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I am stuck with these hard-star exercises for some time now (they are from a book called "Introduction a L'Algebre et L'Analyse Modernes" de M.Zamansky"), if somebody sees the right way, I will be very grateful to know it.

Let L be a subgroup of $\mathbb{Z}^{3}$. Let $ q_{1}\mathbb{Z}$, $(q_{1}\ge 0)$ be the group of all $ x_{1} \in \mathbb{Z}$ with $ (x_{1},0,0) \in L$ and let $\displaystyle u_{1}=(q_{1},0,0)$. Let $q_{2}\mathbb{Z}$, $(q_{2}\ge 0)$ be the group of all $x_{2} \in \mathbb{Z}$ so that there exists $x_{21} \in \mathbb{Z}$ with $(x_{21},x_{2},0) \in L$. If $q_{2}>0$ then $ u_{2}=(q_{21},q_{2},0) \in L$; otherwise $u_{2}=0$. Let $ q_{3}\mathbb{Z}$, $(q_{3}\ge 0)$ be the group of all $x_{3} \in \mathbb{Z}$ so that there are $x_{31},x_{32} \in \mathbb{Z}$ with $ (x_{31},x_{32},x_{3}) \in L$. If $ q_{3} > 0$ then $ u_{3} = (q_{31},q_{32},q_{3}) \in L$ otherwise $u_{3}=0$

1.** If $q_{1}>0, q_{2}> 0 , q_{3}>0$, then $u_{1},u_{2},u_{3}$ can be chosen so that $0\le q_{21}, q_{31}< q_{1}, 0 \le q_{32} < q_{2}$
2.** It is possible to find $q_{1},q_{2},q_{3},q_{21},q_{31},q_{32}$ for $L=\{(x_{1},x_{2},x_{3}) \in \mathbb{Z}^{3}; 2x_{1}+4x_{2}+5x_{3} = 0 mod 7 \}$

I have come this far:

  1. Assume one can not choose $u_{1}, u_{2}, u_{3}$ with $q_{1}, q_{2}, q_{3} > 0$ so that $0\le q_{21}, q_{31}< q_{1} , 0\le q_{32} < q_{2}$. I thought about giving a counterexample with numbers but since one doesn't know what L is that is not really possible…

  2. One looks at $2x_{1}+4x_{2}+5x_{3} = 7$. $2x_{1}$ and $4x_{2}$ are always even. I don't see how to continue. But why is the book asking for $q_{21}, q_{31}, q_{32}$, too, if they can be set to 0?

As you see it is not very much… Maybe for solving this it helps to know the solution of the previous exercise: If L is a subgroup of $\mathbb{Z}^{3}$, linearly independent, linear equations .

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