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Let $G=\langle x, y, z| xyx^{-1}=zy, xzx^{-1}=z, yz=zy\rangle$, 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:

This problem is related to the Ore condition. I want to show that $l$ does not exist, suppose it exists, then I have considered the natural quotient map $\phi: G\to H=G/<z^2>$. 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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  • $\begingroup$ Can you provide more motivation? What have you tried, etc? $\endgroup$ – Cheerful Parsnip May 1 '13 at 15:18
  • $\begingroup$ @ Grumpy, it is related to the 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.. $\endgroup$ – ougao May 1 '13 at 16:11
  • $\begingroup$ Would you mind editing your question to include this? $\endgroup$ – Cheerful Parsnip May 2 '13 at 0:00
  • $\begingroup$ @ Grumpy, I have edit it. Hope it help. $\endgroup$ – ougao May 2 '13 at 3:36
  • $\begingroup$ see the element $l$ given here: mathoverflow.net/questions/132311/… $\endgroup$ – ougao Jul 3 '13 at 3:05
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The above link in the comment contains a solution by ad hoc method.

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