Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question
    
Can you provide more motivation? What have you tried, etc? –  Grumpy Parsnip May 1 '13 at 15:18
    
@ 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.. –  ougao May 1 '13 at 16:11
    
Would you mind editing your question to include this? –  Grumpy Parsnip May 2 '13 at 0:00
    
@ Grumpy, I have edit it. Hope it help. –  ougao May 2 '13 at 3:36
    
see the element $l$ given here: mathoverflow.net/questions/132311/… –  ougao Jul 3 '13 at 3:05

1 Answer 1

up vote 1 down vote accepted

The above link in the comment contains a solution by ad hoc method.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.