So there's a statement in Lang that I would like to understand better.

It's contained in his proof of the following statement:

Every finite abelian p-group is isomorphic to a product of cyclic p-groups. If it is of type $(p^{r_1},...,p^{r_2})$ with $$r_1\geq r_2\geq\cdots\geq r_s>1$$*then the sequence of integers $(r_1,...,r_s)$ is uniquely determined.*

So what I would understand is this part of his proof:

We want to establish $$A=A_1\oplus\cdots\oplus A_s$$ where each $A_i$ are cyclic subgroups of order given in the statement of the problem

Conversely, suppose that $m_1,...,m_s$ are integers $\geq0$ such that $$0=m_1x_1+\cdots+m_sx_s$$Since a_i has period $p^{r_i}$(i=1,...,s), we may suppose that $m_i<p^r{_i}$. Putting a bar onthis equation yields $$0=m_2\overline{x_2}+\cdots+m_s\overline{x_s}$$Thus $m_i=0$ for all $i=2,...,s$. But then$m_1=0$. From this it follows at once that $(A_1+\dots+A_i)\cap A_{i+1}=0$

Question: I just want to make sure I understand the last statement of the proof correctly. So if the only way we can write 0 is if the all coefficients $m_i=0$ , then it must follow that $(A_1+\cdots+ A_i)\cap A_{i+1}=0$, why is that?


Suppose $\;x\in\left(A_1+\ldots+A_i\right)\cap A_{i+1}\;$ , then

$$\begin{cases}x\in A_1+\ldots + A_i\implies& x=m_1a_1+\ldots +m_ia_i\\{}\\x\in A_{i+1}\implies& x=m_{i+1}a_{i+1}\end{cases}$$

and from both equations above we get

$$m_1a_1+\ldots +m_ia_i-m_{i+1}a_{i+1}=0\iff m_k=0\;\;\forall\;k=1,2,...,i+1$$

and in particular, $\;x=m_{i+1}a_{i+1}=0\cdot a_{i+1}=0\;$

  • 2
    $\begingroup$ Okay I see the reasoning now. Thanks! $\endgroup$ – Enigma Sep 26 '14 at 23:06
  • $\begingroup$ Just for a little clarification can I ask how you deduced $m_k=0 \forall k=1,2,...,i+1$? Is your method similar to what Lang does? $\endgroup$ – Enigma Sep 26 '14 at 23:40
  • $\begingroup$ It is the basic assumption, @S.Sheng .You yourself wrote it in you question . :) $\endgroup$ – Timbuc Sep 26 '14 at 23:42
  • 2
    $\begingroup$ Gotcha. I'm just having one of those days where I'm glossing over my own assumptions. Thanks again. $\endgroup$ – Enigma Sep 26 '14 at 23:44

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