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Let $$G=\bigoplus_{n\in\mathbb{Z}}\left(\mathbb{Z}/2\mathbb{Z}\right)_n$$ be a group, and for any $n\in \mathbb{Z}$, denote $\delta_n$ to be the element in $G$ with $n$-th coordinate $1$ and zero at all other coordinates, and for any $g, h\in G$, let $g\oplus h\in G$ to be usual addition of $g$ and $h$ in $G$. (I use this notation in order to differentiate the addition in the group $G$ and the addition in the group ring $\mathbb{Z}G$.)

I guess we can prove that

For any $n, l\in \mathbb{Z}^{+}$, $s_1,s_2,\dots,s_n,\lambda_{i,j}, c\in \mathbb{Z}$, $s_1<s_2<\cdots<s_n$, $c\neq 1$, $\sum_{i=1}^n\sum_{j=1}^l|\lambda_{i,j}|\not=0$ and any $g_{i,j}\in G$ with $\{g_{i,j}|~j=1,\dots, l\}$ are $l$ distinct elements for any fixed $i=1,\cdots, n$. (Note that we do not require $\{g_{i,j}|i=1,\dots,n;j=1,\dots,l\}$ are all distinct.)

We always have:

$$c\sum_{i=1}^{n}\sum_{j=1}^l\lambda_{i,j}g_{i,j}\not=\sum_{i=1}^{n}\sum_{j=1}^l\lambda_{i,j}(g_{i,j}\oplus \delta_{s_i})$$ in the group ring $\mathbb{Z}G$.

Can anyone help prove it or give a counterexample?


Note that for the special case $n=1$, it is not difficult to prove the above inequality holds, but when $n>1$ it becomes too complicated for me to prove it...

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Where have the $s_j$'s gone? I can't see them. –  1015 Mar 22 '13 at 0:55
    
In the right hand side of the inequality, I use the $\delta_{s_i}$. –  ougao Mar 22 '13 at 1:07

2 Answers 2

There's a map of rings $\mathbb{Z}G \to \mathbb{Z}$ given by sending each $g \in G$ to $1$. Under this map, the images of the two sides of your inequality are respectively $c\sum_i \sum_j \lambda_{i,j}$ and $\sum_i \sum_j \lambda_{i,j}$. Since $c \ne 1$, this is all you need.

(I'd wager this is also true for $c = 1$, maybe under certain other conditions on the $g_{i,j}$ and $s_i$. My argument doesn't use anything about the group itself, or any of your conditions other than $c \ne 1$. May I ask why you want to know this, or in what context it came up? Maybe there's a more enlightening answer to be given.)

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@ Paul, but it can happen that $\sum_i\sum_j\lambda_{i,j}$ can be zero. And I'd add the requirement $\sum_{i=1}^n\sum_{j=1}^l|\lambda_{i,j}|\not=0$ to rule out the trivial case. –  ougao Mar 22 '13 at 1:13
    
@ Paul, I want to show some relation can not hold in some group ring, and a natural attempt to disprove this relation reduced to prove the inequality holds in the above question. –  ougao Mar 22 '13 at 1:25
up vote 0 down vote accepted

Omit this question, my guess is not right, the equality could hold, but it might be not easy to see this fact directly from my formulation in this way.

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You asked for either a proof or counterexample in the question, and since your claim here is negative it'd be great if you bolstered this answer with a counterexample. –  anon Mar 22 '13 at 23:17
    
@ anon, I also want to find a explicit example here, but the point is that the motivation to ask this question is that initially I guess I could find a counterexample to an inequality involving some relation, say R, in a group ring, say $\mathbb{Z}\Gamma$(but not the one mentioned in the problem!), and elementary calculation has reduced the inequality to this one, which looks like a nontrivial problem in combinatorics, then I stuck in finding a counterexample, so I asked here, but today, someone remind me the relation R, always holds by modifying a proof for a related problem. –  ougao Mar 23 '13 at 1:15
    
of course, the proof uses some property on the group $\Gamma$ and the key point is that the proof also does not provide a concrete example. So, you know, as often happened in math, one problem can be solved easily when it is stated in a proper form, but may seem to be difficult even in an equivalent but other form, say as a combinatorics problem in this example. –  ougao Mar 23 '13 at 1:20
    
of course, I still expect to see a concrete example to show the equality can hold in the above problem. –  ougao Mar 23 '13 at 1:21

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