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Let $$G=\langle x, y, z\mid xyx^{-1}=zy, xzx^{-1}=z, yz=zy\rangle,$$ denote $l^1(G)^{\times}$ to be the set of units in $l^1(G)$, which we have considered as a ring with multiplication defined by the usual convolution, i.e., $(\sum_{g\in G}\lambda_gg)(\sum_{h\in G}\mu_hh)=\sum_{g, h\in G}\lambda_g\mu_hgh$.

Can we find $l=p_1(y, z)x^{n_1}+\cdots+ p_k(y,z)x^{n_k}\in l^1(G)^{\times}$ such that $$\sum_{i=1}^k2^{n_i}p_i(y,z)(1-z^{n_i}y)=0\;?$$

Here, $\forall~ 1\leq i\leq k, ~p_i(y,z)\in \mathbb{Z}G$ and $n_1<\cdots<n_k\in\mathbb{Z}$ to be determined. Note that the group element $x$ does not appear in $p_i(y, z)$.

Remarks:

1, This problem was asked previously in MSE, but no answer appeared, so I think it might be suitable for MO.

2, This problem is related to a modified Ore condition. I want to show that $l$ does not exist, suppose it exists, then I have considered the natural quotient map $\phi: G\to H=G/\langle z^2\rangle$. Note that it would induce a map $\phi: l^1(G)^{\times}\to l^1(H)^{\times}$, then $\phi(l)\in l^1(H)^{\times}$, but I still could not handle this..

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  • $\begingroup$ why some symbols could not appear? Anyone helps fix this? $\endgroup$
    – Jiang
    May 30, 2013 at 7:38
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    $\begingroup$ @Jiang: I've edited the MathJax code. There are some issues with how it works here, but they should go away when MathOverflow moves to the SE 2.0 network sometime soon: tea.mathoverflow.net/discussion/1416/1/mathoverflow-20 $\endgroup$
    – Zev Chonoles
    May 30, 2013 at 8:00
  • $\begingroup$ How about $l = 2(1-zy) - (1-y)x$? $\endgroup$
    – user91132
    Jun 4, 2013 at 8:39
  • $\begingroup$ @Konstantin, thanks for your comment. For this $l$, we can check that $l^{-1}=sum_{n\geq 0}\frac{1}{2^{n+1}}(1-y)(1-z^ny)^{-1}(1-z^{n+1}y)^{-1}x^n\not\in l^1(G)^{\times}$ $\endgroup$
    – Jiang
    Jun 16, 2013 at 9:53

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Omit this problem, we can take $l=z+\frac{1}{2}(-z-1)x+\frac{1}{4}x^2$, then $l^{-1}=\sum_{n\geq 0, m\geq 0}\frac{1}{2^{n+m}}x^{n+m}z^{-m-1}\in l^1(G)$.

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