Can we fully describe the group of units (=invertible elements) $(KG)^\times$ of the group algebra $KG$ for $K=\mathbf{F}_2$, $G=D_\infty=\langle s,t|s^2=t^2=1\rangle$, the infinite dihedral group?

I'm also interested in other fields $K$ (in which case one can focus on describing the quotient by its central subgroup $K^*$).

If $A=G_{\mathrm{ab}}$ is the Klein group, I'd already be happy in a description of the kernel of $(KG)^\times \to (KA)^\times$, which has finite index (for a finite field $K$).

This question is motivated by this question concerning a more complicated (but torsion-free) virtually abelian group. I'd also be interested in the same question replacing $G$ with the Klein bottle group, which is a non-abelian semidirect product $\mathbf{Z}\rtimes\mathbf{Z}$. Still, in a sense I'd like to take advantage of the semidirect product decomposition $\mathbf{Z}^d\rtimes$(finite). Note that when $G$ is torsion-free abelian, the group of units in $KG$ is reduced to $K^\times\times G$.

"Fully describe" is a bit unclear a priori, but I'd like a characterization of its elements within the group ring, if possible for which we can deduce the answer about the simplest natural questions about its structure (is it virtually abelian? solvable? etc.)


A description of the group of units of the group algebra $\mathbb{F}_2 D_\infty$ can be found in Theorem 4.1 of the following paper:

M. Mirowicz: Units in group rings of the infinite dihedral group, Can. Math. Bull. 34 (1991), 83-89

DOI link (the paper is accessible with no restriction at the moment)

  • $\begingroup$ Nice! If I understand correctly, it describes the group $(\mathbf{F}_2D_\infty)^\times$ as a certain explicit semidirect product $N\rtimes D_\infty$, with $N$ naturally isomorphic to $(C_2^{\mathbf{N}})^{\ast \mathbf{Z}}$, and checks it's not finitely generated. $\endgroup$ – YCor Mar 22 at 20:42
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    $\begingroup$ The paper starts with the observation in general that $RD_\infty$ is isomorphic as $R$-algebra with the subalgebra of matrices $\begin{pmatrix} a & b\\ b^* & a^*\end{pmatrix}$ with $a,b\in RC_\infty=R[t^\pm]$ and $P^*(t)=P(1/t)$. This somewhat generalizes: if $H$ has index $n$ in $G$ then $RG$ can be embedded in $M_n(RH)$. Namely $RG$ embeds canonically into $\mathrm{End}_{\mathrm{mod}-RH}(RG)$ by $a\mapsto f_a$, $f_a(x)=ax$, and $RG$ is a free right $RH$-module of rank $n$. In particular if $G$ is virtually abelian and $R$ is a domain then the group of units in $RG$ is linear. $\endgroup$ – YCor Mar 22 at 20:46
  • $\begingroup$ I agree. The same trick is sometimes used in the theory of polynomial identities of group algebras as well. $\endgroup$ – Salvatore Siciliano Mar 22 at 22:53
  • $\begingroup$ @Ycor since polycyclic groups can all be realised as matrix groups, don't you get the conclusion that if $G$ is virtually polycyclic, then $(RG)^\times$ is linear? (I'm asking since I'm probably jumping over an important point) $\endgroup$ – ARG Mar 23 at 8:46
  • $\begingroup$ @ARG no, it just makes it linear over the non-commutative ring $RH$ for $H$ polycyclic. I really need $G$ virtually abelian to get linearity over something commutative. Actually, for, say, $G$ discrete Heisenberg, I don't know whether every f.g. subgroup of $M_n(\mathbf{C}G) $ is linear. $\endgroup$ – YCor Mar 23 at 9:15

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