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1. (Weak version) If we have an ultraproduct of groups Gn in a class C, and a surjective homomorphism f : Qn→ω Gn → H where H is a finite group, then when is B also in class C?

2. (Strong version) If we have a projective limit G of a surjective inverse direct system of groups Gn in a class C, and a surjective homomorphism f : G → H where H is a finite group, then when is B also in class C?

The strong version would immediately imply the weak version since the ultraproduct of groups is a quotient of the direct product of the same groups. The answer to the strong version of our question is already well known in certain cases. For example, if C is the class of finite abelian groups, then obviously any projective limit would also be abelian, and hence any quotient would also be abelian. By Corollary 4.2.3 of [RZ00], the case of p-groups is proven. Since each nilpotent group is a product of its p-Sylow subgroups, the case of nilpotent groups follows. Finally, the case of solvable groups is proven by Corollary 4.2.4 of [RZ00]. For all these groups, the answer to the strong version of our question is already known to be affirmative. In this paper, we shall prove that the strong version is true for several other cases as well. Theorem 1.3. Let C be the class of finite semisimple groups, or the class of finite central products of finite quasisimple groups, or the class of finite perfect groups with bounded commutator width. Then any finite quotient of a pro-C group must remain in the class C. If C is the class of finite products of finite almost simple groups, then any finite quotient of a pro-C group is a direct product of almost simple groups and solvable groups. In fact, we also showed that statements of this type are true for certain variants of the classes of groups above. Note that Theorem 1.3 is dependent on the classification of finite simple groups. However, what if we move away from the solvable groups or groups that are close to being simple? For example, if we choose C to be the class of finite perfect groups, then even the weak version of our question is false. For example, in Theorem 2.1.11 of [HP89], Holt and Plesken have constructed specific examples of finite perfect groups Gn with huge centers, such that Gn must have large commutator width. Here the commutator width of a group G refers to the least integer n such that every element of G is a product of at most n commutators. If a sequence of perfect groups has increasing commutator width, then it should be expected that their “limit” of any kind would have “infinite commutator width”, i.e., it is probably not going to be perfect after all. Then it would have a non-trivial abelian quotient, which is not in C. This is indeed the case, as shown in Proposition 1.5 in Section 3 of this paper. Another such example is constructed by Nikolov [Nik04], where the groups Gn are even constructed to be finite perfect linear groups of dimension 15. However, it is worth noting that in both examples by Holt and Plesken (Theorem 2.1.11 of [HP89]) and the one by Nikolov, Qn→ω Gn must either have non-solvable quotients, or abelian quotients. There is nothing in between. In pursuit of this phenomenon, Holt1 asked that, if G is a subdirect product of finitely many perfect groups, must its solvable quotients be abelian? This is a “finite” version of our inquiry, and it was answered by Mayr and Ruˇskuc in the negative [MR19]. It turns out that such a quotient is not necessarily abelian, but must always be nilpotent. In a related manner, Thiel2 asked whether all groups can be achieved as a projective limit of perfect groups. An answer was provided by de Cornulier, but his construction was through an injective directed system of infinite perfect groups. In this regard, we would establish the following result in our paper as well.

1https://mathoverflow.net/questions/280887/subdirect-product-of-perfect-groups 2https://mathoverflow.net/questions/289390/inverse-limits-of-perfect-groups

2 Theorem 1.4. For any finite group G, G is a quotient of the projective limit of a surjective inverse directed system of finite perfect groups. Note that Theorem 1.4 is not dependent on the classification of finite simple groups.

1.2 Outline of the paper In Section 2, we shall establish the proof of Theorem 1.3, using mostly elementary arguments combined with the main results on ultraproducts of finite simple groups from [Yan16]. We can break it down into the following subsections:

• In Section 2.1, we establish properties of a class C such that any pro-C is a direct product of groups in C. This shall come in handy in almost all cases below. • In Section 2.2, we establish the semisimple case of Theorem 1.3. • In Section 2.3, we establish the quasisimple case of Theorem 1.3. • In Section 2.4, we establish the almost simple case of Theorem 1.3. • In Section 2.5, we define a slightly more generalized notion of almost semi-simple groups, and prove a similar result to Theorem 1.3. • In Section 2.6, we establish the case of finite perfect groups with bounded commutator width for Theorem 1.3.

We shall prove Theorem 1.4 in Section 3. The construction here is a bit elaborate, so we shall provide a brief breakdown of the arguments here. We shall show that any finite group can be realized as a quotient of Q Pn, a direct product of a family of finite perfect groups. First of all, if groups in the sequence Pn are perfect groups with increasing commutator width, then their direct product is not perfect, and therefore they would have a non-trivial abelian quotient, as shown in the proposition below.

Proposition 1.5. If {Pn}n∈N is a sequence of finite perfect groups with strictly increasing commutator N width, and ω is any non-principal ultrafilter on , then P = Qn→ω Pn is not perfect.

Proof. Since Pn has strictly increasing commutator width, each Pn must have a commutator width of at least n. Pick gn ∈ Pn such that it cannot be written as the product of less than n commutators. If limn→ω gn ∈ P can be written as the product of d commutators, then we must have {n ∈ N : gn is the product of d commutators} ∈ ω. However, this set is finite, while ω is a non-principal ultrafilter, a contradiction. Hence limn→ω gn is not in the commutator subgroup of P . In particular, P is not perfect. We now establish the generic case. For an arbitrary finite group G, we would construct a surjective homomorphism from a “pro-perfect” group onto G. The homomorphism we constructed would in fact factor through an ultraproduct, as described in the statement of Theorem 1.6.

Theorem 1.6. Given any finite group G, there is a sequence of finite perfect groups {Pn}n∈N and an N ultrafilter ω on such that the ultraproduct Qn→ω Pn has a surjective homomorphism onto G. Since an ultraproduct is a quotient of the direct product, therefore Theorem 1.4 is simply a corollary of Theorem 1.6 above. We first reduce the generic case to the case of a special kind of groups. In 1989, Zelmanov solved the restricted Burnside problem in [Zel90, Zel91]. In particular, the following theorem is true. Theorem 1.7. For arbitrary positive integers d, m, there is a unique finite group B(d, m), such that any finite group of exponent m on d generators is isomorphic to a quotient of B(d, m).

3 So to prove Theorem 1.6, it is enough to show that for arbitrary positive integers d, m, the restricted Burnside group B(d, m) can be realized as an abstract quotient of an ultraproduct of finite perfect groups. We shall provide an explicit construction of these groups Pn. The proof can be decomposed into the following steps:

• In Section 3.1, we construct and discuss the restricted Burnside coproduct, defined on finite groups with exponents dividing m. This is a crucial ingredient for Section 3.3. • In Section 3.2, we discuss some functorial properties of the ultraproduct and its interaction with the restricted Burnside coproduct. This is a crucial ingredient for Section 3.3. • In Section 3.3, we reduce the case of the restricted Burnside group B(d, m) to the case of the cyclic group Z/mZ. In particular, if the cyclic group Z/mZ could be realized as an abstract quotient of an ultraproduct of “nice” finite perfect groups, then so can the restricted Burnside group B(d, m).

• In Section 3.4, we construct a sequence of finite perfect groups Gn.

• In Section 3.5, we construct a sequence of finite abelian groups Mn with exponents dividing m, and each Mn has a Gn action.

• In Section 3.6, we show that the groups Pn = Mn ⋊ Gn are perfect with increasing commutator width. • In Section 3.7, we establish some properties on Qn→ω Mn and its “dual”. • Z Z In Section 3.8, we construct an explicit homomorphism ψ : Qn→ω Mn → /m , which could be Z Z extended to a surjective homomorphism from Qn→ω Pn to /m . 1.3 Notations and Conventions Given a perfect group G and an element g ∈ G, we define its commutator length to be the smallest integer d such that it can be written as a product of at most d commutators. We define the commutator width of G to be the smallest integer d such that every element of G has commutator length at most d. Given a group G, for elements x, y ∈ G, subgroups H,K ≤ G, subset S ⊆ G, we adopt the following notations. Here hSi means the subgroup generated by the subset S.

gh := h−1gh, [g,h] := g−1h−1gh, [H,K] := h[g,h] | g ∈ H and h ∈ Ki, H′ := [H,H], ⋆k S := {s1s2 ...sk | s1,...,sk ∈ S}.

In our paper, we adopt the convention that all group actions are from the right. Given a group G acting on another group H via φ : G → Aut(H), the corresponding semi-direct product is written as H ⋊ G. Elements of this semi-direct product group is written as hg where h ∈ H and g ∈ G, while the action of g on h is written as hg. For h ∈ H,g ∈ G, and subgroup K ≤ H, we adopt the following conventions.

g −1 h := φg(h)= g hg, [h,g] := h−1hg, [K, G] := h[h,g] | h ∈ K,g ∈ Gi.

4 Throughout this paper, all finite groups will be equipped with the discrete topology whenever needed. Let C be a class of finite groups equipped with the discrete topology. Then we say G is a pro-C group if it is the inverse direct limit of a surjective system of groups in C. Other than the inverse direct limit, another kind of limit that concerns us is the ultraproduct. Given any index set I, a collection ω of subsets of I is an ultrafilter if the followings are true:

1. If J1,J2 ∈ ω, then J1 ∩ J2 ∈ ω.

2. If J1 ⊆ J2 as subsets of I, and J1 ∈ ω, then J2 ∈ ω. 3. For any subset J of I, exactly one of J and its complement I − J is in ω.

An ultrafilter is principal if the intersection of all elements of ω is non-empty. Given a sequence of groups {Gn}n∈N, and an ultrafilter ω on N, consider the subgroup Nω of the group : N Qn∈N Gn defined as Nω = {(gn)n∈ ∈ Qn∈N Gn : {gn = e} ∈ ω}, where e is the identity element. This is a normal subgroup, and we denote the quotient Qn∈N Gn/Nω as Qn→ω Gn. This is the ultraproduct of the sequence {Gn}n∈N through the ultrafilter ω. For the element (gn)n∈N ∈ Qn∈N Gn, its quotient image in Qn→ω Gn is denoted as limn→ω gn, called the ultralimit of the sequence (gn)n∈N. For further discussions, see [SB81].

5 2 Projective limits of non-abelian groups

In this section, we consider projective limits of finite groups that are close to being non-abelian finite simple. Throughout this section, we use the term semisimple group to refer to a direct product of non-abelian finite simple groups.

2.1 Preliminary on Profinite groups It is well-known that any projective limits of finite semisimple groups must be a direct product of finite simple groups. (E.g., see Lemma 8.2.3 of [RZ00].) In this preliminary section, we try to generalize such a result to some similar classes of groups. Let C be a class of finite groups. Let P(C) be the class of finite products of groups in C. Let us say that C is a central-simple class of groups if all groups in C are non-abelian, and for any homomorphism f : G → H between groups G, H ∈ C, f(G)CH (f(G)) = H implies that f is trivial or f is an isomorphism. Here CH (f(G)) refers to the centralizer of f(G). We claim that the following result is true. Proposition 2.1. Suppose C is a central-simple class of finite groups. Then any pro-P(C) group is a direct product of groups in C. Now let us prove Proposition 2.1. For any group G, let us say a normal subgroup N of G is C-normal if G/N ∈C, and we write N EC G. Lemma 2.2. Suppose C is a central-simple class of finite groups. Then for any G ∈ P(C), the natural homomorphism from G to G/N is an isomorphism. QNEC G

Proof. Suppose G ∈ P(C), and it is the internal direct product G = Qi∈I Gi where Gi ∈ C for all i ∈ I. It is enough to show that the only C-normal subgroups are kernels of the projection maps onto some factor group. We perform induction on the number of factor groups, i.e., on the cardinality of the index set I. If I has only one element, then G is a group in C. If N is a C-normal subgroup of G, then the quotient map q from G to G/N is a surjective homomorphism between groups in C. In particular, we must have q(G)CG/N (q(G)) = G/N, so q is either trivial or an isomorphism. Since the trivial group (which is abelian) is not in C, therefore the quotient map must be an isomorphism. Hence the only C-normal subgroup of G is the trivial subgroup, and the statement is true. Now suppose I has more than one element. Let N be any C-normal subgroup. Consider the quotient map q : G → G/N. For any i ∈ I, since Gi is a direct factor of G, any element of q(Gi) must commute with any element of q(Gj ) for j 6= i. In particular, we have q(Gi)CG/N (q(Gi)) = G/N. So q restricted to Gi must either be trivial or an isomorphism. If q is trivial on all Gi, then G/N is trivial and not in C, which is impossible. Hence q restricts to an isomorphism to some Gi. Furthermore, if q restricted to Gi and q restricted to Gj are both isomorphisms for some i 6= j, then all elements in q(Gi) = G/N and in q(Gj ) = G/N must commute, so G/N is an abelian group and not in C, also impossible. Hence, there is a unique i ∈ I such that q restricted to Gi is an isomorphism, and q kills all other Gj where j 6= i. In particular, q is a projection map onto a factor group Gi. So we are done. Proof of Proposition 2.1. Let G be a pro-P(C) group. Then it is the projective limits of G/U ∈P(C) for all open P(C)-normal subgroups U. But for each open P(C)-normal subgroup U, by Lemma 2.2, we have an isomorphism G/U ≃ G/N QN∈MU where MU is the set of open C-normal subgroups containing U. By taking inverse limits of these isomor- phisms, we have an isomorphism G ≃ QN∈M G/N where MU is the set of open C-normal subgroups of G.

6 2.2 Semisimple groups We start with projective limits of finite semisimple groups. Note that, as we have already noted, this must be a direct product of non-abelian finite simple groups. Such groups are also sometimes called semisimple profinite groups. Let us first describe all maximal normal subgroups of a semisimple profinite group. Given a semisimple profinite group G = Qi∈I Si for a family of non-abelian finite simple groups Si, for any subset J ⊆ I, we define GJ := {(gi)i∈I ∈ G | gi = e if i∈ / J}, i.e., the subgroup of elements whose “support” is inside J. Clearly G is the internal direct product of GJ and GI−J . For any ultrafilter U on I, we define NU := {(gi)i∈I ∈ G | {i ∈ I | gi = e} ∈ U}. Note that the ultraproduct of the groups Si is by definition G/NU . Proposition 2.3. If G = Qi∈I Si for a family of non-abelian finite simple groups Si, and M is any abstract maximal normal subgroup of G, then NU ⊆ M for some ultrafilter U on M. Proof. Note that if G is an arbitrary product of non-abelian finite simple groups, then each factor group of G has commutator width 1 by the classification of non-abelian finite simple groups. So G has commutator width 1, and thus it is perfect. So G/M is also a perfect group. Since M is maximal, G/M is perfect and simple. So by Corollary 2.9 below, if G = Qi∈I Si for a family of non-abelian finite simple groups Si, then the quotient map G → G/M must factor through some ultraproduct Qi→U Si for some ultrafilter U on I. So we are done. Corollary 2.4. If G = Qi∈I Si for a family of non-abelian finite simple groups Si, and M is any abstract maximal normal subgroup of G with finite index, then M = NU for some ultrafilter U on I, and G/M is isomorphic to Si for some i.

Proof. By Proposition 2.3, we know that we can find an ultrafilter U such that NU ⊆ M. But by [Yan16], G/NU is either isomorphic to Si for some i, or it cannot have any non-trivial finite dimensional unitary representation. But G/M is a finite quotient of G/NU , which must have a non-trivial finite dimensional unitary representation. Hence G/NU is either isomorphic to Si for some i, and thus M = NU and G/M is isomorphic to Si for some i. Corollary 2.5. If G = Qi∈I Si for a family of non-abelian finite simple groups Si, then intersections of maximal normal subgroups of finite indices of G is a perfect group.

Proof. Using Corollary 2.4, let these maximal normal subgroups of finite indices be {NUk }k∈K for some

index set K and ultrafilters Uk on I. For each g = (gi)i∈I ∈ G, let supp(g)= {i ∈ I : gi 6= e}. Then g ∈ NUk if and only if supp(g) ∈U/ k. So g ∈ Tk∈K NUk if and only if supp(g) ∈/ Sk∈K Uk. In particular, Tk∈K NUk is generated by GJ where J/∈ Sk∈K Uk. Since each GJ is also a semisimple profinite group, hence a perfect group, therefore the subgroup they generate must also be perfect. So we are done. Now we can establish our desired result. Proposition 2.6. If G is a semisimple profinite group, then any finite quotient of G is a finite semisimple group.

Proof. Say G = Qi∈I Si for a family of non-abelian finite simple groups Si. Suppose for contradiction that our statement is not true, and let B be a minimal counterexample, i.e., B is a finite quotient of G, but it is not semisimple. Let q : G → B be the quotient map. If B has a non-trivial center Z, then by minimality, B/Z is a finite semisimple group. Therefore, Z is an intersection of maximal normal subgroups of B with finite indices. Pulling back to G, we see that q−1(Z) is an intersection of maximal normal subgroups of B with finite indices. Therefore according to Corollary 2.5, it is a perfect group. Hence its image Z must also be a perfect group. But Z is also abelian, hence it must be trivial, a contradiction.

7 So B must be centerless. Furthermore, if B is the direct product of two non-trivial groups, then by minimality of B, the two factor groups must both be semisimple groups. Hence B itself is a semisimple group, a contradiction. So B must be a centerless group, and it is not the direct product of two non-trivial groups. By Corol- lary 2.8 below, this means that q : G → B must factor through an ultraproduct. But by [Yan16] again, this means that B is either infinite, or B is isomorphic to Si for some i. Neither case is possible. To sum up, such a counterexample is impossible. Here are the necessary lemmas for the proofs above. The idea is adapted from Bergman and Nahlus [BN11], who proved a similar statement for Lie algebras. Lemma 2.7 . ([BN11, Lemma 7]) Suppose (Ai)i∈I is a family of sets, B is a set, and f : Qi∈I Ai → B is a set map whose image has more than one element. Then the following are equivalent:

(a) For every subset J ⊆ I, the map f factors either through the projection Qi∈I Ai → Qi∈J Ai, or through the projection Qi∈I Ai → Qi∈I−J Ai.

(b) The map f factors through the natural map Qi∈I Ai → Qi∈I Ai/U, where U is an ultrafilter on the index set I. When these hold, the ultrafilter U is uniquely determined by the map f. Corollary 2.8 (Analogue of [BN11, Lemma 5]). Let B be a non-trivial group. Then the following are equivalent:

(i) Every surjective homomorphism f : A1 × A2 → B from a direct product of two groups A1, A2 onto B factors through the projection of A1 × A2 onto A1 or the projection of A1 × A2 onto A2.

(ii) There are no non-trivial normal subgroups N1,N2 of B such that B = N1N2 and [N1,N2]= {e}. (iii) The center Z(B) of B is trivial, and B is not the direct product of two non-trivial groups.

(iv) For every surjective homomorphism f : Qi∈I Gi → B from a direct product of a family of groups Gi over any index set I onto B, f factors through a quotient map from Qi∈I Gi onto some ultraproduct Qi∈I Gi over some ultrafilter U on I. Proof. First let us show that the negation of (i), (ii), and (iii) are equivalent. Suppose (i) is false. Then let f : A1 × A2 → B be a counterexample. Let N1 = f(A1 ×{e}) and N2 = f({e}× A2). Then N1 and N2 are non-trivial. Since f is surjective, N1 and N2 are normal subgroups of B, and B = N1N2. Finally,

[N1,N2] = [f(A1 ×{e}),f({e}× A2)] = f([A1 ×{e}, {e}× A2]) = {e}.

So (ii) is false. Suppose (ii) is false. Let N1,N2 be a pair of normal subgroups of B and a counterexample to (ii). If N1 ∩ N2 = {e}, then B = N1 × N2, and (iii) is false. Suppose N1 ∩ N2 is non-trivial. For any element a ∈ N1 ∩ N2, since a ∈ N2 and [N1,N2]= {e}, a must commute with every element of N1. Similarly, since a ∈ N1, it must commute with every element of N2. And since B = N1N2, we must have a ∈ Z(B). So Z(B) ⊇ N1 ∩ N2 is non-trivial. So (iii) is false. Now suppose (iii) is false. If B is the direct product of two non-trivial groups, then obviously (i) is false. Suppose Z(B) is non-trivial. Then let f : Z(B) × B → B be the homomorphism induced by the inclusion homomorphism of Z(B) into B and the identity homomorphism of B. This provides a counterexample of (i). Now we shall show that (i) and (iv) are equivalent. (i) is a special case of (iv) when I = {1, 2}. So we only need to show that (i) implies (iv). But this follows easily from Lemma 2.7.

8 Corollary 2.9. If B is a perfect simple group, then for any surjective homomorphism f : Qi∈I Gi → B from a direct product of a family of groups Gi over any index set I onto B, f factors through a quotient map from Qi∈I Gi onto some ultraproduct Qi→U Gi over some ultrafilter U on I. Proof. Since B is perfect and simple, it cannot have a non-trivial center, and it cannot be the direct product of two non-trivial subgroups. So the statement is true by Corollary 2.8.

2.3 Quasisimple groups A group is quasisimple if it is a perfect central extension of a non-abelian finite simple group. Analogously, one can define a “quasi-semisimple group” to be a perfect central extension of a finite semisimple group. It turns out that such a group is simply a central product of quasisimple groups. Lemma 2.10. A finite perfect central extension of a finite semisimple group is a central product of finite quasisimple groups. Proof. Suppose G is a perfect central extension of a finite semisimple group S, which is the internal direct product of non-abelian finite simple subgroups S1,...,Sk. Let the quotient map be q : G → S, let Mi = −1 q (Si), i.e., the pull-back in the following diagram.

Mi Gi

Si S

Let Qi be the unique maximal perfect normal subgroup of Mi. Now each Qi is perfect by construction, and since Mi is finite, Mi/Qi is solvable. Therefore q(Qi)= Si. So, we have the following diagram.

Qi Mi Gi

Si Si S

Finally, since the kernel of q is the center of G, we see that Qi must be a perfect central extension of Si, i.e., it is quasisimple. I now claim that G is the central product of all these Qi. Let Z be the center of G. Since q(Qi) = Si, we see that q restricted to Q1 ...Qk is already surjective onto S ≃ G/Z. Hence we must have G = Q1 ...QkZ. If Q1 ...Qk is a proper normal subgroup of G, then G/(Q1 ...Qk) is non-trivial and abelian, which cannot happen since G is perfect. So G = Q1 ...Qk. It remains to be shown that [Qi,Qj ] is trivial for all i 6= j. Now Qj acts on Qi via conjugation, but [Qi,Qj ] ⊆ Qi ∩ Qj is in the kernel of q, hence inside of the center Z. So Qj acts trivially on Qi/(Qi ∩ Z) ≃ Si. But any automorphism of a finite quasisimple group must be trivial if it acts trivially on the corresponding simple group. So [Qi,Qj ] is in fact trivial. So G is indeed the central product of Q1,...,Qk. In general, a direct product of quasisimple groups might have a quotient that is not quasisimple anymore, but rather a central product of quasisimple groups, as shown in the following proposition. Proposition 2.11. If G is an arbitrary direct product of finite quasisimple groups, then any finite quotient must be a central product of finite quasisimple groups.

Proof. Say G = Qi∈I Qi for a family of finite quasisimple groups Qi. Let Zi be the center of Qi, and set Z = Qi∈I Zi. Then Z is the center of G, and G/Z is a semisimple profinite group.

9 Suppose we have a surjective homomorphism q : G → B. Then B/q(Z) is a finite quotient of the semisimple profinite group G/Z. Hence B/q(Z) is semisimple. Furthermore, since Z is the center of G and q is surjective, q(Z) is inside the center of B. So B is a central extension of the semisimple group B/q(Z). Finally, by classification of finite simple groups, any finite quasisimple group has commutator width at most 2. So G has commutator width at most 2. So G is perfect, and its quotient B is also perfect. So B is a finite perfect central extension of a finite semisimple group. Now, the class of quasisimple groups is not central-simple (proper quotients of a quasisimple group might still be quasisimple). So the result above does not immediately generalize to projective limits that are not direct products. However, by the theory of Schur multipliers, each non-abelian finite simple group has a unique maximal perfect central extension, i.e., the universal covering group for the non-abelian finite simple group. (See, e.g., Proposition 33.4 of [Asc00].) And each quasisimple group must be a quotient of such a group. These universal covering groups (i.e., quasisimple groups with trivial Schur multiplier) are also sometimes called superperfect quasisimple groups. Then we have the following result. Proposition 2.12. The class of finite superperfect quasisimple groups is central-simple. Proof. Suppose G, H are finite superperfect quasisimple groups, and f : G → H is any group homomorphism such that f(G)CH (f(G)) = H. Since the subgroups f(G) and CH (f(G)) both normalize f(G), therefore f(G)CH (f(G)) = H implies that f(G) is normal in H. If f(G) 6= H, then note that all proper normal subgroups of a quasisimple group are central, so f(G) is abelian. But since G is perfect, f(G) must be trivial. If f(G) = H, then G is a central perfect extension of H. But since H is super perfect, it is already a maximal central perfect extension, hence f can only be an isomorphism. Proposition 2.13. If G is a projective limit of finite central products of finite quasisimple groups, then any finite quotient must be a central product of finite quasisimple groups. Proof. Let C be the class of superperfect quasisimple groups. Then for each finite central product Q of finite quasisimple groups {Qi}i∈I , then Q is a quotient of the direct products of these Qi by a central subgroup. Let Qi be the universal perfect central extension of Qi, then Q is a central quotient of Q = Qi∈I Qi ∈P(C). Nowe let G be the projective limit of a directed family of finite central products eof finite quasisimplee groups {Gi}i∈I . Let Gi be the universal perfect central extension of Gi, so that Gi ∈ P(C). Let Zi be the e e kernel of the quotient map from Gi to Gi. e Note that any surjective homomorphism φij : Gi → Gj must induce a surjective homomorphism φij : e Gi → Gj , and φij must also send the center into the center. So the directed family {Gi}i∈I must induce e e e directed families {Gi}i∈I and {Zi}i∈I , and thus we have corresponding induced projective limits G and Z. Note that the projectivee limit on inverse systems of finite groups is an exact functor. (See, e.g., Propositione e 2.2.4 of [RZ00].) Therefore, G is a quotient of G by the closed subgroup Z. Since G is a pro-P(C) group and C is a central-simple class, G must be a directe product of superperfect quasisimplee groups, and thus any finite quotient of G must be a centrale product of finite quasisimple groups. Since G is a quotiente of G, the result holds for G as well. e Now, note that the class of finite central products of finite quasisimple groups is NOT closed under sub-direct product, so it is not a formation of groups.

Example 2.14. Consider the group SL2(5) of 2 by 2 matrices over the field of order 5, whose determinants are 1. It is quasisimple over the finite simple group A5. Then SL2(5) × SL2(5) is a perfect central extension of A5 × A5. Now take the diagonal subgroup of A5 × A5, and take its pre-image in SL2(5) × SL2(5). This is a sub-direct product of finite quasisimple groups, but it is isomorphic to SL2(5) × (Z/2Z), NOT a central product of finite quasisimple groups.

10 So if one prefers to use a formation of groups, we can consider a larger class of groups, closed under sub-direct product. Proposition 2.15. The following three classes of groups are the same: 1. C is the smallest class of groups containing all finite quasisimple groups which is closed under sub-direct products and quotients. 2. C is the class of finite central extensions of finite semisimple groups. 3. C is the class of finite central products of finite quasisimple groups and finite abelian groups.

Proof. Let us call the three classes C1, C2, C3. Since central products are quotients of the direct products, C1 is closed under central products. Also note that any cyclic group Z can be realized as a center for some quasisimple group over PSLn(q) for some n, q. Let Q be such a quasisimple group over a non-abelian finite simple group S with center Z. Then Q2 has a quotient S2, and the pullback of the diagonal subgroup of S2 in Q2 is a sub-direct product of two copies of Q, and it is isomorphic to Q × Z. Hence C1 must also contain all finite abelian groups. So C3 ⊆C1. Suppose G ∈ C2. Then we have a quotient map q : G → S such that S is semisimple, and the kernel of q is the center Z of G. Since S is perfect, the unique largest perfect normal subgroup N in G must have q(N) = S, and the restriction of q to N must have kernel N ∩ Z, which is in the center of N. So N is a perfect central extension of S. In particular, by Lemma 2.10, N is a finite central product of finite quasisimple groups. But since q(N)= S and Z is the kernel of q, we have G = NZ. And since Z is the center, [N,Z] must be trivial. So G is the central product of N and the abelian group Z. So C2 ⊆C3. It remains to show that C1 ⊆ C2. Since C2 contains all finite quasisimple groups, we only need to show that it is closed under sub-direct products and quotients. The class C2 is obviously closed under quotients. We now prove that sub-direct products of groups in C2 are still in C2. Suppose G has two normal subgroups N1,N2 such that N1 ∩ N2 is trivial, and Gi = G/Ni ∈ C2 for i = 1, 2. Then G is a finite group. Let M be the intersection of all maximal normal subgroups of G whose corresponding quotients are non-abelian finite simple groups. Then G/M is a finite semisimple group. We need to show that M is in the center of G. Consider the quotient map qi : G → Gi. Then qi(M) is in the intersection of all maximal normal subgroups of Gi whose corresponding quotients are non-abelian finite simple groups. Since Gi ∈C2, qi(M) is the center of Gi. In particular, for any g ∈ G and x ∈ M, we have xgNi = gxNi. So, since N1 ∩ N2 is trivial, we have {xg} = (xgN1) ∩ (xgN2) = (gxN1) ∩ (gxN2)= {gx}.

So x is in the center of G. So G is a central extension of G/M, and G ∈ C2. So we conclude that C1 ⊆C2. Now we have a corresponding result on projective limits of finite central extensions of finite semisimple groups. Proposition 2.16. If G is a projective limit of a family of finite central extensions of finite semisimple groups, then any finite quotient of G is still a finite central extension of a finite semisimple group.

Proof. Now if G is a projective limit of a family of groups Gi, each a finite central extension of a finite semisimple group Si. Let Zi be the center of Gi. Then we have a corresponding projective limit Z of the abelian groups Zi, and a projective limit S of the semisimple groups Si. Note that the projective limit on inverse systems of finite groups is an exact functor. (See, e.g., Proposition 2.2.4 of [RZ00].) So Z is a closed central subgroup of G and G/Z ≃ S, a semisimple profinite group. Suppose we have a surjective homomorphism q : G → B where B is finite. Then q(Z) is a central subgroup of B, and B/q(Z) is a finite quotient of the semisimple profinite group G/Z. By Proposition 2.6, B/q(Z) is a finite semisimple group. So B is a finite central extension of a finite semisimple group.

11 2.4 Projective limits via almost simple groups We now briefly introduce the concept of almost simple groups. Definition 2.17 (Definition of CR-radicals). Given any finite group G, we define its CR-radical CR(G) to be the largest semisimple normal subgroup. (“CR” stands for “completely reducible”.) The CR-radical uniquely exists for all groups, and a solvable group must have a trivial CR-radical. Definition 2.18 (Definition of Almost Simple Groups). Given any finite group G, we say it is almost simple if CR(G) is a non-abelian finite simple group, and G has no non-trivial solvable normal subgroups. Note that the above definition is equivalent to the usual definition of an almost simple group, which is a subgroup of Aut(S) containing Inn(S) for some non-abelian finite simple group S. Proposition 2.19 (Alternative Definition of Almost Simple Groups). If G is almost simple, then it is isomorphic to a subgroup of Aut(S) containing Inn(S) for some non-abelian finite simple group S. Proof. Let S be the CR-radical of G. Then G acts on S by conjugation, which induces a homomorphism from G to Aut(S). Let N be the kernel of this homomorphism. Assume for contradiction that it is not trivial. Since N is a normal subgroup of G and CR(N) is characteristic in N, therefore CR(N) is a semisimple normal subgroup of G. Therefore CR(N) ⊆ CR(G) = S and by construction, every element of CR(N) commutes with every element of S. Hence CR(N) is in the center of S. But S is a non-abelian finite simple group. So CR(N) is trivial. Therefore, any minimal normal subgroup of N must be a direct product of cyclic subgroups of prime order. In particular, the solvable radical M of N (the unique largest solvable normal subgroup of N) is non-trivial. Since M is characteristic in N, it is a non-trivial solvable normal subgroup of G, a contradiction. So we must conclude that N is trivial. Hence G is isomorphic to a subgroup of Aut(S). And since G contains S, the image of G in Aut(S) would contain Inn(S). By the classification of finite simple groups, we also see that if G is almost simple, then G/CR(G) is solvable. Proposition 2.20. The class of finite almost simple groups is central-simple. Proof. Suppose G, H are finite almost simple groups, and f : G → H is any group homomorphism such that f(G)CH (f(G)) = H. Then f(G) is a normal subgroup of H. Suppose it is not trivial. Let S be the unique minimal normal subgroup of H. Then it is a finite simple group, and any non-trivial normal subgroup of H must contain S = CR(H). So f(G) contains S. However, since H can be seen as a subgroup of Aut(S), the centralizer of S in H is trivial. Hence CH (f(G)) is trivial. So f(G)CH (f(G)) = H implies that f(G)= H. Finally, note that proper quotients of an almost simple group must be solvable. But f(G) = H is not solvable. So f has a trivial kernel, and thus it is an isomorphism. As a result of Proposition 2.20, projective limits of finite products of almost simple groups must be a direct product of almost simple groups. Now, the quotient of such a profinite group might no longer be almost simple. For example, if we take a single almost simple group that is not simple itself, then it will have a solvable quotient. So we need to look at a larger class of groups. Let C be the class of finite groups that are direct products of solvable groups and almost simple groups. Note that this class is closed under central products. Proposition 2.21. If G, H ∈C, then any central product of G, H is still in C.

12 Proof. Note that C is obviously closed under direct products. So we only need to show that it is closed under central quotients. Suppose G ∈C, say G is the internal direct product of almost simple subgroups G1,...,Gk and a solvable subgroup S. Now G acts on its normal subgroup CR(G1) ... CR(Gk) by conjugation. Since G1,...,Gk are all almost simple, the kernel of the induced homomorphism G → Aut(CR(G1) ... CR(Gk)) is exactly S. So the center Z of G is contained inside of S. Hence G/Z ≃ G1 ×···× Gk × (S/Z) ∈C. In the remainder of this section we establish the following: Proposition 2.22. Let G be a pro-C group. Then any finite homomorphic image of G must still be in C. Proof. Consider a potential counterexample. Suppose for contradiction that we have a pro-C group G = Qi∈I Gi, each Gi is either finite solvable or finite almost simple, and φ : G → B is a surjective homomorphism onto a finite group B NOT in C. Since B is finite, if such counterexample exists, then we can in fact let B be a minimal counterexample. Then by Corollary 2.30 below, B cannot exist. Hence we are done.

In the remainder of this section, we preserve the notations of G = Qi∈I Gi and φ : G → B as in the minimal counterexample in the proof above. For any subset J of I, let GJ = {(gi)i∈I : gi = e if i∈ / J}. We use the notation CR(GJ )= {(gi)i∈I : gi = e if i∈ / J and gi ∈ CR(Gi) if i ∈ J}. Note that GJ /CR(GJ ) is isomorphic to Qi∈J (Gi/CR(Gi)), which is a pro-solvable group. We use CR(G) to denote CR(GI ).

Lemma 2.23 (Dichotomy between complementary factors). For any subset J of I, either φ(GJ ) = B or φ(GI−J )= B. In particular, one is B and the other one is inside the center of B.

Proof. Note that G is the internal direct product of GJ and GI−J , hence B = φ(G) is the central product of φ(GJ ) and φ(GI−J ). Note that φ(GJ ) and φ(GI−J ) are also finite homomorphic images of products of groups in C. So if both are proper subgroups, then by minimality of B, both are in C. Hence by Proposition 2.21, B as their central product is also in C. A contradiction. So one of φ(GJ ) and φ(GI−J ) is B. But since B is the central product of these two groups, the other one must be in the center.

In particular, let J = {i ∈ I : Gi is solvable}. Then B is the central product of φ(GJ ) and φ(GI−J ). but since B/∈ C while φ(GJ ) must be solvable, we see that B = φ(GI−J ). So we may safely throw away all solvable factor groups Gi from G when we construct our counterexample. WLOG, from now on we may simply assume that all factor groups Gi of G are almost simple groups.

Proposition 2.24 (Dichotomy via ultrafilter). There is an ultrafilter ω on I, such that φ(GJ )= B if and only if J ∈ ω, and φ(GJ ) is in the center Z(B) of B if and only if J/∈ ω.

Proof. We define ω to be the collection of all subsets J such that φ(GJ ) = B. Then it is enough to show that this is an ultrafilter. We already see that for any J, exactly one of J and I − J is in ω.

If J1 ⊂ J2 and J1 ∈ ω, then GJ1 ⊆ GJ2 . Hence B = φ(GJ1 ) ⊆ φ(GJ2 ). So J2 ∈ ω as desired.

If J1,J2 ∈ ω, then φ(GJ1 )= φ(GJ2 )= B. Then, as a result, φ(GI−J1 ), φ(GI−J2 ) are both in the center of B. Then

φ(GI−(J1∩J2))= φ(G(I−J1)∪(I−J2))= φ(GI−J1 GI−J2 )= φ(GI−J1 )φ(GI−J2 ).

So φ(GI−(J1∩J2)) is also in the center of B. So φ(GJ1∩J2 )= B and J1 ∩ J2 ∈ ω. So ω is indeed an ultrafilter.

We fix this ultrafilter ω from now on. We now show that, in fact, CR(B) is an ω-ultraproduct of CR(Gi) for i ∈ I, and the corresponding quotient map is exactly φ restricted to CR(G).

13 Corollary 2.25 (CR(B) is simple). φ(CR(G)) = CR(B), and this quotient map yields CR(B) as an ω- ultraproduct of CR(Gi) for i ∈ I. In particular, CR(B) is a non-abelian finite simple group isomorphic to CR(Gi) for some i ∈ I. Proof. Note that CR(G) is a direct product of non-abelian finite simple groups. Hence any finite homo- morphic image must also be a direct product of non-abelian finite simple groups. So φ(CR(G)) is inside of CR(B). In particular, CR(B) cannot be trivial. This is because B/CR(B) is a finite quotient of G/CR(G), which is solvable, but B/∈C. Now B/φ(CR(G)) is a finite quotient of G/CR(G), which is a pro-solvable group. So B/φ(CR(G)) is solvable. In particular, CR(B)/φ(CR(G)) is solvable, and thus it must be trivial. So φ(CR(G)) = CR(B). Now for any J ∈ ω, φ(GJ )= B, and by similar reasoning, we see that φ(CR(GJ )) = CR(B). While for J/∈ ω, φ(CR(GJ )) ⊂ φ(GJ ) is in the center of B. But it must also be a direct product of non-abelian finite simple groups. Hence it is trivial. In particular, the restriction of φ to CR(G) must in fact factor through the ω-ultraproduct. Such an ultraproduct must either have no non-trivial finite quotient, or it is isomorphic to one of the factor groups of CR(G). Since CR(B) is non-trivial, then CR(G) has a non-trivial finite quotient. So in our case, Qω CR(Gi) is isomorphic to the finite simple group CR(Gi) for some i ∈ I, and CR(B) = φ(CR(G)) is a non-trivial quotient of this finite simple group. Hence CR(B) is a finite simple group. Now we establish a description of the center of B. Note that B acts on CR(B) by conjugation, and this induces a natural homomorphism from B to Aut(CR(B)). Let N be the kernel of this map. Then B/N is by definition almost simple.

Proposition 2.26 (Description of pre-images of N). For any element (gi)i∈I ∈ G, we have φ((gi)i∈I ) ∈ N if and only if {i ∈ I : gi = e} ∈ ω.

Proof. Suppose we have an element (gi)i∈I where {i ∈ I : gi = e} ∈ ω. Then for any hi ∈ CR(Gi), −1 −1 we have {i ∈ I : gi hi gihi = e} ∈ ω. Note that, since we picked (hi)i∈I ∈ CR(G), the element −1 −1 (gi hi gihi)i∈I is also in CR(G), and φ restricted to CR(G) is the quotient map onto the ω-ultraproduct. −1 −1 Hence φ((gi hi gihi)i∈I ) = e. In particular, φ((gi)i∈I ) commutes with φ((hi)i∈I ). Since φ(CR(G)) = CR(B), φ((hi)i∈I ) could be any element of CR(B). Therefore φ((gi)i∈I ) commutes with all elements of CR(B), and it must be in N. So we have proven necessity. Now we pick any element of N, say φ((gi)i∈I ) ∈ N. Note that each CR(Gi) is generated by at most ′ ′ two elements, say hi and hi. Consider the elements (hi)i∈I and (hi)i∈I . Their images through φ are in φ(CR(G)) = CR(B), so we know φ((gi)i∈I ) must commute with φ((hi)i∈I ), φ((hi)i∈I ). In particular, we have −1 −1 −1 ′ −1 ′ −1 −1 φ((gi hi gihi)i∈I ) = φ((gi hi gihi)i∈I ) = e. However, the element (gi hi gihi)i∈I is in CR(G), and φ −1 −1 restricted to this subgroup is the quotient map onto the ω-ultraproduct. Therefore φ((gi hi gihi)i∈I )= e −1 −1 −1 −1 implies that {i ∈ I : gi hi gihi = e} ∈ ω. Similarly, {i ∈ I : gi hi gihi = e} ∈ ω as well. Taking their ′ intersection, we see that {i ∈ I : gi commutes with hi,hi} ∈ ω. ′ But if gi commutes with hi,hi, this means gi commutes with all elements of CR(Gi). But since Gi is an almost simple group, this imples that gi = e. Hence {i ∈ I : gi = e} ∈ ω. Corollary 2.27. N is the center of B. Proof. Let Z be the center of B. Since B/N is almost simple, which is centerless, we must have Z ⊆ N. Now pick any element of N, say φ((gi)i∈I ) ∈ N. Then {i ∈ I : gi = e} ∈ ω. Let J = {i ∈ I : gi = e}, then (gi)i∈I ∈ GI−J , where I − J/∈ ω. Hence φ((gi)i∈I ) ∈ φ(GI−J ) is in the center of B. So N ⊆ Z. Now we proceed to show that N is in fact a direct factor of B, which would result in a contradiction to the minimality of B, finishing our proof of Proposition 2.22. Proposition 2.28. The derived subgroup B′ has a trivial intersection with N.

14 ′ Proof. Pick any element of B , say φ(([ai,bi])i∈I ). Suppose this is also inside N, then {i ∈ I :[ai,bi]= e} ∈ ω. ′ ′ ′ Let J = {i ∈ I :[ai,bi]= e}. We define ai = ai if i∈ / J and ai = e if i ∈ J. Similarly, we define bi = bi if ′ ′ ′ ′ ′ i∈ / J and bi = e if i ∈ J. Then [ai,bi] = [ai,bi] always. So φ(([ai,bi])i∈I )= φ(([ai,bi])i∈I ). ′ ′ ′ ′ However, we clearly have (ai)i∈I , (bi)i∈I ∈ GI−J . So φ((ai)i∈I ), φ((bi)i∈I ) are in the center N. So we have ′ ′ ′ ′ φ(([ai,bi])i∈I ) = [φ((ai)i∈I ), φ((bi)i∈I )] = e. Hence B′ has a trivial intersection with N. For the moment, we use the notation Bk := {gk | g ∈ B} and N k := {gk | g ∈ N}. Proposition 2.29. For any integer k, NB′ ∩ BkB′ ⊆ N kB′.

′ k k ′ k ′ k Proof. First let us show that NB ∩ B ⊆ N B . Suppose φ((xi)i∈I )[φ((ai)i∈I ), φ((bi)i∈I )] ∈ NB ∩ B , say

k φ((xi)i∈I ) = [φ((ai)i∈I ), φ((bi)i∈I )]φ((yi)i∈I ).

−1 Here yi = xi[ai,bi] and φ((yi)i∈I ) ∈ N. Then since φ((yi)i∈I ) ∈ N, the index set J = {i ∈ I : yi = e} is an element of ω. ′ ′ ′ Define ai,bi, xi to be ai,bi, xi respectively if i∈ / J, and to be e if i ∈ J. Then we have

′ k ′ ′ φ((xi)i∈I ) = [φ((ai)i∈I ), φ((bi)i∈I )]φ((yi)i∈I ).

′ ′ ′ ′ But φ((ai)i∈I ), φ((bi)i∈I ) ∈ N since ω-almost-all coordinates of (ai)i∈I and (bi)i∈I are trivial. Hence their ′ k ′ commutator is also trivial. So φ((yi)i∈I ) = φ((xi)i∈I ). We also have φ((xi)i∈I ) ∈ N since ω-almost-all ′ k coordinates of (xi)i∈I are trivial. Therefore φ((yi)i∈I ) ∈ N . So we have k k ′ φ((xi)i∈I ) = [φ((ai)i∈I ), φ((bi)i∈I )]φ((yi)i∈I ) ∈ N B . Now, suppose gh ∈ NB′ ∩ BkB′ where g ∈ Bk and h ∈ B′. Then g ∈ NB′h−1 = NB′. So we have

g ∈ NB′ ∩ Bk ⊆ N kB′.

Hence gh ∈ N kB′B′ = N kB′. Corollary 2.30. B cannot exist. Proof. We claim that N is a direct factor of B. Consider NB′/B′ in the abelian group B/B′. Then for any integer k, (NB′/B′) ∩ (B/B′)k ⊆ (NB′ ∩ BkB′)/B′ ⊆ N kB′/B′ = (NB′/B′)k. By the Kulikov criteria [Kul45], this means that NB′/B′ is a direct summand of the abelian group B/B′. Let its complement subgroup be H/B′ for some subgroup H of B containing B′. So NB′ ∩ H = B′ and (NB′)H = B. Then N ∩ H ⊆ N ∩ NB′ ∩ H = N ∩ B′. So N ∩ H is trivial. But we also have B = (NB′)H = NH. Hence B is the internal direct product of N and H. But now N is abelian, and H ≃ B/N is almost simple. So B ∈C, a contradiction.

15 2.5 Almost semisimple groups When studying a class of groups, it is sometimes useful if the class of groups is in fact a formation of groups. Definition 2.31. A class of groups is a formation if it is closed under quotients and subdirect products. Note that the class of finite products of finite almost simple groups and finite solvable groups is not closed under subdirect product. Hence it is not a formation of groups.

Example 2.32. Think of the symmetric group S5 as the semi-direct product A5 ⋊ Z/2Z, and the symmetric group S6 as the semi-direct product A6 ⋊ Z/2Z. Now, let Z/2Z act on both A5 and A6 simultaneously, then we have a group (A5 × A6) ⋊ Z/2Z, which is a sub-direct product of the almost simple groups S5 and S6. However, the resulting group is not a direct product of almost simple groups and solvable groups. So if we want to study a formation of groups, we need to consider a larger class of groups. Definition 2.33. We define a group G to be almost semisimple if G has no solvable normal subgroups. These groups are natural generalizations of almost simple groups. Proposition 2.34. If G is an almost semisimple group, then G acts on its CR-radical CR(G) faithfully. In particular, we may think of G as a subgroup of Aut(S) containing S for some semisimple group S. Proof. Note that G acts on its CR-radical CR(G) by conjugation. Let N be the kernel of the induced homomorphism G → Aut(CR(G)). Suppose for contradiction that it is not trivial. Let M be a minimal non-trivial normal subgroup of N. Then M is either semisimple or abelian. If M is semisimple, then M ⊆ CR(N). Since CR(N) is characteristic in N, it is also a semisimple normal subgroup of G, so CR(N) ⊆ CR(G) and CR(N) ⊆ N. However, since CR(G) is semisimple, therefore it acts on itself faithfully. So the intersection N ∩ CR(G) is trivial, which implies that CR(N) is trivial as well. Then CR(N) cannot contain M, a contradiction. Suppose M is abelian. Then the unique maximal solvable normal subgroup of N is non-trivial. But this subgroup is also characteristic in N, hence it is a non-trivial solvable normal subgroup of G, which is impossible as we require G to be almost semisimple. Contradiction. So N must be trivial. Now to have a formation of groups, we also want our class to be closed under quotient. Hence we need to throw solvable groups into the mix as well. Proposition 2.35. If G/CR(G) is solvable, then G is a subdirect product of a solvable group and an almost semisimple group. Proof. Let sol(G) be the solvable radical of G (i.e., the unique largest solvable normal subgroup of G). Then G/sol(G) is an almost simple group, while G/CR(G) is solvable. Finally, the intersection of sol(G) and CR(G) must be both CR and solvable, so it is trivial. So we are done. Now let C be the class of groups G such that G/CR(G) is solvable. It is clearly closed under taking quotients. Proposition 2.36. C is closed under subdirect product.

Proof. Suppose G has two normal subgroups N1,N2 such that G/N1, G/N2 ∈ C. We can find normal subgroups M1,M2 of G that are the pre-images of CR(G/N1), CR(G/N2). Then G/M1, G/M2 is solvable. As a result, since G/(M1 ∩ M2) is a subdirect product of solvable groups G/M1, G/M2, G/(M1 ∩ M2) is also solvable, and its subgroups M1/(M1 ∩ M2) and M2/(M1 ∩ M2) are also solvable. Let us now show that N1(M1 ∩ M2)= M1. To see this, note that M1/(N1(M1 ∩ M2)) is a quotient of the solvable group M1/(M1 ∩ M2), hence it is solvable. However, it is also the quotient of M1/N1 ≃ CR(G/N1), therefore it is a CR group. Since it is both solvable and CR, it has to be trivial. So we have N1(M1 ∩ M2)= M1, and similarly N2(M1 ∩ M2)= M2 as well.

16 G solvable

solvable

N1(M1 ∩ M2) M1 solvable M2 N2(M1 ∩ M2)

solvable CR CR N1 M1 ∩ M2 N2 CR CR

N1 ∩ M2 N2 ∩ M1

{e}

So we have

(M1 ∩ M2)/(N1 ∩ M2) = (M1 ∩ M2)/(N1 ∩ (M1 ∩ M2)) ≃ N1(M1 ∩ M2)/N1 = M1/N1 ≃ CR(G/N1).

So (M1 ∩ M2)/(N1 ∩ M2) and similarly (M1 ∩ M2)/(M1 ∩ N2) are both CR groups. We also see that (N1 ∩ M2) ∩ (M1 ∩ N2) ⊆ N1 ∩ N2 is trivial. Hence M1 ∩ M2 is the subdirect product of CR groups (M1 ∩ M2)/(N1 ∩ M2) and (M1 ∩ M2)/(M1 ∩ N2). So M1 ∩ M2 must be a CR group itself. In particular, M1 ∩ M2 ⊆ CR(G). On the other hand, since G/(M1 ∩M2) is solvable, CR(G) ⊆ M1 ∩M2. So we see that M1 ∩M2 = CR(G), and G/CR(G) is solvable. Proposition 2.37. If G is a pro-C group, then any finite homomorphic image of G must still be in C.

Proof. Suppose G is formed via a surjective inverse directed system {Gi}i∈I . Let Si be the CR-radical of Gi. For any surjective homomorphism φij : Gi → Gj , then Si must be mapped into Sj . Hence this induces homomorphisms Si → Sj and Gi/Si → Gj /Sj . So the inverse directed system {Gi}i∈I would induce the corresponding inverse directed systems {Si}i∈I and {Gi/Si}i∈I . Let S be the projective limit of the system {Si}i∈I , then S is a semisimple profinite group and a closed normal subgroup of G. Note that the projective limit on inverse systems of finite groups is an exact functor. (See, e.g., Proposition 2.2.4 of [RZ00].) As a result, G/S is the projective limit of the system {Gi/Si}i∈I , and therefore it is pro-solvable. Now suppose B is a finite quotient of G. Let q be the quotient map from G to B. Then since S is a semisimple profinite group, q(S) must be a semisimple normal subgroup of B. Hence q(S) ≤ CR(B). So S is in the kernel of the quotient map from G to B/CR(B). This induces a surjective homomorphism from G/S to B/CR(B). Since G/S is pro-solvable, we see that B/CR(B) is solvable. Hence B is in C.

2.6 Perfect groups with bounded commutator width We devote a small section to perfect groups with bounded commutator width.

Proposition 2.38. Let Pn be a surjective inverse direct system of perfect groups, and suppose they all have commutator width at most d. Let G be their projective limit. Then G is also perfect with commutator width d. Proof. Since the class of finite perfect groups with commutator width at most d is quotient closed, therefore any continuous finite quotient of G must be perfect with commutator width at most d. Let X be the set of commutators in G, then X⋆d is mapped surjectively onto all continuous finite quotients. In particular, X⋆d is dense in G.

17 2d However, consider the map w : G → G sending (g1,h1,...,gd,hd) to the element [g1,h1] ... [gd,hd]. Since w must be continuous, and G2d is compact, therefore its image X⋆d = w(G2d) is compact. Since G is Hausdorff, X⋆d is a dense closed subset of G. Hence X⋆d = G. So G is perfect with commutator length at most d.

18 3 Perfect Groups with Unbounded Commutator Width

In this section, we shall prove Theorem 1.6 according to the layout in Section 1.2.

3.1 The restricted Burnside coproduct For any positive integer m, consider the category C of finite groups with exponent dividing m. Then Theorem 1.7 showed that B(d, m) is the free object on d generators in this category. Proposition 3.1. The category C is a reflexive subcategory of the category of finitely generated groups (i.e., the inclusion functor has a left adjoint). Proof. Let us create this reflexive functor to the inclusion functor. Given any finitely generated group G, then it is the quotient of the free group Fd for some positive integer d. Let the kernel of this quotient map q : Fd → G be NG. According to Theorem 1.7, the free group Fd has a normal subgroup NB such that any group homomor- phism from Fd to a group in C must factor through Fd/NB. Let Nm be the normal subgroup generated by NG and NB, and we define Gm = F/Nm. Since Gm is a quotient of the restricted Burnside group B(d, m), therefore Gm ∈C as desired. We claim that for any H ∈C, any homomorphism f : G → H, then there is a unique induced homomor- phism fm : Gm → H. This would show that the functor G 7→ Gm is indeed a left adjoint to the inclusion functor. To see the claim, note that f ◦ q : Fd → H must have both NG and NB in its kernel. So Nm is in this kernel. Hence this map factors through a unique homomorphism fm : Gm → H.

Corollary 3.2. In the category C, finite product and finite coproduct exist. And the coproduct of B(d1,m) and B(d2,m) is isomorphic to B(d1 + d2,m). Proof. Since the category of finitely generated groups has coproducts, therefore any reflexive subcategory has coproducts. The other statements are trivial.

We define a functor Bd : C→C that maps each group G in C to the coproduct of G with itself d-times, and maps each morphism φ : G → H to the corresponding morphism on the d-times iterated coproducts.

Proposition 3.3. Bd preserves all colimits. Proof. Note that the coproduct functor F : Cd →C is left adjoint to the diagonal functor G : C→Cd, while the diagonal functor G is left adjoint to the product functor. Therefore the functor Bd = F ◦ G is left adjoint and preserves colimits.

Corollary 3.4. Bd(φ) is surjective if φ is surjective.

Lemma 3.5. Let ι1,...,ιd : G →Bd(G) be the coprojection morphisms. Then they are injective. Further- more, ι1(G),...,ιd(G) are independent subgroups generating Bd(G). Proof. Since the category C has zero morphisms, therefore all coprojections are split monomorphisms, and the images ι1(G),...,ιd(G) are independent subgroups. Since the restricted Burnside coproduct Bd(G) is a quotient of the free product of d copies of G, therefore the subgroups ι1(G),...,ιd(G) generates Bd(G).

We now examine the interaction of Bd with group actions. In the following, M is a finite group with exponents dividing m, and M1,...,Md are the images of the coprojection morphisms from M into Bd(M). We use e1,...,ed to denote the generators of B(d, m)= Bd(Z/mZ).

19 Proposition 3.6. Suppose G is a group acting on a finite group M, and the exponent of M divides m. Then G has an induced action on Bd(M). If the coprojection images of M in Bd(M) are M1,...,Md, then these subgroups are G-invariant, and the G-action on Bd(M) restricted to each Mi is identical to the G-action on M.

Proof. Suppose we have φ : G → Aut(M). As Bd is functorial and M is a finite group with exponent dividing m, we have a map Bd : Aut(M) → Aut(Bd(M)). So Bd ◦ φ is the induced G-action on Bd(M). Let ι1,...,ιd : M →Bd(M) be the coprojection morphisms so that Mi = ιi(M). For any φ ∈ Aut(M), then ιi ◦ φ = Bd(φ) ◦ ιi by definition of the coproduct. So

Bd(φ)(Mi)= Bd(φ) ◦ ιi(M)= ιi ◦ φ(M)= ιi(M)= Mi.

So Mi is Bd(φ)-invariant, and the Bd(φ)-action on Bd(M) restricted to Mi is identical to the φ-action on M. Proposition 3.7. Suppose G is a group acting on a finite group M, and the exponent of M divides m, and M ⋊ G is perfect. Then for the induced action of G on Bd(M), and Bd(M) ⋊ G is perfect.

Proof. Now Bd(M) is generated by G-invariant subgroups M1,...,Md as in Proposition 3.6. However, since G acts on Mi the same way as its action on M, therefore Mi ⋊ G as a subgroup of Bd(M) ⋊ G is isomorphic to M ⋊ G, which is perfect. As these perfect subgroups Mi ⋊ G generate the whole group, Bd(M) ⋊ G is perfect. Proposition 3.8. Suppose G is a group acting on a finite group M, and the exponent of M divides m. Then for any G-invariant homomorphism φ : M → Z/mZ, the induced homomorphism Bd(φ) is G-invariant.

g Proof. Since φ is G-invariant, for any x ∈ M and g ∈ G, we have φ(x)= φ(x ). Let ι1,...,ιd : M →Bd(M) be the coprojection morphisms so that Mi = ιi(M). Then

g g Bd(φ) ◦ ιi(x )= ιi(φ(x )) = ιi(φ(x)) = Bd(φ) ◦ ιi(x).

So Bd(φ) restricted to each Mi is G-invariant. But since these Mi generates the domain Bd(M), therefore we see that Bd(φ) is G-invariant.

3.2 The ultraproduct functor N The ultraproduct of groups can be thought of as a functor from Grp to Grp. It has the following categorical properties:

1. The ultraproduct functor preserves all finite limits. (In particular, the ultraproduct of subgroups is a subgroup of the ultraproduct, and the ultraproduct of intersections is the intersection of ultraproducts.) 2. The ultraproduct functor preserves short exact sequences. (In particular, it preserves normal sub- groups, monomorphism, epimorphism, and quotients.) 3. The ultraproduct functor preserves semi-direct products. 4. The ultraproduct of abelian groups is an abelian group. 5. The ultraproduct of groups of exponent dividing m is a group of exponent dividing m. 6. The ultraproduct of a sequence of the same finite group is the finite group itself.

Given any sequence of finite groups Mn of exponents dividing a fixed integer m, we want to under- Z Z stand homomorphisms from Qn→ω Mn to /m . Note that many of such homomorphisms are elements of ∗ Qn→ω Mn, as shown in the example below.

20 Example 3.9. One way to construct such a homomorphism is to first pick a homomorphism φn : Mn → Z Z Z Z /m for each n, and take the ultralimit limn→ω φn, which is a homomorphism from Qn→ω Mn to Qn→ω /m = Z/mZ. ∗ Z Z If we use Mn to denote the abelian group of homomorphisms from Mn to /m (i.e., the “dual group”), ∗ then these limn→ω φn correspond to elements of the ultraproduct Qn→ω Mn. Z Z ∗ Unfortunately, not all homomorphisms from Qn→ω Mn to /m are elements of Qn→ω Mn. Here is Z Z ∗ another kind of construction: we take all homomorphisms from Qn→ω Mn to /m in Qn→ω Mn as above, ∗ and take their “consensus” according to some ultrafilter U on Qn→ω Mn. Definition 3.10. ∗ Let U be any ultrafilter on the set Qn→ω Mn, then we define the U-consensus to be the homomorphism φU = limφ→U φ such that φU (x) = limφ→U φ(x) for all x ∈ Qn→ω Mn, Z Z Then it is straight forward to verify that φU : Qn→ω Mn → /m is indeed a homomorphism. Z Z It turns out that, when all groups Mn are abelian, then any homomorphism φ : Qn→ω Mn → /m is ∗ equal to φU for some ultrafilter U on Qn→ω Mn, as in the above example.

Proposition 3.11. Fix a sequence of finite abelian groups Mn of exponents dividing a fixed integer m. Then Z Z ∗ any homomorphism φ : Qn→ω Mn → /m is equal to φU for some ultrafilter U on Qn→ω Mn. Proof. This is a special case of Proposition 3.12 below, when the group actions are all trivial. In fact, we can prove something stronger. Proposition 3.12 below shows that we can in fact do this even ∗ with respect to a given group action. To do this, we need a definition of G-invariant ultrafilter on Qn→ω Mn. Suppose we have a finite group Gn acting on Mn for each n. Then there is an induced action of G = Qn→ω Gn on Qn→ω Mn. ∗ ∗ For each g ∈ G, let Sg ⊆ Qn→ω Mn be the collection of all g-invariant elements of Qn→ω Mn. Then we say the ultrafilter U is G-invariant if and only if Sg ∈U for all g ∈ G. Then we have the following result.

Proposition 3.12. Fix a sequence of finite abelian groups Mn of exponents dividing a fixed integer m. Suppose for each n, we have a finite group Gn acting on Mn. So we have an action of G = Qn→ω Gn on Qn→ω Mn. Z Z Then any G-invariant homomorphism φ : Qn→ω Mn → /m is equal to φU for some G-invariant ∗ ultrafilter U on Qn→ω Mn. ∗ ∗ Proof. For each g ∈ G, let Sg ⊆ Qn→ω Mn be the collection of all g-invariant elements of Qn→ω Mn. For ∗ ∗ each x ∈ Qn→ω Mn, let Sx ⊆ Qn→ω Mn be the collection of all ψ ∈ Qn→ω Mn such that ψ(x)= φ(x). Then U is G-invariant if and only if it contains all Sg for all g ∈ G, and we have φ = φU if and only if U contains Sx for all x ∈ Qn→ω Mn. So our proposition is the same as the claim that there is an ultrafilter on ∗ Qn→ω Mn containing all Sg and all Sx. To prove this claim, we need to show that the collection of all Sg and all Sx has finite intersection property. This is done by Lemma 3.14.

First, let us show that the collection of all Sx has the finite intersection property.

Lemma 3.13. Let Mn be a sequence of finite abelian groups with exponents dividing m. For any x1,...,xk ∈ Z Z ∗ Qn→ω Mn and any homomorphism φ : Qn→ω Mn → /m , we can find a homomorphism ψ ∈ Qn→ω Mn such that ψ(xi)= φ(xi) for all i. Proof. Note that finite abelian groups with exponents dividing m are the same as Z/mZ-modules. Picking Z Z Z Z k elements x1,...,xk ∈ Qn→ω Mn is the same as picking a /m -modules homomorphism f : ( /m ) → Z Z Qn→ω Mn. Say we let e1,...,ek be the standard generators of /m , so that we have f(ei)= xi. k Let xi = limn→ω xi,n, then we can define Z/mZ-modules homomorphism fn : (Z/mZ) → Mn such that k fn(ei)= xi,n. Then f = limn→ω fn in the sense that f(v) = limn→ω fn(v) for all v ∈ (Z/mZ) .

21 Let N = Ker(f). Since f(v) = 0 if and only if v ∈ N, the sets {n ∈ N | fn(v)=0} for v ∈ N are inside k the ultrafilter ω, and the sets {n ∈ N | fn(v) 6= 0} for v ∈ (Z/mZ) − N are also inside the ultrafilter ω. Since (Z/mZ)k is a finite set, we have the following result.

N N N {n ∈ | Ker(fn)= N} = ( \ {n ∈ | fn(v)=0}) ∩ ( \ {n ∈ | fn(v) 6=0}) ∈ ω. v∈N v∈(Z/mZ)k−N

So Ker(fn)= N for ω-almost all n.

fn

′ k q k fn (Z/mZ) (Z/mZ) /N Mn

′ f φ◦f ψn Z/mZ

Now the homomorphism φ ◦ f : (Z/mZ)k → Z/mZ factors through the quotient map q : (Z/mZ)k → k ′ k (Z/mZ) /N, and induces a homomorphism f : (Z/mZ) /N → Z/mZ. For ω-almost all n, the map fn : Z Z k ′ Z Z k ( /m ) → Mn also factors through q as well, and induces an injective homomorphism fm : ( /m ) /N → Mn. Note that by Baer’s criterion (see, e.g., [Lam12]), Z/mZ is an injective module over itself. Since for ′ Z Z k Z Z ω-almost all n, fn : ( /m ) /N → Mn is injective, therefore we have an extension ψn : Mn → /m such ′ ′ ∗ that ψn ◦ fn = f . Clearly ψn ∈ Mn. In particular, ψn ◦ fn = φ ◦ f for ω-almost all n. ∗ Let ψ = limn→ω ψn ∈ Qn→ω Mn. Let us verify that ψ is what we need.

ψ(xi) = lim ψn(xi,n) = lim ψn(fn(ei)) = lim φ ◦ f(ei)= φ ◦ f(ei)= φ(xi). n→ω n→ω n→ω So we are done.

Lemma 3.14. Let Mn be a sequence of finite abelian groups with exponents dividing m. Let g1,n,...,gk,n be automorphisms of Mn for each n. Then gi = limn→ω gi,n is an automorphism of Qn→ω Mn for each i. Z Z For any x1,...,xk ∈ Qn→ω Mn and any g1,...,gk-invariant homomorphism φ : Qn→ω Mn → /m , ∗ we can find a g1,...,gk-invariant homomorphism ψ ∈ Qn→ω Mn such that ψ(xi)= φ(xi) for all i.

Proof. Suppose xi = limn→ω xi,n. For each n, let Nn be the subgroup generated by [Mn,g1,n],..., [Mn,gk,n]. For simplicity of notation, let N = Qn→ω Nn. Let qn : Mn → Mn/Nn be the quotient map. Then if φ is g1,...,gk-invariant, then Qn→ω Nn ⊆ Ker(φ). So we have an induced homomorphism ′ Z Z ′ ′ ∗ φ : Qn→ω Mn/Nn → /m . By Lemma 3.13, we can find an element ψ = limn→ω ψn ∈ Qn→ω(Mn/Nn) such that for all i,

′ ′ ′ ′ φ (xi mod N)= ψ (xi mod N)= ψ ( lim qn(xi,n)) = lim (ψ ◦ qn)(xi,n). n→ω n→ω n

′ ′ Let ψ = limn→ω ψn ◦ qn, then ψ(xi)= φ (xi mod N)= φ(xi) as desired. Furthermore, each homomorphism ′ ψn ◦ qn is invariant under g1,n,...,gk,n. Thus ψ is invariant under g1,...,gk. ∗ Now we want to apply the restricted Burnside coproduct on the homomorphisms φ ∈ Qn→ω Mn and φU ∗ for ultrafilters U on Qn→ω Mn. Here are the constructions. ∗ Z Z For each element limn→ω φn ∈ Qn→ω Mn, note that each φn : Mn → /m induces a homomorphism Bd(φn): Bd(Mn) →Bd(Z/mZ)= B(d, m). As a result, we make the following definition.

22 Definition 3.15. ∗ Given φ = limn→ω φn ∈ Qn→ω Mn, we define its restricted Burnside power to be the homomorphism Bd(φ) := lim Bd(φn). n→ω

This is a homomorphism from Qn→ω Bd(Mn) to B(d, m). Definition 3.16. ∗ For any ultrafilter U on the set Qn→ω Mn, we may define

Bd(φU ) := lim Bd(φ). φ→U such that for all x ∈ Qn→ω Bd(Mn), we have

Bd(φU )(x) = lim Bd(φ)(x). φ→U

The image limφ→U Bd(φ)(x) is well-defined because B(d, m) is a finite group. Z Z Then under these definitions, the properties of homomorphisms from Qn→ω Mn to /m will carry over Z Z to homomorphisms from Qn→ω Bd(Mn) to Bd( /m ) = B(d, m). Two particularly notable properties are surjectivity and invariance under a group action.

Proposition 3.17. Fix a sequence of finite groups Mn of exponents dividing m. For any ultrafilter U on ∗ Z Z the set Qn→ω Mn, if φU is surjective onto /m , then Bd(φU ) is also surjective onto B(d, m).

Proof. Let ι : Z/mZ → B(d, m) be the i-th coprojection map. Let ιn : Mn →Bd(Mn) be the corresponding i-th coprojection map. We claim that Bd(φU ) ◦ (limn→ω ιn)= ι ◦ φU . Z Z If the claim is true, then we can pick any x ∈ Qn→ω Mn that is a pre-image of 1 ∈ /m under the map φU . Then Bd(φU ) ◦ (limn→ω ιn)(x)= ι ◦ φU (x)= ι(1) is the i-th generator of B(d, m). So each generator of B(d, m) is in the image of Bd(φU ), hence Bd(φU ) is surjective. We now prove our claim. For any homomorphism φn : Mn → Z/mZ, by definition of the restricted Burnside coproduct functor, we have ι ◦ φn = Bd(φn) ◦ ιn.

φn Mn Z/mZ

ιn ι

Bd(φn) Bd(Mn) B(d, m)

∗ Taking ultralimit n → ω, we have ι ◦ φ = Bd(φ) ◦ (limn→ω ιn) for any φ = limn→ω φn ∈ Qn→ω Mn.

φ Z Z Qn→ω Mn /m

limn→ω ιn ι

Bd(φ) Qn→ω Bd(Mn) B(d, m)

Note that the right side of the diagram stays unchanged because the groups on the right side are all finite. So we have the calculation

Bd(φU ) ◦ ( lim ιn) = lim [Bd(φ) ◦ ( lim ιn)] = lim ι ◦ φ = ι ◦ ( lim φ)= ι ◦ φU . n→ω φ→U n→ω φ→U φ→U

Here the ultralimit operator limφ→U and the homomorphism ι commute because ι is a homomorphism between finite groups independent of φ.

Proposition 3.18. Fix a sequence of finite groups Mn of exponents dividing m. Suppose we have a finite group Gn acting on Mn for each n. So we have an action of G = Qn→ω Gn on Qn→ω Mn. ∗ For any G-invariant ultrafilter U on the set Qn→ω Mn, then both φU and Bd(φ)U are G-invariant.

23 ∗ Proof. Suppose U is G-invariant. For any g ∈ G, let Sg be the subset of g-invariant elements of Qn→ω Mn. g g Then Sg ∈U for all G. So for each x ∈ Qn→ω Mn, we have φU (x ) = limφ→U φ(x ) = limφ→U φ(x)= φU (x). Now for each φ ∈ Sg, say φ = limn→ω φn and g = limn→ω gn. Then φn is gn-invariant for ω-almost all n. But by Proposition 3.8, this means Bd(φn) is gn-invariant for ω-almost all n. So Bd(φ) = limn→ω Bd(φn) is g-invariant. So Bd(φ) is g-invariant for all φ ∈ Sg, i.e., for U-almost all φ. So for each x ∈ Qn→ω Bd(Mn), we have g g Bd(φ)U (x ) = limφ→U Bd(φ)(x ) = limφ→U Bd(φ)(x)= Bd(φ)U (x). Since g is an arbitrary element of G, both φU and Bd(φ)U are G-invariant.

3.3 Reduction to the cyclic case Recall that our goal is to realize the restricted Burnside group B(d, m) as a quotient of an ultraproduct of finite perfect groups. In this subsection, we shall use the fact B(d, m) = Bd(Z/mZ) to reduce our goal to this: it is enough to realize the cyclic group Z/mZ as a quotient of an ultraproduct of “nice” finite perfect groups. In particular, we shall show that it is enough to have the following lemma. We delay its proof to later subsections. Lemma 3.19. Pick any non-principal ultrafilter ω on N and any positive integer m. Then we can find a sequence of finite groups {Gn}n∈N, each Gn acting on the corresponding finite abelian group Mn whose exponent divides m, such that they satisfy the following conditions.

1. Mn ⋊ Gn is perfect for each n. Z Z 2. There is a Qn→ω Gn-invariant surjective homomorphism ψ : Qn→ω Mn → /m . Note that the conditions here guarantee the following corollary. Corollary 3.20. Any cyclic group Z/mZ can be realized as an abstract quotient of an ultraproduct of finite perfect groups.

Proof. Pick any non-principal ultrafilter ω on N and any positive integer m. Let {Mn}n∈N and {Gn}n∈N and ψ be the sequences of groups and the homomorphism in Proposition 3.19. ⋊ Since ψ is n→ω Gn-invariant, by Lemma 3.21 below, it has an extension ψ : ( n→ω Mn) ( n→ω Gn) → Q e Q Q Z/mZ. So ψ is surjective, and as ultraproduct preserves semi-direct product, the domain of ψ is canonically e ⋊ e isomorphic to Qn→ω(Mn Gn), an ultraproduct of finite perfect groups. Lemma 3.21. Suppose we have a group G acting on another group M, and a G-invariant homomorphism ψ : M → B to some group B. Then there is an extension ψ : M ⋊ G → B of ψ. e Proof. Let N = [M, G]. Then since ψ is G-invariant, N ⊆ Ker(ψ). So the map ψ factors into a quotient map q : M → M/N and a map ψ′ : M/N → B. For any x, y ∈ M and g,h ∈ G, we have [x, g]h = [xh,gh] ∈ N and [x, g]y = [xy,g][y,g]−1 ∈ N, so N is a G-invariant normal subgroup of M ⋊ G. So N ⋊ G is a subgroup of M ⋊ G. Furthermore, for any x ∈ N,g ∈ G, y ∈ M,h ∈ G, then in the group M ⋊ G we have the following calculation:

(xg)y = xygy = xygyg−1g = xy[y,g−1]g ∈ N ⋊ G.

(xg)h = xhgh = xx−1xhgh = x[x, h]gh ∈ N ⋊ G. Hence we see that N ⋊ G is a normal subgroup of M ⋊ G. But for each x ∈ M and g ∈ G, then [x, g] = (x−1g−1x)g is in the normal closure of G. So N ⋊ G is in the normal closure of G, and hence it is the normal closure of G.

24 Let q be the quotient homomorphism from M ⋊ G to (M ⋊ G)/(N ⋊ G) ≃ M/N. Then we have a commutativee diagram:

′ q ψ N M M/N B

≃ qe N ⋊ GM ⋊ G (M ⋊ G)/(N ⋊ G)

So we have a map ψ : M ⋊ G → B extending ψ : M → B. e Assuming Lemma 3.19, which builds a homomorphism onto Z/mZ, now we show that it induces a similar construction which shall induce a homomorphism onto B(d, m)= Bd(Z/mZ).

Lemma 3.22. Pick any non-principal ultrafilter ω on N and any positive integers d, m. Let {Mn}n∈N and {Gn}n∈N and ψ be the sequences of groups and the homomorphism in Lemma 3.19. Then they satisfy the following conditions:

1. Bd(Mn) ⋊ Gn is perfect for each n.

2. We can find a Qn→ω Gn-invariant surjective homomorphism from Qn→ω Bd(Mn) to B(d, m).

Proof. By Proposition 3.7, Bd(Mn) ⋊ Gn is perfect for each n. Let G = Qn→ω Gn. Since ψ is G-invariant, by Proposition 3.12, we can find a G-invariant ultrafilter U ∗ on Qn→ω Mn such that ψ = φU . Then by Proposition 3.18, Bd(φ)U is also G-invariant. Furthermore, by Proposition 3.17, since φU = ψ is surjective, Bd(φ)U is also surjective. So we are done. Corollary 3.23. For any positive integer d, m, the restricted Burnside group B(d, m) can be realized as an abstract quotient of an ultraproduct of finite perfect groups.

Proof. Pick any non-principal ultrafilter ω on N and any positive integer m. Let {Mn}n∈N and {Gn}n∈N and ψ be the sequences of groups and the homomorphism in Lemma 3.19. Let G = Qn→ω Gn. For any positive integer d, by Lemma 3.22, we can then find the G-invariant surjective homomorphism Bd(φ)U : Qn→ω Bd(Mn) → B(d, m). By Lemma 3.21, Bd(φ)U has an extension ⋊ ψ : n→ω Bd(Mn) Gn → B(d, m). e Q Note that Bd(Mn)⋊Gn is perfect for each n, so B(d, m) is indeed an abstract quotient of an ultraproduct of finite perfect groups. Hence Theorem 1.4 is proven.

3.4 Constructing the group Gn

We now move on to the proof of Lemma 3.19, by an explicit construction of the groups Gn and Mn and the homomorphism ψ. Pick any two distinct prime numbers p, q coprime to m, and p ≥ 5. For each n ∈ N, the alternating group F Fn p Fn Fn p Ap acts on the q-vector space ( q ) by permuting the p factor spaces q . Let Vn be the subspace of ( q ) Fn of tuples (v1,...,vp) with v1,...,vp ∈ q such that P vi = 0. Note that Vn is Ap-invariant.

Proposition 3.24. Vn ⋊ Ap is a perfect group with commutator width at most 2. σ Fn Proof. For any σ ∈ Ap, we have (v1,...,vp) = (vσ(1),...,vσ(p)) for any v1,...,vp ∈ q . Now if v1+···+vp = 0, then vσ(1) + ··· + vσ(p) = 0 as well. Hence Vn is an Ap-invariant subspace. σ Since p ≥ 5 and it is a prime, it must be odd. So let σ be the p-cycle in Ap such that (v1,...,vp) = Fn (vp, v1, v2,...,vp−1) for any v1,...,vp ∈ q . Note that the action by σ induces an automorphism of Vn, σ and if (v1,...,vp) = (v1,...,vp) in Vn, then we must have vi = vj for all i, j. Since p, q are coprime,

25 0 = v1 + ··· + vp = pvi implies that vi = 0 for all i. So σ induces an automorphism with only trivial fixed point. Now the map v 7→ [v, σ] is a linear automorphism of Vn, and its kernel is exactly the set of fixed points of σ, which must be trivial. So this map is bijective. So for any v ∈ Vn, we can find w ∈ Vn such that v = [w, σ]. So all elements of Vn are commutators of Vn ⋊ Ap. Now for the group Vn ⋊ Ap, all elements in Ap are commutators since p ≥ 5, and all elements in Vn are commutators. Therefore Vn ⋊ Ap is perfect with commutator width at most 2.

Remark 3.25. The bound here is tight. Suppose p =5. Let σ = (12)(34) and pick any v ∈ Vn, and consider vσ ∈ Vn ⋊ Ap. Suppose vσ = [v1σ1, v2σ2], then [σ1, σ2]= σ. Then σ1(5) = σ2(5) = 5 by Lemma 3.26 below. −1 Now [v1σ1, v2σ2] = (L1(L2 − I)v1 + L1L2(L1 − I)v2)σ, where Li is the linear automorphism on Vn −1 induced by σi. But since both L1,L2 fix the fifth component, L1(L2 − I) and L1L2(L1 − I) must kill the fifth component. So we must have vσ 6= [v1σ1, v2σ2] when v has non-zero fifth component.

Lemma 3.26 (Remark Lemma). If [σ1, σ2] = (12)(34) in A5, then σ1(5) = σ2(5) = 5.

Proof. The proof is basically a direct enumeration of all possible cycle-types of σ1. −1 ′ −1 Suppose σ1 is a 5-cycle, then this implies that a product of two 5-cycles σ = σ1 and σ = σ2 σ1σ2 is (12)(34). Suppose σ, σ′ are two 5-cycles such that σσ′ = (12)(34). Then σ′(σ(5)) = 5. By symmetry, WLOG, say σ(5) = 1. Since σ′(σ(1)) = 2 and σ′ has no fixed point, we see that σ(1) 6= 2. We also know that σ(1) 6=1 since σ has no fixed point and σ(1) 6= 5 since σ is a 5-cycle. Hence σ(1) = 3 or 4. By symmetry, WLOG say σ(1) = 3. Now σ′(σ(3)) = 4, so σ(3) 6= 4. But we also have σ(3) 6= 3 or 5 since σ is a 5-cycle, and σ(3) 6= 1 since σ(5) = 1. This means σ(3) = 2. Now σ(2) 6=2, 3, 5 since they are already taken by other σ-values, and σ′(σ(2)) = 1 implies σ(2) 6=1. So σ(2) = 4. And finally, this means σ(4) = 5. So σ = (13245). It is then easy to see that σ′ = (14235)−1. −1 But now we see that (13245) and (14235) are NOT conjugate in A5. However, by definition σ = σ1 and ′ −1 σ = σ2 σ1σ2 are conjugate in A5, a contradiction. So σ1 cannot be a 5-cycle. Similarly, σ2 also cannot be a 5-cycle. Now suppose σ1 = σ is a 3-cycle. Then (12)(34) is the product of two 3-cycles. Note that the union of the supports of the two 3-cycles must obviously include {1, 2, 3, 4}. However, if two 3-cycles have only one element in common in their support, then their product is a 5-cycle. So the two 3-cycles must have two elements in common. So they must both have support inside {1, 2, 3, 4}. WLOG suppose the other fixed point of σ is 4. Then σ(3) = 1 or 2. WLOG suppose σ(3) = 1. Then σ = (123) and σ′ = (143) = (134)−1. So σ2 = (234), (142) or (13)(24). We see that in this case, indeed we have σ1(5) = σ2(5) = 5. −1 ′ −1 Finally, suppose neither σ1 nor σ2 is a 3-cycle. Then both σ = σ1 and σ = σ2 σ1σ2 must be of cycle type (2, 2, 1). Then σσ′ = (12)(34) = [(12)(34)]−1 = (σ′)−1σ−1 = σ′σ. Suppose σ(5) 6= 5, say WLOG σ(5) = 1. Then since σσ′ = (12)(34), we see that σ′(5) = 1 as well. So σ = (15)(ab) and σ′ = (15)(cd), where ′ a,b,c,d ∈{2, 3, 4}. But then σσ is a 3-cycle, a contradiction. So σ(5) = 5, and therefore σ1(5) = σ2(5) = 5 as desired. ⋊ We shall set Gn = Vn Ap. Let V = Qn→ω Vn and G = Qn→ω Gn.

Proposition 3.27. G = V ⋊ Ap. Proof. Note that the ultraproduct of semi-direct products is the semidirect product of ultraproducts. And since Ap is finite, we have a canonical identification Qn→ω Ap = Ap. Fn We now establish some properties of G to be used later. Note that q has a natural “inner prod- Fn F uct”. If v, w ∈ q , let them be v = (v1,...,vn), w = (w1,...,wn) for v1,...,vn, w1,...,wn ∈ q, then we Fn define hv, wi = P viwi. Note that this defines a non-degenerate symmetric bilinear form on q . This Fn p then induces a symmetric bilinear form on ( q ) such that h(v1,...,vp), (w1,...,wp)i = Phvi, wii for Fn v1,...,vp, w1,...,wp ∈ q .

26 − (σ 1) σ Fn p Proposition 3.28. hv , wi = hv, w i for all v, w ∈ ( q ) and σ ∈ Ap.

− (σ 1) σ Proof. h(v1,...,vp) , (w1,...,wp)i = P vσ−1 (i)wi = P viwσ(i) = h(v1,...,vp), (w1,...,wp) i. Fn p Proposition 3.29. The induced bilinear form on ( q ) restricts to a symmetric non-degenerate bilinear form on Vn.

Proof. We only needs to verify that it is non-degenerate. Suppose for some v ∈ Vn, hv, wi = 0 for all w ∈ Vn. Fn ′ Fn ′ ′ Then if v = (v1,...,vp) for v1,...,vp ∈ q , pick any w ∈ q , then hv, (w , −w , 0,..., 0)i = 0 implies that ′ ′ ′ Fn Fn hv1, w i = hv2, w i for all w ∈ q . Since h−, −i is non-degenerate on q , we have v1 = v2. Similarly we have vi = vj for all i, j. But since v ∈ Vn, we have P vi = 0, so pvi = 0. Finally, since p, q are coprime, we have vi = 0 for all i. So v = 0.

Lemma 3.30. The symmetric non-degenerate bilinear forms on Vn induce a symmetric non-degenerate bilinear form on the ultraproduct V = Qn→ω Vn.

Proof. Taking the ultralimit of h−, −i : Vn ×Vn → Fq, we obtain a symmetric bilinear map h−, −i : V ×V → Fq. Let us verify that it is indeed non-degenerate. Suppose limn→ω vn is non-zero in V . Then vn 6= 0 for ω-almost all n. For each vn 6= 0, since h−, −i is non-degenerate, we can find wn ∈ Vn such that hvn, wni= 6 0. If vn = 0, we pick wn arbitrarily. So hvn, wni= 6 0 for ω-almost all n by construction. Then since our bilinear form takes value in a finite set Fq, we have hlimn→ω vn, limn→ω vni = limn→ωhvn, wni 6= 0. So h−, −i is indeed non-degenerate. We now study subspaces of V with finite codimension, which would be useful later. We first make a definition of a “closed” subspace. Definition 3.31. Given a vector space V with a symmetric non-degenerate bilinear form, we say v, w ∈ V are perpendicular if hv, wi =0. We define the orthogonal complement of a subset S ⊆ V to be the set S⊥ := {v ∈ V : hv, wi = 0 for all w ∈ S}, i.e., the subspace of elements perpendicular to all elements of S. We say a subspace W of V is closed if W = S⊥ for some subset S. Lemma 3.32. If S ⊆ T for two subsets S,T in a vector space V with a symmetric non-degenerate bilinear form, then T ⊥ ⊆ S⊥ and S ⊆ S⊥⊥. Proof. If v ∈ T ⊥, then v is perpendicular to all elements of T , which includes all elements of S. Hence v ∈ S⊥. If v ∈ S, then it is by definition perpendicular to all elements of S⊥, so s ∈ S⊥⊥. Lemma 3.33. Given a vector space V with a symmetric non-degenerate bilinear form, then we have the following results. 1. A subspace W is closed if and only if W = W ⊥⊥. 2. If W is a closed subspace with finite codimension c, then W ⊥ has dimension c. 3. Finite dimensional subspaces are closed. Proof. Suppose W = S⊥, then S ⊆ S⊥⊥ = W ⊥, and hence W ⊥⊥ ⊆ S⊥ = W . On the other hand, W ⊆ W ⊥⊥. So we see W = W ⊥⊥. Now suppose W is a subspace with finite codimension c. Pick linearly independent vectors v1,...,vc ∈ V ⊥ ⊥ ⊥ such that W, v1,...,vc span V . Then since the bilinear form is non-degenerate, W ∩{v1} ∩···∩{vc} must be the trivial. Since each vi is non-zero and the bilinear form is non-degenerate, the map w 7→ hv, wi is a surjective ⊥ ⊥ linear map from V to Fq with kernel {vi} , so {vi} is a closed subspace of codimension 1.

27 ⊥ ⊥ ⊥ Since W, v1,...,vc are linearly independent, the subspaces W , {v1} ,..., {vc} intersect each other ⊥ ⊥ ⊥ ⊥ transversally. Since W ∩{v1} ∩···∩{vc} has dimension zero, W has dimension c. ⊥ ⊥ ⊥ Finally, suppose W is finite dimensional. Say it has a basis v1,...,vk. Then W = {v1} ∩···∩{vk} , so it is a closed subspace of codimension k. So dim W ⊥⊥ = k. Since we also have dim W = k and W ⊆ W ⊥⊥, we conclude that W = W ⊥⊥, i.e., it is a closed subspace. ⋊ We now go back to our discussion of the group G = V Ap where V = Qn→ω Vn.

Proposition 3.34. If W is a closed Ap-invariant subspace of V of codimension c, then for ω-almost all n, we can find closed Ap-invariant subspace Wn of Vn of codimension c such that W = Qn→ω Wn. ⊥ ⊥ Proof. By Lemma 3.33, W is c-dimensional. Let v1,...,vc be a basis for W , say vi = limn→ω vi,n ∈ V . Let Wn be the orthogonal complement of v1,n,...,vc,n. Since v1,...,vc are linearly independent, and there are only finitely many possible linear combinations among c vectors, we see that v1,n,...,vc,n are linearly independent for ω-almost all n. So Wn has codimension c. Let us show that Qn→ω Wn is exactly the orthogonal complement of {v1,...,vc}. Fix 1 ≤ i ≤ c. For each w = limn→ω wn ∈ Qn→ω Wn, then since wn ∈ Wn for all n, we see that wn ⊥ vi,n. So hw, vii = limn→ωhwn, vi,ni = 0. Conversely, suppose w = limn→ω wn ∈ V is orthogonal to all vi. Then limn→ωhwn, vi,ni = 0 for all i. Since there are only finitely many possible i, therefore for ω-almost all n, hwn, vi,ni = 0 for all i, and hence wn ∈ Wn. So w ∈ Qn→ω Wn. In conclusion, we have seen that W = Qn→ω Wn. Now we need to show that Wn is Ap-invariant for ω-almost all n. Suppose for contradiction that this is not the case. Then for ω-almost all n, Wn is NOT Ap-invariant, and we can find wn ∈ Wn such that σn σn wn ∈/ Wn for some σn. So limn→ω wn ∈ W but limn→ω wn ∈/ W . σn σ σ However, since Ap is a finite set, let σ = limn→ω σn. Then limn→ω wn = limn→ω wn = (limn→ω wn) ∈ W , a contradiction. So for ω-almost all n, Wn must be an Ap-invariant subspace. Proposition 3.35. Suppose W is a closed subspace of V with finite codimension. Then we can find a closed Ap-invariant subspace W ′ ⊆ W with finite codimension in V .

⊥ p Proof. Let v1,...,vk be a basis for W . Then the orbit Ø(vi) for each vi under the A action has at most p! k(p!) 2 elements. So the union of all these orbits has at most 2 elements, and they span a finite dimensional ⊥ ⊥ Ap-invariant subspace U containing W . So U is an Ap-invariant closed subspace. Note that U is finite dimensional, so by Lemma 3.33, it must be a closed subspace. So U ⊥ has codimension k(p!) dim U, which is at most 2 . Finally, here is a lemma we need for future use.

Lemma 3.36. If Wn ⊆ Vn is a subspace of codimension c, and n>c(p − 1). Then we can find w ∈ Wn such that its orbit Ø(w) in Vn under the Ap action has exactly p elements.

Proof. Since Vn is finite dimensional, all subspaces are closed. Let Wn be the orthogonal complement of Fn p Fn v1,...,vc ∈ Vn. Let vi = (vi,1,...,vi,p) ∈ Vn ⊆ ( q ) , where vi,j ∈ q and Pj vi,j = 0 for each i. Fn Now consider the span of these vi,j in q . Since Pj vi,j = 0 for each i, this subspace has dimension Fn at most c(p − 1). Let U be the orthogonal complment of this subspace in q . Since n>c(p − 1), U is non-trivial. Pick any non-zero u ∈ U. Consider w = (u, . . . , u, (1 − p)u) ∈ Vn. Since u ⊥ vi,j for all i, j, we see that w ⊥ vi for all i. Hence w ∈ Wn. Now Ap acts on Vn by permuting the p components. Since p, q are coprime, we must have (1 − p)u = u − pu 6= u. Hence Ø(w) has exactly p elements.

28 3.5 Constructing the Z/mZ-modules Mn and an action by Gn

We now move on to the construction of Mn in Lemma 3.19. Note that the category of finite abelian groups whose exponent divides m is the same as the category of finite Z/mZ-modules. So we construct Mn as Z/mZ-modules. We use the same p, q, and Gn as in the last section. q q Consider the free Z/mZ-module (Z/mZ) . Let B be the submodule of (Z/mZ) of tuples (t1,...,tq) Z Z with t1,...,tq ∈ /m such that P ti = 0. Consider the module automorphism f : B → B such that f : (t1,...,tq) 7→ (tq,t1,...,tq−1). Proposition 3.37. B is a free module, and f only has the trivial fixed-point and has order q.

q Proof. Obviously f = id the identity automorphism, and if f((t1,...,tq)) = (t1,...,tq) for an element (t1,...,tq) ∈ B, then we see that all ti are the same. But since P ti = 0 and q,m are coprime, we see that ti = 0 for all i. Hence f only has the trivial fixed-point. Let us now show that B is a free module. Set b1 = (1, −1, 0,..., 0), and set bi+1 = f(bi) for each i ∈ Z/qZ. We claim that b1,...,bq−1 form a basis. For any b = (t1,...,tq) ∈ B, then since tq = −t1 −···− tq−1, we have

b = t1b1 + (t1 + t2)b2 + ··· + (t1 + ··· + tq−1)bq−1.

So b1,...,bq−1 are spanning. Now suppose P tibi = (0,..., 0). Then since the first component of P tibi is zero, and only b1 contributes to this component, we see that t1 = 0. Moving on inductively, we see that ti = 0 for all i, and hence b1,...,bq−1 are linearly independent. Corollary 3.38. f − id is also an automorphism of B, where id is the identity automorphism of B.

Vn−{0} Let Mn = B , the Z/mZ-module of all set-functions from Vn −{0} to B. Then clearly Mn is a free Z/mZ module. v hv,wi We let Vn act on Mn such that x (w)= f (x(w)) for each x ∈ Mn, v ∈ Vn, w ∈ Vn −{0}. This is well defined because hv, wi ∈ Fq = Z/qZ and f has order q. − σ (σ 1 ) We also let Ap act on Mn such that x (v)= x(v ) for each x ∈ Mn, v ∈ Vn −{0}, σ ∈ Ap.

Proposition 3.39. We have a group action of Gn = Vn ⋉ Ap on Mn induced from the group actions of Vn and Ap on Mn. Proof.

− − −1 −1 −1 (σ 1) −1 −1 (σ 1) σ σ xσ vσ(w)= xσ v(w(σ ))= f hv,w i(xσ (w(σ ))) = f hv,w i(x(w)) = f hv ,wi(x(w)) = x(v )(w).

So the group action is well-defined.

3.6 Mn ⋊ Gn is perfect

We not only want this to be perfect, but we also want Mn ⋊ Gn to have increasing commutator width. ⋊ Otherwise, Qn→ω Mn Gn will also have bounded commutator length and become a perfect group, in which ⋊ Z Z case any ψ : Qn→ω Mn Gn → /m would be trivial.

Proposition 3.40. The group Mn ⋊ Gn is perfect.

Proof. We already know that Gn = Vn ⋊ Ap is perfect. So we just need to show that Mn is also in the commutator subgroup of Mn ⋊ Gn. We are going to show that Mn = [Mn, Vn]. Pick any v ∈ Vn −{0} and any b ∈ B, and let xv→b ∈ Mn be the function such that xv→b(v) = b and ′ ′ xv→b(v ) = 0 for all v 6= v. Note that such elements generate Mn. So we only need to show that xv→b for all v ∈ Vn −{0} and all b ∈ B are in [Vn,Mn].

29 ′ Pick any w ∈ Vn such that w is NOT perpendicular to v, say hw, vi = 1. Then for any v 6= v, we have

′ w ′ [xv→b, w](v ) = (xv→b − xv→b)(v )= f(0) − 0=0. And we also have w [xv→b, w](v) = (xv→b − xv→b)(v)= f(b) − b = (f − id)(b).

Here id is the identity automorphism on B. So [xv→b, w]= xv→(f−id)(b). Since f −id is also an automorphism, we see that for any b′ ∈ B, we can find b ∈ B such that b′ = (f −id)b, so xv→b′ = [xv→b, w] ∈ [Vn,Mn]. So we are done. We make the following useful definition.

Definition 3.41. For each x ∈ Mn, we define the null set for x as

null(x) := {0}∪{v ∈ Vn −{0} | x(v)=0}.

We include zero for convenience. Under certain conditions, we want null(x) to be some subspace of Vn.

Lemma 3.42. For any σ ∈ Ap v ∈ Vn and x ∈ Mn, we have

[x, vσ] =[xv, σ] + [x, v], {v}⊥ ⊆null([x, v]).

And for each Ap-orbit S of Vn −{0}, we have

X[x, σ](w)=0. w∈S Proof. [x, vσ] = (xv)σ − x = (xv)σ − xv + xv − x = [xv, σ] + [x, v]. If hv, wi = 0, then

[x, v](w)= xv(w) − x(w)= f hv,wi(x(w)) − x(w)= f 0(x(w)) − x(w)=0.

Finally, for each Ap-orbit S of Vn −{0},

− σ (σ 1 ) X[x, σ](w) = (X x (w)) − (X x(w)) = (X x(w )) − (X x(w)) = (X x(w)) − (X x(w)) =0. w∈S w∈S w∈S w∈S w∈S w∈S w∈S

⋊ n(p−1) Proposition 3.43. Mn Gn has a commutator width of at least (p!) .

Proof. We know that Mn = [Mn, Gn]. We first investigate the diameter of Mn with respect to the generators [g, x] for g ∈ Gn, x ∈ Mn. Fix a positive integer k. Suppose x ∈ Mn has x = [x1, v1σ1]+ ··· + [xk, vkσk] for some σ1,...,σk ∈ Ap, v1,...,vk ∈ Vn and x1,...,xk ∈ Mn. Let Wn be an Ap-invariant subspace of Vn orthogonal to all vi. Thus by Proposition 3.35, it has codi- p! 2n(p−1) mension at most k 2 . If k < p! , then Wn is not trivial. Furthermore, since Wn is Ap-invariant, Wn is a collection of Ap-orbits. By Lemma 3.42, we have

vi x = [x1, v1σ1]+ ··· + [xk, vkσk]= X[xi , σi]+ X[xi, vi].

30 ⊥ Now for each i, we have Wn ⊆{vi} ⊆ null([xi, vi]). Hence Wn ⊆ null( [xi, vi]). vi P Furthermore, for each Ap-orbit S of Vn −{0}, we have Pw∈S[xi , σi](w) = 0. Therefore, for each Ap-orbit S inside of Wn, we have

vi X X[xi , σi](w)=0, w∈S i

X X[xi, vi](w)=0. w∈S i

So together, we see that Pw∈S x(w) = 0. In particular, let us pick any x ∈ Mn such that Pw∈S x(w) 6= 0 for each Ap-orbit S of Vn −{0}. Then x = [x1, σ1v1]+ ··· + [xk, σkvk] implies that the corresponding Wn must be trivial. In particular, we must 2n(p−1) have k ≥ p! . 2n(p−1) So the collection of [g, x] for g ∈ Gn, x ∈ Mn generates Mn with a diameter of at least p! . Thus the n(p−1) commutator width of our group is at least (p!) by Lemma 3.44 below. Lemma 3.44. Let M be an abelian group, and suppose we have a group G acting on M. If the group M ⋊G has commutator width at most d, then the elements [x, g] for all x ∈ M,g ∈ G generate M with a diameter of at most 2d. Proof. We write the group operation on M multiplicatively for the sake of this argument. Suppose M ⋊ G has commutator width at most d. Then pick any xg ∈ M ⋊ G, it must be the product of d commutators, say

d ′ ′ xg = Y[xigi, xigi]. i=1

′ Then by taking all the gi,gi to the right, for some ri,si,pi, qi in G, we have

d d −1 ri ′ −1 si pi ′ qi ′ xg = Y((xi ) ((xi) ) xi (xi) )(Y[gi,gi]). i=1 i=1 For the two sides to be equal, we must have

d −1 ri ′ −1 si pi ′ qi x = Y((xi ) ((xi) ) xi (xi) ). i=1

′ ri ′ si −1 −1 By replacing xi, xi by xi , (xi) , and replacing pi, qi by ri pi,si qi, we have

d −1 ′ −1 pi ′ qi x = Y(xi (xi) xi (xi) ). i=1 Note that M is abelian. Then

d d −1 ′ −1 pi ′ qi ′ x = Y(xi (xi) xi (xi) )= Y([xi,pi][xi, qi]). i=1 i=1

In particular, since x is artibrary, M must be generated by elements [x, g] for all x ∈ M,g ∈ G with a diameter of at most 2d.

In particular, as n increases, we see that the commutator width of Mn ⋊ Gn is unbounded as desired.

31 ∗ 3.7 Ultraproducts of Mn and Mn ∗ ∗ Let V = Qn→ω Vn, M = Qn→ω Mn, and let M = Qn→ω Mn. We shall establish some properties of them to be used later. Note that each Mn is the collection of set functions from Vn −{0} to B. Hence given a sequence (xn)n∈N, their ultralimit limn→ω xn ∈ M is a set function from Qn→ω(Vn −{0})= V −{0} to B. So M is canonically a submodule of BV −{0}.

V −{0} Remark 3.45. We do NOT have M = B . Note that all Mn are finite, so the cardinality of their ultraproduct M by a non-principal ultrafilter is that of the continuum. Similarly, note that all Vn are also finite, so the cardinality of V is also that of the continuum. Then BV −{0} will have cardinality strictly larger than that of the continuum. In particular, M 6= BV −{0}. This means we can treat each x ∈ M as a set-function from V −{0} to B. So we can make the following definition. Definition 3.46. For each x ∈ M, we define its null set as

null(x) := {0}∪{v ∈ V −{0} | x(v)=0}.

Lemma 3.47. If x = limn→ω xn, then

null(x)= Y null(xn) ⊆ V. n→ω

Proof. Note that for each v = limn→ω vn ∈ V −{0}, then x(v) = 0 if and only if

{n ∈ N : xn(vn)=0} ∈ ω.

But note that xn(vn) = 0 implies that vn ∈ null(xn), so we have

{n ∈ N : vn ∈ null(xn)} ∈ ω.

So we have v ∈ Y null(xn). n→ω Since this is true for all v ∈ V −{0}, we have

null(x) ⊆ Y null(xn). n→ω The other direction is obvious. Proposition 3.48. For any x ∈ M and v ∈ V , then

{v}⊥ ⊆ null([x, v]).

Proof. Say v = limn→ω vn and x = limn→ω xn. Then

v vn [x, v]= x − x = lim (x − xn) = lim [xn, vn]. n→ω n n→ω

⊥ Since {vn} ⊆ null([xn, vn]) by Lemma 3.42, we see that

⊥ ⊥ {v} = Y {vn} ⊆ Y null([xn, vn]) = null([x, v]). n→ω n→ω

32 Proposition 3.49. For any x ∈ M and v ∈ V and w ∈ V −{0}, then

xv(w)= f hv,wi(x(w)).

Proof. Say v = limn→ω vn, w = limn→ω wn and x = limn→ω xn. Then

v vn hvn,wni x (w) = lim xn (wn) = lim f (xn(wn)). n→ω n→ω But since we have

{n ∈ N : hvn, wni = hv, wi} ∈ ω,

{n ∈ N : xn(wn)= x(w)} ∈ ω.

So their intersection is still in ω, so we have

hvn,wni hv,wi lim f (xn(wn)) = f (x(w)). n→ω

Now we define the “support for φ” as the dual concept of “null set for x”.

∗ Definition 3.50. For each φn ∈ Mn, we say a subset S of Vn is a supporting set for φn if for all xn ∈ Mn such that null(xn) ⊇ S, we have φn(xn)=0. Define supp(φn) to be the intersection of all supporting sets for φn.

∗ Proposition 3.51. supp(φn) is also a supporting set for all φn ∈ Mn.

Proof. First, the whole space Vn is a supporting set for φn. For any two supporting sets for φn, say S1,S2, suppose S1 ∩ S2 ⊆ null(x) for some x ∈ Mn. Let x1 ∈ Mn be the function that agrees with x on S1 − S2, but sends all other inputs to zero. Then S2 ⊆ null(x1) and hence φn(x1) = 0. Now we have

S1 ∩ S2 ⊆null(x),

S1 ∩ S2 ⊆null(x1).

Therefore we have S1 ∩ S2 ⊆ null(x − x1). But we also see that S1 − S2 ⊆ null(x − x1).

So we can conclude that S1 ⊆ null(x − x1) and hence φn(x − x1) = 0. Combining results above, we have

φn(x)= φn(x − x1)+ φn(x1)=0.

So finite intersections of supporting sets are supporting sets. Finally, since Vn is finite, therefore supp(φn) is a finite intersection of supporting sets for φn, so it is a supporting set for φn.

Proposition 3.52. For any vn ∈ Vn, then vn ∈ supp(φn) if and only if there exists xn ∈ Mn such that the followings are true

null(xn)= Vn −{vn},

φn(xn) 6=0.

33 Proof. Suppose vn ∈ supp(φn). Then Vn −{vn} is NOT a supporting set for φn. So there exists xn ∈ Mn such that we have the following

null(xn) ⊇ Vn −{vn},

φn(xn) 6=0.

In particular, φn(xn) 6= 0 implies that xn 6= 0. Therefore null(xn) 6= Vn, and we must in fact have null(xn)= Vn −{vn} as desired. Conversely, suppose vn ∈/ supp(φn). Then for all xn ∈ Mn with null(xn)= Vn −{vn}, we have

supp(φn) ⊆ null(xn).

And thus φn(xn)=0.

∗ Now consider φ = limn→ω φn ∈ M . We make the following definition. Definition 3.53. We say a subset S of V is a supporting set for φ if φ(x)=0 whenever null(x) ⊇ S. Define supp(φ) to be the intersection of all supporting sets for φ. Proposition 3.54. If φ = limn→ω φn, then supp(φ) is a supporting set for φ and is equal to Qn→ω supp(φn).

Proof. For x = limn→ω xn ∈ M, suppose

Y supp(φn) ⊆ null(x) n→ω . Then by Proposition 3.47, we have

Y supp(φn) ⊆ Y null(xn). n→ω n→ω Then for ω-almost all n, we have supp(φn) ⊆ null(xn).

Thus φn(xn)=0 for ω-almost all n. So φ(x) = 0. So the set Qn→ω supp(φn) is a supporting set for φ. Conversely, for each vn ∈ supp(φn), then there exists xn ∈ Mn such that

null(xn) ⊇ Vn −{vn},

φn(xn) 6=0.

Then for v = limn→ω vn ∈ Qn→ω supp(φn) and x = limn→ω xn, we have null(x) ⊇ V −{v}, φ(x) 6=0.

So NO subsets of V −{v} can be a supporting set for φ. Hence any supporting set for φ must contain v. Since we started with arbitrary vn ∈ supp(φn), the statement above is true for arbitrary v ∈ Qn→ω supp(φn). As a result, Qn→ω supp(φn) is the unique smallest supporting set for φ.

34 3.8 The existence of U and ψ Z Z ∗ We now construct the homomorphism ψ : Qn→ω Mn → /m by picking the right ultrafilter U on M , and set ψ to be the U-consensus homomorphism φU . To facilitate computations, we shall first pick out bases for all the free Z/mZ-modules involved. For B, set b1 := (1, −1, 0,..., 0), and set bi+1 := f(bi) for each i ∈ Z/qZ. Then we know that b1,...,bq−1 form a basis. We fix this basis for B for this section. For each v ∈ Vn −{0} and each integer 1 ≤ i ≤ q − 1, let xv→i ∈ Mn be the function that sends v to bi and all other inputs to zero.

Proposition 3.55. The elements xv→i for each v ∈ Vn and each integer 1 ≤ i ≤ q − 1 form a basis for Mn. Proof. Straightforward verification.

We shall use the basis xv→i as a standard basis for Mn. For each v ∈ Vn −{0}, let φv : Mn → Z/mZ such that xv→1 is sent to 1, and all other standard basis elements xw→i for (w,i) 6= (v, 1) are sent to zero.

σ Lemma 3.56. φv(x )= φvσ (x) for all x ∈ Mn, v ∈ Vn −{0}, and σ ∈ Ap.

− (σ 1) σ Proof. Note that (xv→i) = xvσ →i by definition. So the xv→1-component of x is the same as the σ xvσ →1-component of x. So φv(x )= φvσ (x).

Proposition 3.57. For any v ∈ Vn −{0}, we have

1. null(xv→i)= Vn −{v}.

2. supp(φv)= {v}.

3. For any homomorphism φ : Mn → Z/mZ, then v ∈ supp(φ) if and only if φ(xv→i) 6=0 for some i.

Proof. We have null(xv→i)= Vn −{v} by definition. For the second statement, note that φv(xv→1) = 1, so v ∈ supp(φv ). And for any x ∈ Mn with {v}⊆ null(x), then all xv→i-components of x are zero. So φv(x)=0. So {v} is the smallest supporting set for φv. For the last statement, if φ(xv→i) 6= 0 for some i, note that the null set for xv→i is Vn −{v}. Then all subsets of Vn −{v} cannot be a supporting set for φ. So we must have v ∈ supp(φ). Conversely, if v ∈ supp(φ), then by Proposition 3.52, for some x ∈ Mn with null(x) = Vn − {v}, we have φ(x) 6= 0. But the identity null(x) = Vn −{v} implies that x must be a linear combination of xv→1,...,xv→q−1, and thus φ(xv→i) 6= 0 for some xv→i. Now we start to build the homomorphism ψ. We want to guarantee that ψ is surjective, so it must some specific element of M to the generator 1 of Z/mZ. We define zn to be the element in Mn such that

zn := X xv→1. v∈Vn−{0}

We then take an ultralimit and set

z := lim zn ∈ M = Mn. n∈∞ Y n→ω

Our goal is to construct ψ := φU such that it sends z to 1, and it is G-invariant. Let us now reduce this to the task of finding a nice ultrafilter U on M ∗. ∗ ∗ For any closed subspace W ⊆ V , let MW be the collection of all φ ∈ M such that supp(φ) ⊆ W . ∗ ∗ σ Let E be the collection of all φ ∈ M such that φ(z)= 1, and φ(x)= φ(x ) for all x ∈ M and σ ∈ Ap.

35 ∗ ∗ Proposition 3.58. Suppose U is an ultrafilter on M such that MW ∈U for all closed subspace W of finite ∗ codimension in V , and E ∈U. Then U is G-invariant and φU : M → Z/mZ is surjective and G-invariant.

∗ ∗ Proof. Note that for all φ ∈ E , φ(z) = 1. Since E ∈ U, we conclude that φU (z) = 1. So the image of φU contains a generator of Z/mZ, and thus φU is surjective. ∗ For any g ∈ G = V ⋊ Ap, say g = vσ. Let Sg be the set of all g-invariant elements φ ∈ M . Let us show that Sg ∈U. ⊥ ∗ ∗ ∗ Set W = {v} , and consider any φ ∈ E ∩ MW . Then since φ is in E , it is σ-invariant. ∗ Since φ is in MW , for any x ∈ M, we have supp(φ) ⊆ W ⊆ null([x, v]).

Thus we have 0= φ([x, v]) = φ(xv − x). Therefore φ(x)= φ(xv). Hence φ is v-invariant. ∗ ∗ In conclusion, φ is both σ-invariant and v-invariant, therefore it is g-invariant. So E ∩ MW ⊆ Sg. But ∗ ∗ since E ∩ MW ∈U, therefore we see that Sg ∈U. So U is G-invariant, and consequently, φU is also G-invariant. In particular, the proof of Lemma 3.19 depends entirely on the existence of an ultrafilter containing E∗ ∗ ∗ and MW ∈ U for all closed subspace W of finite codimension in V . So we need to show that E and these ∗ MW have the finite intersection property.

Lemma 3.59. Let Wn be an Ap-invariant subspace of Vn of codimension c, such that n>c(p − 1). Then ∗ there exists φ ∈ Mn such that for all x ∈ Mn and σ ∈ Ap, we have

φ(zn)=1,

supp(φ) ⊆ Wn, φ(x)= φ(xσ ).

Proof. Since n>c(p − 1), by Lemma 3.36, we can find w ∈ Wn whose Ap-orbit Ø(w) has exactly p elements. Let φ = Pv∈Ø(w) φv. Since Wn is Ap-invariant, we have Ø(w) ⊆ Wn, and thus

supp(φ)=Ø(w) ⊆ Wn.

By Lemma 3.56, for all x ∈ Mn, v ∈ Vn −{0}, and σ ∈ Ap, we have

σ φv(x )= φvσ (x).

So we see that for all x ∈ Mn and σ ∈ Ap, we have

σ σ φ(x)= X φv(x)= X φvσ (x)= X φv(x )= φ(x ). v∈Ø(w) v∈Ø(w) v∈Ø(w)

Finally, note that φv(zn) = 1 for all v ∈ Vn −{0}. So we have

φ(zn)= |Ø(w)| = p.

But note that p,m are coprime, so p is invertible in Z/mZ. So by replacing φ with a suitable multiple, we can guarantee that φ(zn) = 1.

∗ ∗ Proposition 3.60. For any closed subspace W of finite codimension in V , then MW ∩ E is non-empty.

36 Proof. Note that if W is closed in V with finite codimension, then by Proposition 3.35, we can find W ′ ⊆ W ∗ ∗ which is closed, Ap-invariant, and has finite codimension in V . We also necessarily have MW ′ ⊆ MW . So, ′ replacing W by the smaller subspace W , we can assume that W is also Ap-invariant. By Proposition 3.34, since W is Ap-invariant and closed with finite codimension in V , therefore W = Qn→ω Wn where each Wn is Ap-invariant and closed with bounded codimension in Vn for all n. Say the ∗ codimension are all at most c. Then for all n>c(p − 1), we can find φn ∈ Mn such that for all x ∈ Mn and σ ∈ Ap, we have

φn(zn)=1,

supp(φn) ⊆ Wn, σ φn(x)= φn(x ).

Then since ω is not principal, we only need to consider the cases where n>c(p − 1) as above. So this ∗ defines φ = limn→ω φn ∈ M such that for all x = limn→ω xn ∈ M and σ ∈ Ap,

φ(z) = lim φn(zn)=1, n→ω

supp(φ)= Y supp(φn) ⊆ Y Wn = W, n→ω n→ω σ σ φ(x) = lim φn(xn) = lim φn(x )= φ(x ). n→ω n→ω n

∗ ∗ So φ ∈ MW ∩ E . So we are done. ∗ ∗ Proposition 3.61. There is an ultrafilter U on M such that MW ∈ U for all closed subspace W of finite codimension in V , and E∗ ∈U.

∗ Proof. We only need to show that these subsets of M have finite intersection property. Pick W1,...,Wk closed subspaces of finite codimension in V , we want to show that E∗ ∩ ( k M ∗ ) is non-empty. Ti=1 Wi k Let W = Ti=1 Wi. This is a finite intersection of closed subspaces of finite codimensions, so it is itself closed with finite codimension. Note that

k ∗ φ ∈ \ MWi if and only if supp(φ) ⊆ Wi for all i, i=1 k if and only if supp(φ) ⊆ \ Wi = W, i=1 ∗ if and only if φ ∈ MW .

So we have k M ∗ = M ∗ . Ti=1 Wi W ∗ ∗ So we only need to show that E ∩ MW is non-empty. But since W is closed with finite codimension, this is non-empty by Proposition 3.60. So we have the desired finite intersection property, and the desired ultrafilter exists. We now finish the proof of Lemma 3.19. Proof of Lemma 3.19. Let U be the ultrafilter in Proposition 3.61. Then it satisfies all the requirements of Proposition 3.58, so φU is surjective and G-invariant. Furthermore, by Proposition 3.40, the groups Mn ⋊ Gn are finite perfect groups as desired.

37 References

[Asc00] Michael Aschbacher. Finite group theory, volume 10. Cambridge University Press, 2000. [BN11] George M Bergman and Nazih Nahlus. Homomorphisms on infinite direct product algebras, espe- cially lie algebras. Journal of Algebra, 333(1):67–104, 2011. [CLP15] Valerio Capraro, Martino Lupini, and Vladimir Pestov. Introduction to sofic and hyperlinear groups and Connes’ embedding conjecture, volume 1. Springer, 2015. [ES04] G´abor Elek and Endre Szab´o. Sofic groups and direct finiteness. J. Algebra, 280(2):426–434, 2004. [HP89] Derek F Holt and W Plesken. Perfect groups. Oxford, 1989. [HR17] Derek Holt and Sarah Rees. Some closure results for C-approximable groups. Pacific Journal of Mathematics, 287(2):393–409, 2017. [Kul45] L Ya Kulikov. On the theory of abelian groups of arbitrary cardinality. Mat. Sb, 16(2):129–162, 1945. [Lam12] Tsit-Yuen Lam. Lectures on modules and rings, volume 189. Springer Science & Business Media, 2012. [MR19] Peter Mayr and Nik Ruˇskuc. Generating subdirect products. Journal of the London Mathematical Society, 100(2):404–424, 2019. [Nik04] Nikolay Nikolov. On the commutator width of perfect groups. Bulletin of the London Mathematical Society, 36(1):30–36, 2004. [NST18] Nikolay Nikolov, Jakob Schneider, and Andreas Thom. Some remarks on finitarily approximable groups. Journal de l’Ecole´ polytechnique-Math´ematiques, 5:239–258, 2018. [RZ00] Luis Ribes and Pavel Zalesskii. Profinite groups. In Profinite Groups, pages 19–77. Springer, 2000. [SB81] Hanamantagouda P Sankappanavar and Stanley Burris. A course in universal algebra, volume 78. Citeseer, 1981. [Tho12] Andreas Thom. About the metric approximation of Higman’s group. 2012. [Yan16] Yilong Yang. Ultraproducts of quasirandom groups with small cosocles. Journal of Group Theory, 19(6):1137–1164, 2016. [Zel90] Efim Isaakovich Zel’manov. Solution of the restricted Burnside problem for groups of odd exponent. Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 54(1):42–59, 1990. [Zel91] Efim Isaakovich Zel’manov. A solution of the restricted Burnside problem for 2-groups. Matem- aticheskii Sbornik, 182(4):568–592, 1991. [Zil14] Boris Zilber. Perfect infinities and finite approximation. In Infinity and truth, pages 199–223. World Scientific, 2014.

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