$N$$. $$N$$ is maximal among Abelian characteristic subgroups; it is also self-centralizing, and is a critical subgroup.

Groups of unipotent upper-triangular matrices
Let $$U_m(p)$$ denote the group of upper-triangular matrices with $$1$$s on the diagonal over the field of $$p$$ elements. The following turn out to be true:


 * When $$m = 2k$$ is even, there is a unique maximum among Abelian characteristic subgroups, and this is also an Abelian subgroup of maximal order. This is the subgroup comprising all block matrices of the form:

$$\begin{pmatrix} I & A \\ 0 & I \end{pmatrix}$$

where all the blocks are $$k \times k$$ in size. The group is isomorphic to the elementary Abelian group of order $$p^{k^2}$$.
 * When $$m = 2k + 1$$ is odd, there are two Abelian subgroups of maximum order, related by an outer automorphism. These are the rectangle groups.