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...

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

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.)

  • $\begingroup$ @ 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. $\endgroup$ – ougao Mar 22 '13 at 1:13
  • $\begingroup$ @ 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. $\endgroup$ – ougao Mar 22 '13 at 1:25

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.

  • $\begingroup$ 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. $\endgroup$ – anon Mar 22 '13 at 23:17
  • $\begingroup$ @ 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. $\endgroup$ – ougao Mar 23 '13 at 1:15
  • $\begingroup$ 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. $\endgroup$ – ougao Mar 23 '13 at 1:20
  • $\begingroup$ of course, I still expect to see a concrete example to show the equality can hold in the above problem. $\endgroup$ – ougao Mar 23 '13 at 1:21

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