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 $\{z_1,...,z_n\}$ be the canonical Basis of $\mathbb{Z}^n$, such that $z_i$ equals the vector $(0,\dotsc,0,i,0,\dotsc,0)$ with a 1 in the $i$th position. I want to show that the $\mathrm{SL}(n,\mathbb Z)$ is generated by the subgroup $\mathrm{SL}(n-1,\mathbb Z)$ corresponding to the subbasis $\{z_1,\dotsc,z_n-1\}$ and the product of the following two matrices:

  1. The Matrix A which transpose the two vectors $z_{n-1}$ and $z_n$.

  2. The Matrix B which changes $z_n$ into the element $-z_n$.

It's clear that $\det(AB)=1$, since $\det(A)=\det(B)=-1$. So it follows that $\{\mathrm{SL}(n1,\mathbb Z)\backslash\mathrm{SL}(n-1,\mathbb Z)\}$.

So how can I show that?

Thanks for help.

share|improve this question
4  
You'll need to modify your conjecture, because there is the intermediate subgroup of things of the form $\pmatrix{A&b\cr 0& d}$ where $A$ is in $SL(n-1,\mathbb Z)$ and $b$ is $(n-1)\times 1$, and $d\in \{\pm1\}$. –  paul garrett Jan 18 '12 at 19:39
    
Thanks. I modified my question. –  Peter Jan 18 '12 at 20:19
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.