An infinite finitely presented residually finite 2-group

Proof. Fix d. First we construct phases for literal nonzero words, without claiming invariance under replacements. We work backward through the levels j=d,d-1,\ldots,0. At level d+1, use the trivial phase modulo the empty product 1, with threshold 1. Suppose that phases with the subword property have been constructed at level j+1 modulo

above a threshold n_{j+1}\geq1.

For every token t\in\mathcal T_j, choose one word u_t satisfying (2.2), and use this same choice at every occurrence of t. For the decomposition (2.1), set

Choose

Then every nonzero word w of length at least n_j has such a decomposition and

Moreover, U(w) is nonzero in A_{j+1}. Indeed, the substitution homomorphism and the token equalities give

so [U(w)]=0 would imply [w]=0. Define the literal start phase by

Multiplication by L_j makes this independent of the representative chosen for the residue \sigma_{j+1}(U(w)) modulo L'.

We verify the subword property at level j. Let v be a subword of w of length at least n_j, starting at offset b. Its cuts are exactly the cuts of w whose following letters lie in v. Consequently, its complete tokens are a consecutive subsequence of the complete tokens of w. Our fixed token choices make U(v) a literal subword of U(w). If a is its upper-level offset, the lower-level distance to its first complete token can be computed in two ways:

Both upper words meet the upper-level threshold. Applying the inductive subword property and then (2.7) yields

This completes the backward induction.

At level zero we now have literal phases modulo L, with the subword property for lengths at least n_0. Only invariance under replacements remains. Choose

Consider one replacement inside a nonzero word of length at least N. It affects at most k letters, so an unchanged prefix or an unchanged suffix has length at least n_0. That common subword has the same offset before and after the replacement, because the replacement preserves length. Applying the literal subword property to both words shows that their phases agree. Every intermediate word in a nonzero replacement class is nonzero. Chaining these equalities proves that the start phase is constant on each class. Word length is constant on each class as well, so the end phase is also invariant. The subword property already proved continues to hold after increasing the threshold from n_0 to N. Finally, every L_j is an odd integer greater than one, so the products L_0\cdots L_d are odd and unbounded.□

Proof. If the concatenation is nonzero, apply (2.4) to its prefix u and its suffix v. This gives \sigma(uv)=\sigma(u) and \sigma(v)=\sigma(uv)+|u|=\tau(u); adding |v| gives the end-phase identity.□

Proof. All roots in S\times T commute by (3.2). Root addition collects a word into one factor per coordinate pair. The corresponding entry matrices have pairwise zero products, so the matrix image reads each collected coefficient directly. This proves injectivity and identifies the group law with addition.

For one root supported on S, conjugation on an outgoing strip follows from (3.3): it performs the corresponding elementary row addition. The same relations on the reverse strip perform the inverse elementary column operation. Iterating over the factors of a word gives the stated left and right actions, with their prescribed order over the possibly noncommutative algebra B.□

Proof. Let z\in J(B) and choose a finite set S containing the support of a representative of z. To show that z commutes with a specified root, enlarge S to contain both indices of that root. Choose h\notin S. By Lemma 3.1, z acts trivially on the strips S\times\{h\} and \{h\}\times S, since its matrix image is the identity. For i,j\in S with i\ne j, the identity

therefore implies that z commutes with x_{ij}(a). These roots generate the group.□

Proof. For a transposition of coordinates a,b, use

On a column strip from \{a,b\} to an outside coordinate, its three elementary row additions exchange the two entries over \F_2. Lemma 3.1 gives this as an exact group action. The reverse strip is exchanged in the same way. Roots with both indices outside \{a,b\} commute with p. The remaining two root positions can be written, with c\notin\{a,b\}, as

The strip actions exchange these as well. Products of transpositions give arbitrary finite permutations. Any injective relabeling of a finite set extends to a permutation of a larger finite set, so the last assertion follows from Lemma 3.2.□

Proof. The subgroups H=\langle A,Z\rangle and K=\langle X,B'\rangle commute elementwise by the four hypotheses. Conjugating [X,Z]=1 by Y gives

Separating the commuting H and K factors yields [A^{-1},Z][X,B']=1. Thus [X,B']=[A^{-1},Z]^{-1} lies in H\cap K, which is central in both H and K. Centrality in H then gives [A,Z]=[A^{-1},Z]^{-1}=[X,B'].□

Proof. Additivity and the commuting relation follow factor by factor. For the consecutive commutator, the only nontrivial elementary commutators are

Each resulting root commutes with both input blocks. Expanding the commutator and using additivity at each position therefore gives the entries \sum_t a_{st}a'_{tv} of AA'.

An elementary operation supported inside outer block i acts on the outgoing strip to block j by the corresponding row operation; an operation inside block j acts by the inverse column operation. These are exact strip actions by Lemma 3.1. An operation inside any other outer block centralizes the block root. Gaussian elimination over \F_2 writes P as a product of elementary matrices: every nonzero pivot is 1, and a row transposition is a product of three row additions. Lift those factors inside each outer block. Applying the strip actions successively gives the asserted conjugation. For b=1, P=1 and the assertion is immediate.□

Proof. The gap of a generator of U is t-s>0. Two such roots either commute or have a commutator root whose gap is the sum of their gaps, by the consecutive-root relations in the two possible orders. They cannot be opposite roots. For g\geq1, let U_g be generated by roots of gap at least g; thus U_1=U and U_b=1. Conjugating a generator of U_g by a generator of U adds only roots of larger gap, so U_g is normal in U. Modulo U_{g+1}, the group U_g is generated by the commuting roots of gap g.

It follows, successively for g=1,\ldots,b-1, that every element of U is an ordered product with one factor at each coordinate pair, in increasing gap; repeated coefficients at a given pair are added. This is a finite collection process, since there are finitely many coordinate pairs and U_b=1. If its matrix image is the identity, the entries of gap one are precisely the corresponding collected coefficients, so they vanish. After coefficients of smaller gap have vanished, the same assertion holds at gap g: any product of two remaining entry matrices has gap greater than g. Induction makes every coefficient zero and proves faithfulness.

A root with equal inner indices cannot be opposite to a positive-gap root. Its commutator with a positive-gap root is either trivial or has the same positive gap. This proves that it normalizes U.

For a polynomial entry a(t), the matrix a(T) is upper triangular with diagonal entries a(\lambda). Split its block root into its commuting diagonal and strictly upper factors. The diagonal factors normalize U, so the strictly upper factors in the complete word can be collected on the right. Diagonal factors at different inner indices commute, giving

The left side and all the words v(\lambda) have identity matrix image. Hence u has identity matrix image. Faithfulness of U gives u=1 and proves (4.1).□

Proof. For each factor, split its cyclic evaluation as D_hF_h, where D_h is its evaluation at S_b and F_h consists of the entries that wrap past inner index b-1. The splitting is exact because all roots in one outer rectangle commute. Since d<b, a term of degree a\leq d goes from s to s+a and can wrap only once. Every root in F_h therefore has source in \{b-\delta,\ldots,b-1\} and target in \{0,\ldots,\delta-1\}.

For r\geq1, set

When b>2r\delta, the roots from inner indices in H_r to inner indices in L_r, with arbitrary outer indices, form an abelian strip V_r that injects into matrices by Lemma 3.1. Figure 1 displays these two bands. We shall show that conjugation by one D_h enlarges their widths by at most \delta.

More precisely, for r\leq m we claim that

Fix a root of V_r. The constituents of D_h all have the same outer source i and outer target j\ne i, and inner displacements between 0 and d. They commute with one another. A constituent that commutes with the initial root also commutes with every conjugate of it by the other constituents, and so can be omitted from this conjugation.

An interacting constituent changes the source endpoint only when its target is the old source; the new source is at most d positions to the left. It changes the target endpoint only when its source is the old target; the new target is at most d positions to the right. These statements follow directly from the consecutive-root relations, including their reverse order. An opposite-root interaction cannot occur. Initially the source inner index minus the target inner index is greater than b-2r\delta. After one endpoint has moved, the difference is still greater than

Thus no constituent, whose forward displacement is at most d, can be opposite to either the initial root or a root obtained after one endpoint move.

Nor can displacements accumulate repeatedly at one endpoint during this conjugation. A changed source has outer index i, whereas every constituent has target outer index j\ne i; the source cannot change again. A changed target has outer index j, whereas every constituent has source outer index i; the target cannot change again. Both endpoints may change, but each changes at most once. A root with both endpoints changed has outer indices i,j and commutes with every constituent. All roots produced lie in V_{r+1}, proving (4.2). This reasoning applies to a product of strip roots as well.

It remains to collect the corrections without conjugating them by one another. Put D_{>h}=D_{h+1}\cdots D_m and D_{>m}=1. The exact group identity

where the final product is in increasing order of h, does this. Every F_h starts in V_1, and every suffix has at most m factors. By (4.2), all final corrections belong to V_{m+1}. Therefore

Both evaluated words have identity matrix image. The same is true of u, and the disjointness of H_{m+1} and L_{m+1} makes V_{m+1} faithful. Hence u=1.□

Proof. Choose one finite root word for z, and apply Lemma 4.3 to its graded word. Since S_b is strictly upper triangular, Lemma 4.2 gives, in the stable group,

The fixed injection extends each \iota_s to a homomorphism on the stable group. Since z(0) is trivial there by hypothesis, every displayed coordinate image is trivial. The empty word satisfies the conclusion for all b.□

Proof. Let z be such an element. If b is a power of two, then C_b^b=\Id and characteristic two gives (C_b-\Id)^b=0. A basis adapted to the successive kernels of the nilpotent map C_b-\Id makes C_b upper triangular with all diagonal entries 1. Thus some P\in\GL_b(\F_2) has this effect. By Lemma 4.1, conjugation by a Steinberg lift of the repeated basis change takes W_z(C_b) to W_z(P^{-1}C_bP). Centrality of J(R) makes these elements equal. Lemma 4.2 then gives

The last equality uses Lemma 3.3: every coordinate image of the central element z equals z in the stable group. Choose a power of two b large enough for Lemma 4.4. Then z^b=1.□

Proof. Apply Lemma 3.4 to

Then A=[X,Y]=x_{ij}(ab) and B'=[Y,Z]=x_{hi}(bc). The Steinberg relations give [A,X]=[B',Z]=1. They also give [X,Z]=1 because ca=0, and [A,B']=1 because (bc)(ab)=b(ca)b=0. This proves (5.2).

Taking b=1 proves independence of the auxiliary index. Express the commutator using an index h different from i,j. Each of its two factors then commutes with x_{ij}(a) and with x_{ji}(c): the only potential nonzero consecutive-root coefficient is ca. Thus the commutator commutes with both defining roots. Since each root has square 1, expanding [x_{ij}(a)^2,x_{ji}(c)] shows that its square is 1. Finally, multiplication of the four elementary matrices, using ca=0, gives \Id+acE_{ii}.□

Proof. The phase mismatch gives qp=0, so Lemma 5.1 applies. We first show that the commutator does not change when the cut between p and q moves, provided both new parts still have length at least N.

For a move to the right, write the original word as abc, with a=p and bc=q, and with |c|\ge N. The word c is nonzero and has \tau(c)=\tau(q). Hence ca=0, and (5.2), together with auxiliary-index independence, identifies the commutators for the cuts a\mid bc and ab\mid c. This equality remains valid if ab=0, in which case the commutator is 1. Otherwise \sigma(ab)=\sigma(a), so the new cut retains the mismatch. For a move to the left, write p=ab and q=c with |a|\ge N. Now a is nonzero with \sigma(a)=\sigma(p), so ca=0 again, and the same identity compares ab\mid c with a\mid bc. If the new suffix is zero we again obtain 1; otherwise its end phase is unchanged. This proves invariance under every allowed movement of the cut. If a part becomes zero, the equalities already obtained show that the original commutator was 1, and the argument can stop.

Next consider a single defining replacement inside a word of length n\ge M. The allowed cut positions are the integers in [N,n-N]. A rule of length at most k can straddle at most k-1 of these positions, whereas (5.1) supplies more than k allowed positions. We can therefore move the cut so that the entire replacement lies in one part. That part has the same value in R before and after the replacement. If it is nonzero, its phases are unchanged because phases depend only on the word class. The other part is unchanged. The commutator consequently survives the replacement with the same value and mismatch, unless a zero part has already proved the original commutator to be 1.

If the concatenation is nonzero, every part of every equivalent word is nonzero. Cut movement and the preceding replacement argument show that the commutator depends only on the class of the concatenation. Conversely, a nonzero class w as in the statement can be cut into two parts of lengths at least N. The subword assertion of Lemma 2.3 makes their start and end phases \sigma(w) and \tau(w), respectively. Define H_i(w) by any such cut. The independence just proved makes this definition unambiguous.

If the concatenation represents zero, the criterion (2.3) gives a sequence of replacements ending at a word that contains a zero-declared subword. Follow this sequence with the commutator. Either a part has already become zero, or the cut can be moved once more to put the final zero-declared subword wholly in one part. That part is zero in R, and the commutator is 1. This proves the zero case as well. The order and matrix-image assertions now follow from Lemma 5.1.□

Proof. Write U=U(S,P). First consider an off-diagonal generator x_{ij}(a) of U and an off-diagonal root x_{ef}(b) whose entry is either wrong-row or consistent. If j=e and i\ne f, their commutator is x_{if}(ab). The product is zero when b is wrong-row, since \tau(a)=P(j)\ne\sigma(b). When b is consistent and ab\ne0, the product inherits \sigma(a) and \tau(b), so it is a wrong-row entry from i to f. If i=f and j\ne e, the reverse product ba is zero, because \tau(b)=P(i)\ne\sigma(a), and the roots commute.

If the two positions are opposite, so that e=j and f=i, the same mismatch \sigma(a)\ne\tau(b) allows us to use (5.3). The commutator is H_i(ab) when ab\ne0, and is 1 when ab=0. In the first case ab is a wrong-row entry from i to itself. In particular the commutator is 1 whenever b is wrong-row. All remaining pairs of positions commute by the defining Steinberg relations. This proves both the required commutations between off-diagonal generators of U and their required behavior under consistent roots.

Now take a diagonal generator H_i(w). Represent it as a commutator through an auxiliary coordinate h\notin S. To compare it with a root x_{ef}(b) on S, choose h outside its indices as well. If neither index is i, the supports are disjoint and the elements commute. If one index is i, the root lies in a strip between \{i,h\} and its complement. By Lemma 3.1, conjugation by H_i(w) acts on that strip by its matrix \Id+wE_{ii}. This matrix is its own inverse, since w^2=0 by the phase mismatch. Consequently, for f\ne i and e\ne i,

The incoming product bw is always zero, because \tau(b)=P(i)\ne\sigma(w). The outgoing product wb is zero when b is wrong-row; when b is consistent and wb\ne0, it is a wrong-row entry from i to f. These formulas prove that diagonal generators commute with off-diagonal generators of U, and that their commutators with consistent roots belong to U.

Two diagonal generators at different coordinates can be represented using disjoint pairs of coordinates, and hence commute. For two at the same coordinate i, write the second as

using t distinct from the auxiliary coordinate for the first. Here \sigma(a)=\sigma(w')\ne P(i) and \tau(c)=\tau(w')=P(i). The phase rule therefore gives wa=0 and cw=0. Applying the same strip action to the two displayed factors shows that H_i(w) fixes both, and thus commutes with H_i(w'). The use of parts of length at least N is sufficient for these two applications of the phase rule.

We have proved that all generators of U commute. They have square 1 by root addition and Lemma 5.2. Their matrix images are \Id+wE_{ij}, indexed by positions and nonzero word classes satisfying (5.4). Any two added matrices multiply to zero: a potentially nonzero product requires matching positions, and then the end phase of the first entry is the designated row phase of the second, which differs from its start phase. Thus a product of distinct generators has matrix image

The word classes form a basis of R, so this is the identity only when the sum is empty. This proves injectivity on U. Finally, the commutator calculations show that conjugation by any consistent root sends every generator of U into U. Such a root is its own inverse, so it normalizes U.□

Proof. The empty word presents no issue, so assume m\ge1, and let S contain the finitely many original indices and at least three coordinates. Put A=\F_2\oplus R_{\ge M}\subseteq R. We shall compare several evaluations of the word (6.1). Choose numerical columns u_s\in\F_2^L and rows v_s\in(\F_2^L)^* such that v_su_s=1 for every s\in\Z/L. On the word basis of the tail, set

and extend linearly to A. This is a unital algebra homomorphism A\to\Mat_L(R). Indeed, a nonzero product ww' has matching middle phases and inherits the outside phases, so

If ww'=0, the product of the two entry matrices is zero as well, regardless of the middle numerical pairing. Scalar products are immediate.

Applying \rho to the entries in (6.1), and then using the block roots of Lemma 4.1, produces an element of J(R). To justify this assertion, the original matrix word is the identity already over the subalgebra A, since all its entries belong to A and A\hookrightarrow R is injective. Its image under \rho is therefore the identity matrix. This uses only the matrix identity; no injectivity assertion about \St(A)\to\St(R) is needed.

Write \mathbf e_s for the standard numerical columns and \mathbf e_s^* for their rows. Start with u_s=\mathbf e_0 and v_s=\mathbf e_0^* for every s. The evaluated word is z supported on the coordinates (i,0), which is the same stable element by Lemma 3.3 and centrality. We change this evaluation in three steps.

Add entries in unused columns. Keep all u_s=\mathbf e_0, and change the rows to

Every pairing v_tu_s is still 1. Each block root acquires at most one additional factor, going from a coordinate (i,0) to an unused coordinate (j,t) with t\ne0. These factors belong to the strip from S\times\{0\} to S\times(\Z/L\setminus\{0\}). The original factors, supported on S\times\{0\}, normalize this strip by Lemma 3.1. We can therefore collect the additional factors into a single strip element. Both the old and new evaluated words have identity matrix image, so that strip element has identity matrix image. Strip injectivity makes it trivial. The new evaluation is still z.

Change the numerical basis. The rows v_t just defined form a basis. Let Q be the numerical matrix with these rows, so that v_tQ^{-1}=\mathbf e_t^*. Also Q\mathbf e_0=\sum_s\mathbf e_s, because v_t\mathbf e_0=1 for every t. Conjugating all entry matrices by Q therefore changes the columns and rows to

Apply the exact block-conjugation assertion of Lemma 4.1 with P=Q^{-1}. This simultaneous basis change is induced by conjugation in \St(R). The evaluated word belongs to the central kernel, so it remains z.

Remove the factors in the wrong rows. Give coordinate (i,s) the designated phase P((i,s))=s. A block root with entry w now factors as

Its factors commute. The factor with r=\sigma(w) is the root in (6.2); all the other factors are wrong-row generators of U(S\times\Z/L,P). Lemma 5.3 says that the retained, consistent roots normalize this subgroup. Collecting the other factors expresses the evaluated word as the right side of (6.2) times an element u\in U(S\times\Z/L,P). The retained word itself has identity matrix image: apply (6.3) with u_s=\mathbf e_s and v_t=\mathbf e_t^*. Hence u has identity matrix image. Injectivity on the wrong-row subgroup gives u=1, proving (6.2).□

Proof. For a homogeneous word w of degree d=|w|, the cyclic-shift evaluation of x_{ij}(w) is the block root with entry wC_L^d. Since \tau(w)-\sigma(w)=d in \Z/L, its factors are

We claim that any two such factors arising from different offsets a\ne a' commute, even when they come from different roots of the original word.

Write their entries as w,w'. A match between the column coordinate of the first factor and the row coordinate of the second requires \tau(w)+a=\sigma(w')+a'. Since a\ne a', this implies \tau(w)\ne\sigma(w'), and hence ww'=0. A match in the reverse direction similarly implies w'w=0. Thus the Steinberg relations prove the claim unless the two roots are opposite. In that case both products vanish, and the reverse match also gives \sigma(w)\ne\tau(w'). Lemma 5.2, whose length hypotheses hold because |w|,|w'|\ge M\ge N, then gives [x_{uv}(w),x_{vu}(w')]=1. This proves the claim in every case.

We may consequently collect the product by offsets, preserving within each offset the order of the original word. The offset-zero word is precisely the relocated word of Lemma 6.1, and every other offset word is its coordinate permutation. They all give z, since z\in J(R). There are L offsets, proving (6.4).□

Proof. For any two-sided ideal K of a unital ring, the Steinberg presentation shows that the kernel of \St(R)\to\St(R/K) is the normal closure of the roots x_{ij}(a) with a\in K. Indeed, after these roots are killed, root addition makes x_{ij}(a) depend only on the class of a modulo K, and the remaining relations are exactly the Steinberg relations over R/K. For K=R_{\ge2M}, root addition expands every such entry in the homogeneous word basis. It is therefore enough to handle a conjugate g x_{ij}(w)g^{-1} with |w|\ge2M.

Choose a representative word w=ab with |a|,|b|\ge M. Both parts are nonzero because w is nonzero. Let S contain i,j and every coordinate used by a fixed root word for g, and choose h\notin S. The consecutive-root relation gives

The two conjugated factors lie respectively in the strips from S to \{h\} and from \{h\} to S. More explicitly, write A_g=\pi(g)|_S. By Lemma 3.1,

Since R_{\ge M} is a two-sided ideal, every displayed entry still belongs to R_{\ge M}. Expand the two strip elements into roots, expand each entry in the word basis, and then expand their commutator. All resulting root entries have length at least M. Doing this for the finitely many conjugates in a normal-closure expression proves the lemma.□

Proof. Let z lie in the displayed intersection. Its matrix image becomes the identity modulo every R_{\ge d}. The grading has \bigcap_dR_{\ge d}=0, so its matrix image is the identity over R and z\in J(R). Taking d=1 also shows that z maps to 1 in \St(\F_2).

By Lemma 4.4, W_z(C_b)=1 for every sufficiently large integer b. Choose an odd phase modulus L from Lemma 2.3 beyond this threshold. This choice uses a fixed representative of z and precedes every choice of a long-entry representative. Now choose its phase threshold N and an integer M\ge2N+k+1.

The element z is trivial in the truncation at degree 2M. Lemma 6.3 supplies a representative with nonzero word entries of lengths at least M. Applying Proposition 6.2 to this representative yields

On the other hand, Proposition 4.5 gives z^{2^a}=1 for some a\ge0, since z\in J(R) has trivial reduction to \St(\F_2). The integers L and 2^a are coprime. Thus z=1.□

Proof. The extension is central by Proposition 7.1(ii), because its kernel acts trivially on the strip. Both its source and target are perfect. Indeed, an internal root is a consecutive-root commutator, and a strip root has the same expression using an intermediate index among the first r coordinates.

Consider an arbitrary central extension T\to S_{r,l}. The roots available in S_{r,l} are e_{ij}(a) with j\leq r and i\ne j, where i may also be one of the last l coordinates. Choose arbitrary lifts y_{ij}(a)\in T. We first show that lifts of roots appearing in a Steinberg commuting relation actually commute:

Choose h\leq r distinct from i,j,k,t. In matrices,

and both factors on the right commute with e_{kt}(b). In T, their commutators with y_{kt}(b) are therefore central. Conjugating each factor by y_{kt}(b) changes it only by a central factor, so fixes their commutator. Since this commutator differs from y_{ij}(a) by a central element, (7.3) follows. The bound r\geq6 supplies every spare index used below as well.

Define corrected lifts by

They are independent of h and satisfy

To prove this, choose a fourth distinct index f\leq r and apply Lemma 3.4 to

All its commuting hypotheses follow from (7.3), since [X,Y] and [Y,Z] lift, respectively, e_{if}(ab) and e_{hj}(b). The resulting identity equates the left side of (7.5) with [y_{if}(ab),y_{fj}(1)]: replacing a lift by a central multiple does not change a commutator. Setting b=1 proves independence in (7.4); the same identity then proves (7.5) for arbitrary b.

The corrected lifts are additive in their parameters. In (7.4), use y_{ih}(a)y_{ih}(a') as a lift of e_{ih}(a+a') and expand its commutator with y_{hj}(1). Equation (7.3) allows the commutator factors to pass the ih factors, giving

Here \widehat y_{ij}(0)=1, since any lift of an identity matrix is central. Together with (7.3) and (7.5), this proves all Steinberg relations for the corrected lifts on the available roots.

On the first r coordinates they therefore define a homomorphism \St_r(B)\to T. On the strip they define a homomorphism from the additive group \Mat_{l\times r}(B) to T. These homomorphisms respect the semidirect-product action: conjugation of a strip root by an internal root is determined by a commuting relation, except when their indices form a consecutive path, in which case it is determined by (7.5). These are exactly the matrix conjugation rules. Thus they give a homomorphism \widetilde S_{r,l}\to T over S_{r,l}. It is unique, since two such homomorphisms differ by a homomorphism from the perfect group \widetilde S_{r,l} to the central, abelian kernel of T\to S_{r,l}. This proves universality.

Taking l=0 identifies H_2(\E_r(B);\Z) with J_r(B). The inclusion \E_r(B)\to S_{r,l} lifts by adjoining the zero strip and is the identity on the kernels J_r(B). Naturality of the universal central extension now proves (7.2).□

Proof. First let V be a finite labeled collection of nonzero vectors spanning a vector space of dimension m. Different labels remain different vertices even if their vectors agree. Write C_\bullet(V) for the augmented complex on its ordered independent tuples. We prove the claimed vanishing by induction on the number of labels, allowing m=0: the empty collection has its only chain group in degree -1, and the asserted range is then empty.

Delete a label v, and let V'=V\setminus\{v\}. In the relative complex

the generators are the independent tuples containing v. Filter Q_\bullet increasingly by the number a of labels before v. Deleting v is zero in the relative complex; deleting a label before it lowers the filtration; deleting a label after it preserves the filtration. Thus the associated graded complex for a fixed a is a direct sum, over independent ordered prefixes (u_1,\ldots,u_a,v), of complexes in which only the suffix varies.

For one such prefix, project all remaining labeled vectors to the quotient by \langle u_1,\ldots,u_a,v\rangle, and discard the zero projections. The projected collection spans dimension m-a-1 and has fewer labels than V. Its augmented ordered-tuple complex, shifted up by a+1 degrees, is precisely the corresponding summand of the associated graded complex, up to the constant boundary sign (-1)^{a+1}. By induction its homology vanishes through degree

This also covers a zero-dimensional quotient: its sole homology in degree -1 shifts to degree a=m-1, outside the asserted range. The filtration is finite, so H_i(Q_\bullet)=0 for i\leq m-2.

If V' still spans dimension m, induction and the long exact sequence of the pair give the required vanishing for V. Otherwise V' spans dimension m-1, and v is independent of its entire span. Prepending v to every tuple of V' gives a chain homotopy from its inclusion into C_\bullet(V) to zero; in augmented degree -1 it sends 1 to (v). The induced maps H_i(C_\bullet(V'))\to H_i(C_\bullet(V)) are therefore zero. They are also surjective for i\leq m-2, by the relative vanishing just proved. The homology of C_\bullet(V) vanishes in this range.

Finally, any cycle in C_\bullet(m) uses finitely many columns. Adjoin the standard basis columns to this finite collection. Their reductions, with column labels retained, form a finite collection spanning \F_2^m. The finite result supplies a bounding chain in the asserted range, proving the Lemma.□

Proof of Theorem 7.2. We now combine the stabilizer calculation with the ordered-frame complex to prove Theorem 7.2. Fix m\geq13 and let \E_m(B) act on ordered frames by left multiplication. For frame lengths l\leq5, this action is transitive. To see this, complete a given frame to an invertible matrix, placing it in the last l columns, and put r=m-l. By Proposition 7.1(iii), the map \GL_r/\E_r\to\GL_m/\E_m is an isomorphism. Right multiplication by a suitable complementary \GL_r block therefore brings the completion into \E_m without changing its last l columns.

The pointwise stabilizer of the last l standard columns is exactly S_{m-l,l} from (7.1). Indeed, its general form has A\in\GL_{m-l} and arbitrary lower-left strip. The strip is elementary, and the injectivity of \GL_{m-l}/\E_{m-l}\to\GL_m/\E_m says that this matrix belongs to \E_m precisely when A\in\E_{m-l}. Write S_l=S_{m-l,l} for the rest of the proof. All these complementary ranks are at least eight, so Lemma 7.3 applies.

Take a free right \Z[\E_m]-resolution P_\bullet\to\Z of the trivial module and form the first-quadrant double complex

This is the usual equivariant homology double complex; see [1, Chapter VII]. Its total homology in degree two is H_2(\E_m;\Z) via augmentation. For completeness, adjoining the augmentation column p=-1 gives the same tensor product with the augmented frame complex. Each P_q is free and hence flat, so Lemma 7.4 makes the rows exact in the stated range. The augmented total complex has zero homology in degrees one and two. The long exact sequence separating its augmentation column from (7.6) consequently gives the asserted isomorphism.

Take vertical homology first. For 0\leq p\leq4, the permutation module on ordered (p+1)-frames has one orbit and stabilizer S_{p+1}. Shapiro’s Lemma identifies the first page as

The first differential is the alternating sum of the face maps, each expressed by a stabilizer inclusion and a coordinate change that returns the remaining frame to the chosen standard frame.

The row q=1 is zero in these columns, because every S_l is perfect. In row q=0, all entries are \Z and each face map is the identity. Hence the first differential is zero for odd p and the identity for even p. In particular,

In row q=2, the two face maps from column one to column zero coincide. Indeed, the inclusion \E_{m-2}\to S_2 induces an isomorphism on H_2 by Lemma 7.3, so it suffices to compare the faces on this complementary block. The first standard frame is (e_{m-1},e_m). Deleting its first column gives e_m. Deleting its second column gives e_{m-1}, which is returned to e_m by swapping the last two coordinates. This swap is elementary in characteristic two and centralizes the first m-2 coordinates. The two induced maps on that block are therefore identical. Their alternating sum vanishes, and

No higher differential leaves position (0,2). The only possible incoming differentials are from (2,1) on the second page and (3,0) on the third page; both vanish by (7.8). The other positions of total degree two, (1,1) and (2,0), also vanish. The spectral sequence thus shows that the edge map

is an isomorphism. This is the map induced by inclusion: the zero-th column corresponds to the standard column e_m, and its map into the total complex followed by augmentation is the usual homology map for its stabilizer inclusion.

Combining with (7.2), we have proved that the natural stabilization map

is an isomorphism for m\geq13. The natural identifications by universal central extensions turn this into the isomorphism J_{m-1}(B)\to J_m(B).

Finally, an element killed by \St_n(B)\to\St(B) already has identity matrix image, since coordinate embeddings of matrix groups are injective. It therefore lies in J_n(B). If n\geq12, the successive isomorphisms on these kernels show that it cannot become trivial at any later rank. The stable group is the direct limit, so this proves the injectivity asserted in Theorem 7.2.□

Proof. Put B_d=R/R_{\ge d}. This ring is finite: it is the direct sum of finitely many finite-dimensional vector spaces over \F_2. Moreover B_d=\F_2\oplus I_d with I_d^d=0, so every finite square matrix over I_d is nilpotent. The centrality assertion in Proposition 7.1 therefore applies to B_d. The kernel of

is central, and the matrix group on the right is finite. Hence the center of \St_{12}(B_d) has finite index. Schur’s theorem states that a group with center of finite index has finite derived subgroup [15]; see also [14, Theorem 10.1.4]. Since \St_{12}(B_d) is perfect, it is itself finite.

Let g\ne1 in \St_{12}(R). Theorem 7.2 makes its image in \St(R) nonidentity, and Theorem 6.4 detects that image in \St(B_d) for some d. The natural square

commutes on each root generator. Thus the image of g in \St_{12}(B_d) is nonidentity. No injectivity assertion for the bottom arrow is needed. This proves separation by the finite groups just constructed.□

Proof of Theorem 1.1. The algebra R is infinite and finitely presented as a unital \F_2-algebra. Accordingly \Gamma=\St_{12}(R) has an ordinary finite presentation by [13, Theorem 4.1]; this is also a case of [8, Theorem 3]. The group \Gamma is infinite, since its matrix image contains distinct matrices e_{12}(a) for all a\in R. It is residually finite by Proposition 8.1.

Define

where the map is induced by the degree-zero projection of R. The case d=1 of Proposition 8.1 shows that this target is finite. Thus G has finite index in \Gamma. It is infinite and has an ordinary finite presentation by the Reidemeister–Schreier theorem [9, Chapter II, Section 4]. Restricting the separating finite-quotient maps of \Gamma proves residual finiteness of G.

Fix g\in G. Its matrix image is \Id+A for some A\in M_{12}(I). Proposition 2.1 gives an integer r\ge1 with A^r=0. Choose a\ge0 with 2^a\ge r. In characteristic two,

Consequently g^{2^a}\in J_{12}(R), and its reduction to \St_{12}(\F_2) is still trivial. Its stable image belongs to J(R) and reduces to 1 in \St(\F_2), so Proposition 4.5 kills that image by a 2-power 2^b. The injectivity in Theorem 7.2 then yields g^{2^{a+b}}=1 in \Gamma, hence in G. This proves every assertion of Theorem 1.1.□