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I'm calculating a quotient group $A/B$ where $A:=\left\{ ax+by+\sum c_i z_i \middle|\, a,b,c_i\in\mathbb{Z}, \sum c_i =0 \right\}$ $B:=\left\{ x\sum a_i +y\sum b_i +\sum (a_{i-1}-a_i + b_{i-1}-b_i) z_i \middle|\, a_i,b_i\in\mathbb{Z} \right\}$ $x,y,z_i$ are formal variables, sums are all $i=0,\ldots,n-1$ where $n$ is fixed, and addition on indices is modulo $n$ (i.e., $a_{(n-1)+1}=a_0$).

The answer I expect is $A/B\approx\mathbb{Z}_n$, but I'm not seeing how to get there. Ideas?

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    @DonAntonio: When I say "addition on indices is modulo $n$", I mean $a_{(n-1)+1} = a_0$. Likewise, $a_{0-1} = a_n$, so the sum needn't start from $i=1$.2012-11-04

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For each $i$, let $u_i=z_{i+1}-z_i$. Then $A$ is generated by $x$, $y$, and the classes $u_i$ for $i=0,\ldots,n-1$, and has the relation $\sum u_i = 0$. In this notation, $B=\{x\sum a_i + y\sum b_j + \sum (a_i+b_i) u_i \mid a_i,b_i\in \mathbb{Z}\,\forall i\in\mathbb{Z}/n\mathbb{Z}\}$.

Pick an integer $k$ and let $a_k=1$, $a_{k+1}=-1$, $a_j=0$ for $j\neq k$ and $b_j=0$. Then $u_{k+1}-u_k\in B$. Hence $\overline{u_k}=\overline{u_{k+1}}$ as elements of $A/B$. Therefore the value is $\overline{u_k}$ is independent of $k$; we will denote this value by $\bar{u}$.

You can easily check that $\bar{x}=\bar{y}=-\bar{u}$. Then $\bar{u}$ generates $A/B$. We have only the relation $n\bar{u}=\sum \overline{u_j}=0$; the relations $\bar{x}\sum a_i + \bar{y}\sum b_j + \sum (a_i+b_i) \overline{u_i}=0$ become $-\bar{u} \sum a_i -\bar{u} \sum b_j + \sum (a_i+b_i) \bar{u}=0$, which is trivial. The group generated by $\bar{u}$ with the relation $n\bar{u}=0$ is of course $\mathbb{Z}/n\mathbb{Z}$.

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    Right, because $\sum d_i u_i = \sum (d_{i-1} - d_i)z_i$. You meant $u_k - u_{k+1}\in B$? Same conclusion. $a_k=1$ gives $x+u_k\in B$, so $\bar{x}=-\bar{u}$, etc. Clever.2012-11-04