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        STABLY FREE MODULES OVER $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }^{m}]$ ARE FREE
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In the paper [1] I claimed to show that there are no non-trivial stably free modules over integral group rings of the groups $\,(C_{p}\rtimes C_{q})\times C_{\infty }^{m}$ . Unfortunately there are a number of erroneous statements in [1] which vitiate the attempted proof. To explain where these occur, recall that in [1] two Milnor fibre squares $\,(\clubsuit )\,$ and $\,(\heartsuit )\,$ were introduced as follows:

Here $\unicode[STIX]{x1D6E4}=C_{\infty }^{m}$ , and ${\mathcal{T}}_{q}={\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})$ is the ring of quasi-triangular $q\times q$ matrices where $A=\mathbf{Z}[\unicode[STIX]{x1D701}_{p}]^{C_{q}}$ is the subring of the cyclotomic integers $\mathbf{Z}[\unicode[STIX]{x1D701}_{p}]$ fixed under the Galois action of $C_{q}$ and $\unicode[STIX]{x1D70B}\in \text{Spec}(A)$ is the unique prime over $p$ .

The most obvious errors [1, Corollary 3.4] include a misdescription of the unit group $U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])$ , and the possibility of non-trivial rank-one stably free modules over ${\mathcal{T}}_{2}[\unicode[STIX]{x1D6E4}]$ . A slightly less obvious but more significant error concerns the possibility of lifting units from $\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]$ to ${\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]$ . In consequence we must amend the original statement of [1] as follows.

Theorem A. Let $S$ be a stably free module of rank $n$ over $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }^{m}]$ where $m\geqslant 2$ . Then:

  • if $q$ is an odd prime, $S$ is free provided $n\neq 2$ ; and

  • if $q=2$ , $S$ is free provided $n\geqslant 3$ .

Nevertheless, when $m=1$ the original statement continues to hold:

Theorem B. Any stably free module over $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }]$ is free.

Rather than try to patch up the proof in [1] piecemeal we give a more straightforward approach which isolates the real difficulty and avoids it where possible. We first establish four propositions.

Proposition 1. $U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ is finite; in fact

$$\begin{eqnarray}|U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])|\leqslant |U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q})|.\end{eqnarray}$$

Proof. As $q$ is a divisor of $p-1$ , we have

$$\begin{eqnarray}\mathbf{F}_{p}[C_{q}]\;\cong \;\underbrace{\mathbf{F}_{p}\times \cdots \times \mathbf{F}_{p}}_{q}.\end{eqnarray}$$

Consequently

(*) $$\begin{eqnarray}\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]\cong \underbrace{\mathbf{F}_{p}[\unicode[STIX]{x1D6E4}]\times \cdots \times \mathbf{F}_{p}}_{q}[\unicode[STIX]{x1D6E4}].\end{eqnarray}$$

Observe that $U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ contains a copy of $\unicode[STIX]{x1D6E4}^{(q)}=\underbrace{\unicode[STIX]{x1D6E4}\times \cdots \times \unicode[STIX]{x1D6E4}}_{q}$ , namely the diagonal matrices

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FE}_{1},\ldots ,\unicode[STIX]{x1D6FE}_{q})=\left(\begin{smallmatrix}\unicode[STIX]{x1D6FE}_{1}\\ & \unicode[STIX]{x1D6FE}_{2} & \\ & & \ddots \\ & & & \unicode[STIX]{x1D6FE}_{q}\end{smallmatrix}\right),\end{eqnarray}$$

where $\unicode[STIX]{x1D6FE}_{i}\in \unicode[STIX]{x1D6E4}$ . Combining this with the obvious inclusion $U({\mathcal{T}}_{q})\subset U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ gives an injection $U({\mathcal{T}}_{q})\times \unicode[STIX]{x1D6E4}^{(q)}{\hookrightarrow}U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ . Hence we now have a surjection

(**) $$\begin{eqnarray}U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q})\times \unicode[STIX]{x1D6E4}^{(q)}{\twoheadrightarrow}U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]).\end{eqnarray}$$

The ring isomorphism (*) now gives an isomorphism of unit groups

$$\begin{eqnarray}U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])\cong \underbrace{U(\mathbf{F}_{p}[\unicode[STIX]{x1D6E4}])\times \cdots \times U(\mathbf{F}_{p}}_{q}[\unicode[STIX]{x1D6E4}]\,).\end{eqnarray}$$

Now $\unicode[STIX]{x1D6E4}=C_{\infty }^{m}$ is a t.u.p. group so $\mathbf{F}_{p}[\unicode[STIX]{x1D6E4}]$ has only trivial units (cf. [2, Appendix C]). Hence

$$\begin{eqnarray}\displaystyle U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]) & \cong & \displaystyle \underbrace{(U(\mathbf{F}_{p})\times \unicode[STIX]{x1D6E4})\times \cdots \times (U(\mathbf{F}_{p})\times \unicode[STIX]{x1D6E4})}_{q}\nonumber\\ \displaystyle & \cong & \displaystyle \underbrace{U(\mathbf{F}_{p})\times \cdots \times U(\mathbf{F}_{p})}_{q}\times \unicode[STIX]{x1D6E4}^{(q)}\nonumber\end{eqnarray}$$

so that, by (*), there are bijections

$$\begin{eqnarray}\displaystyle U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q})\times \unicode[STIX]{x1D6E4}^{(q)} & \leftrightarrow & \displaystyle U(\mathbf{F}_{p}[C_{q}])\times \unicode[STIX]{x1D6E4}^{(q)}/U({\mathcal{T}}_{q})\times \unicode[STIX]{x1D6E4}^{(q)}\nonumber\\ \displaystyle & \leftrightarrow & \displaystyle U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q}).\nonumber\end{eqnarray}$$

From (**) we obtain a surjection $U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q}){\twoheadrightarrow}U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U$ $({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ . The stated result now follows as $U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q})$ is finite.◻

Proposition 2. Let $p$ be an odd prime and $q$ be a divisor of $p-1$ . Then, for all $n\geqslant 3$ ,

$$\begin{eqnarray}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])=U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])\cdot E_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]).\end{eqnarray}$$

Proof. Given rings $A,B$ such that $\text{GL}_{n}(A)=U(A)E_{n}(A)$ and $\text{GL}_{n}(B)=U(B)E_{n}(B)$ , we have $\text{GL}_{n}(A\times B)=U(A\times B)E_{n}(A\times B)$ . The result thus follows from (*) by induction on $q$ , the case $q=1$ being Suslin’s theorem [3], namely that

$$\begin{eqnarray}\text{GL}_{k}(\mathbf{F}[\unicode[STIX]{x1D6E4}])=U(\mathbf{F}[\unicode[STIX]{x1D6E4}])\cdot E_{k}(\mathbf{F}[\unicode[STIX]{x1D6E4}])\end{eqnarray}$$

for any field $\mathbf{F}$ and any integer $k\geqslant 3$ .◻

Proposition 3. $\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ is finite for $n\geqslant 3$ ; in fact

$$\begin{eqnarray}|\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])|\leqslant |U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q})|.\end{eqnarray}$$

Proof. Evidently $U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])E_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])\subset \text{GL}_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ and so there is a natural surjection $\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])E_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]){\twoheadrightarrow}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ . Also, the ring homomorphism $\natural :{\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}]\rightarrow \mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]$ is surjective and so induces surjections $\natural _{\ast }:E_{k}({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])\rightarrow E_{k}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])$ for all $k\geqslant 2$ . By Proposition 2 we may write

$$\begin{eqnarray}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])=U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])E_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]).\end{eqnarray}$$

We obtain a surjection $U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]){\twoheadrightarrow}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{n}$ $({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ and so the stated result now follows from Proposition 1.◻

Proposition 4. Let $p$ be an odd prime and $q$ be a divisor of $p-1$ . Then, for all $n\geqslant 1$ ,

$$\begin{eqnarray}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times C_{\infty }])=U(\mathbf{F}_{p}[C_{q}\times C_{\infty }])\cdot E_{n}(\mathbf{F}_{p}[C_{q}\times C_{\infty }]).\end{eqnarray}$$

Proof. We follow the same line of argument as Proposition 2 with the exception that, in establishing the induction base, we do not use Suslin’s theorem. Instead we note that, as $\mathbf{F}_{p}[C_{\infty }]$ is a Euclidean domain, we may use the Smith normal form to show that $\text{GL}_{k}(\mathbf{F}_{p}[C_{\infty }])=U(\mathbf{F}_{p}[C_{\infty }])\cdot$ $E_{k}(\mathbf{F}_{p}[C_{\infty }])$ .◻

As in [1], we denote the set of isomorphism classes of locally free $\mathbf{Z}[C_{p}\rtimes C_{q}]$ -modules of rank $k$ by ${\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )$ . By Milnor’s classification, this corresponds to the two-sided quotient

$$\begin{eqnarray}{\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )=\text{GL}_{k}(\mathbf{Z}[C_{q}])\backslash \text{GL}_{k}(\mathbf{F}_{p}[C_{q}])/\text{GL}_{k}({\mathcal{T}}_{q}).\end{eqnarray}$$

Likewise, the locally free $\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}]$ -modules of rank $k$ correspond to the quotient

$$\begin{eqnarray}{\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )=\text{GL}_{k}(\mathbf{Z}[C_{q}\times \unicode[STIX]{x1D6E4}])\backslash \text{GL}_{k}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{k}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]).\end{eqnarray}$$

In particular, if neither $\mathbf{Z}[C_{q}]$ nor ${\mathcal{T}}_{q}$ admits non-trivial stably free modules of rank $k$ , then any stably free module of rank $k$ over $\mathbf{Z}[C_{p}\rtimes C_{q}]$ is locally free. Consequently, the set ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[C_{p}\rtimes C_{q}])$ of stably free modules of rank $k$ over $\mathbf{Z}[C_{p}\rtimes C_{q}]$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )$ . Similarly, ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}])$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$ if neither $\mathbf{Z}[C_{q}\times \unicode[STIX]{x1D6E4}]$ nor ${\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]$ admits non-trivial stably free modules of rank $k$ .

There are obvious mappings of fibre squares $i:(\clubsuit ){\hookrightarrow}(\heartsuit )$ and $r:(\heartsuit )\rightarrow (\clubsuit )$ such that $r\circ i=\text{Id}$ . Consequently, there is a commutative ladder of mappings

where $s_{k,1}$ and $\unicode[STIX]{x1D70E}_{k,1}$ are the obvious stabilization mappings. We note that the mappings $i_{k}$ are injective in view of the fact that $r\circ i=\text{Id}$ .

The argument is now divided into two cases: $q$ is odd, and $q=2$ . First, suppose $q$ is an odd prime dividing $p-1$ . As noted in [1], in $(\clubsuit )$ , the rings $\mathbf{Z}[C_{q}]$ and ${\mathcal{T}}_{q}$ both have property SFC. Consequently, ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[C_{p}\rtimes C_{q}])$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )$ for all $k\geqslant 1$ . Similarly, in the fibre square $(\heartsuit )$ , the rings $\mathbf{Z}[C_{q}\times \unicode[STIX]{x1D6E4}]$ and ${\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]$ also have SFC and once again ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}])$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$ for all $k\geqslant 1$ . The essence of the argument now consists of the following five statements.

  1. (I) For all $n$ , ${\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )$ is finite and $s_{n,1}:{\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{n+1}(\clubsuit )$ is bijective.

  2. (II) $i_{1}:{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{1}(\heartsuit )$ is bijective.

  3. (III) $i_{n}:{\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{n}(\heartsuit )$ is bijective for all $n\geqslant 3$ .

  4. (IV) $\unicode[STIX]{x1D70E}_{n,1}:{\mathcal{L}}{\mathcal{F}}_{n}(\heartsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{n+1}(\heartsuit )$ is injective provided $n\neq 2$ .

  5. (V) If $m=1$ (that is, $\unicode[STIX]{x1D6E4}\;=\;C_{\infty }$ ) then $i_{2}:{\mathcal{L}}{\mathcal{F}}_{2}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{2}(\heartsuit )$ is bijective.

To prove (I) we note that, as $C_{p}\rtimes C_{q}$ is finite, the finiteness of ${\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )$ follows from the Jordan–Zassenhaus theorem, together with Milnor’s classification of projectives. Moreover, as $\mathbf{Z}[C_{p}\rtimes C_{q}]$ satisfies the Eichler condition, the Swan–Jacobinski theorem shows that each $s_{k,1}:{\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{k+1}(\clubsuit )$ is bijective. It follows from Proposition 1 that $|{\mathcal{L}}{\mathcal{F}}_{1}(\heartsuit )|\leqslant |{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )|$ . Thus (II) is true as $i_{1}$ is injective and ${\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )$ is finite. Likewise it follows from Proposition 3 that $|{\mathcal{L}}{\mathcal{F}}_{n}(\heartsuit )|\leqslant |{\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )|$ for $n\geqslant 3$ . Thus (III) is true as $i_{n}$ is injective and ${\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )$ is finite; (IV) now follows from (I), (II) and (III) by diagram chasing using the fact that $i_{2}$ is injective. Finally, (V) follows by the same argument as (III) on substituting Proposition 4 for Proposition 2.

To proceed with the proof of Theorem A, put $\unicode[STIX]{x1D70E}_{n,k}=\unicode[STIX]{x1D70E}_{n+k-1,1}\circ \ldots \circ \unicode[STIX]{x1D70E}_{n,1}$ whenever $k\geqslant 1$ . It follows from (IV) that $\unicode[STIX]{x1D70E}_{n,k}$ is injective provided $n\geqslant 3$ . A straightforward diagram chase using (I), (II) and (III) also shows that each $\unicode[STIX]{x1D70E}_{1,k}$ is injective. Now suppose that $S$ is a stably free module of rank $n\neq 2$ over $\unicode[STIX]{x1D6EC}=\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}])$ and denote its class in ${\mathcal{L}}{\mathcal{F}}_{n}(\heartsuit )$ by $[S]$ . Then $S\oplus \unicode[STIX]{x1D6EC}^{k}\cong \unicode[STIX]{x1D6EC}^{n+k}$ for some $k\geqslant 1$ so that $\unicode[STIX]{x1D70E}_{n,k}[S]=\unicode[STIX]{x1D70E}_{n,k}[\unicode[STIX]{x1D6EC}^{n}]$ . As $\unicode[STIX]{x1D70E}_{n,k}$ is injective, $S\cong \unicode[STIX]{x1D6EC}^{n}$ . Consequently, when $n\neq 2$ there are no non-trivial stably free modules of rank $n$ over $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }^{m}]$ , and this proves the first part of Theorem A.

In the case $q=2$ (i.e. dihedral groups) we cannot claim ${\mathcal{T}}_{2}[\unicode[STIX]{x1D6E4}]$ has property SFC. To see why, consider the square

As $(A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}]$ is commutative, we have $\text{GL}_{2}((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])=U((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])\cdot \text{SL}_{2}((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])$ . The unit group $U((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])$ lifts back to ${\mathcal{T}}_{2}((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])$ . However, it is not clear whether we can lift the elements of $\text{SL}_{2}((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])$ . Thus, it is conceivable that ${\mathcal{T}}_{2}(A,\,\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}]$ has non-trivial stably free modules of rank 1. Nevertheless, using Suslin’s theorem as before, it is clear that ${\mathcal{T}}_{2}(A,\,\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}]$ admits no non-trivial stably free module of rank ${\geqslant}2$ . Consequently, we observe that ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}])$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$ for all $k\geqslant 2$ . We now proceed as above.

Finally, the proof of Theorem B follows exactly the same lines except that now, in the case where $m=1$ and $\unicode[STIX]{x1D6E4}=C_{\infty }$ , we see from (V) that $\unicode[STIX]{x1D70E}_{2,1}$ is also injective. Consequently, each $\unicode[STIX]{x1D70E}_{2,k}$ is injective. Thus, there are no non-trivial stably free modules of any rank over $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }]$ .

References

1. Evans, J. D. P., Stably free modules over Z[(C p C q ) × C m ]) are free. Mathematika 63 2017, 451461.
2. Johnson, F. E. A., Syzygies and Homotopy Theory, Springer (London, 2011).
3. Suslin, A. A., On the structure of the special linear group over polynomial rings. Math. USSR, Izvestiya 11(2) 1977, 221238.