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I want to show, that $GL_n(\mathbb{Z}/b\mathbb{Z})$ operates transitively on $X = \{ (v_1, \ldots, v_n) \in (\mathbb{Z}/b\mathbb{Z})^n \ | \ v_1\mathbb{Z}/b\mathbb{Z} + \ldots + v_n\mathbb{Z}/b\mathbb{Z} = \mathbb{Z}/b\mathbb{Z}\} $
($b$ is an integer.)


  1. $A \in GL_n(\mathbb{Z}/b\mathbb{Z}), x \in X \Rightarrow Ax \in X$
  2. For all $x,y \in X$ exists $A \in GL_n(\mathbb{Z}/b\mathbb{Z})$ such that $x = Ay$.

I have no idea, where to start.


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closed as too localized by José Figueroa-O'Farrill, Ben Webster Apr 25 '10 at 12:51

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First, this question is not really appropriate for MO. Second, the general linear group does not act on any given affine hyperplane. The subgroup which does is the affine linear group in one dimension lower. This acts transitively because of the translations, as in Robin's answer. – José Figueroa-O'Farrill Apr 25 '10 at 11:07
MO isn't intended for questions of the type that would be HW in an undergraduate class (whether they are or not), so your question has been closed. Some other sites that might work better for you are listed in the FAQ. – Ben Webster Apr 25 '10 at 12:54
up vote 1 down vote accepted

Any transformation $$(v_1,\ldots,v_n)\mapsto (v_1,\ldots,v_{j-1},v_j+av_k,v_{j+1},\ldots,v_n)$$ for $j\ne k$ is achievable by means of some such matrix. It suffices to reduce an admissible vector to $(1,0,\ldots,0)$ by means of a sequence of such reductions. I would do it in three stages

  1. Make $v_n$ into a unit in $\mathbb{Z}/b\mathbb{Z}$;

  2. Make $v_1=1$;

  3. Make all $v_j=0$ for $j>1$.

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Ok, I see how to do this transformation, but I don't understand, why it suffices to reduce an admissible vector to (1;0;:::;0). – sci Apr 25 '10 at 11:25
You guys are just doing this guy's homework for them, right? I am nowadays very suspicious of anyone who comes here asking an easy UG-level question and, crucially, who only has 1 reputation point. – Kevin Buzzard Apr 25 '10 at 12:44
I see now. Sorry, it was really not so hard. – sci Apr 25 '10 at 14:37

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