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Assume that the following diagram of abelian groups has exact rows and columns. Can you determine the missing entries and maps? Give short reasoning. $$ \require{AMScd} \begin{CD} {} @. 0 @. 0 @. 0 {} {} \\ @. @VVV @VVV @VVV \\ 0 @>>> {} @>>> {} @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> {} @>>> \mathbb{Z} @>>> {} @>>> 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> \mathbb{Z}/3\mathbb{Z} @>>> {} @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 \\ @. @VVV @VVV @VVV \\ {} @. 0 @. 0 @. 0 {} {} \end{CD} $$ (Original image of this diagram here.)

This is what I tried: $$ \require{AMScd} \begin{CD} {} @. 0 @. 0 @. 0 {} {} \\ @. @VVV @VVV @VVV \\ 0 @>>> 3\mathbb{Z} @> a >> 6\mathbb{Z} @> b >> \mathbb{Z}/2\mathbb{Z} @>>> 0 \\ @. @V g VV @V h VV @V i VV \\ 0 @>>> \mathbb{Z} @> c >> \mathbb{Z} @> d >> \mathbb{Z}/2\mathbb{Z} @>>> 0 \\ @. @V j VV @V k VV @V l VV \\ 0 @>>> \mathbb{Z}/3\mathbb{Z} @> e >> \mathbb{Z}/6\mathbb{Z} @> f >> \mathbb{Z}/2\mathbb{Z} @>>> 0 \\ @. @VVV @VVV @VVV \\ {} @. 0 @. 0 @. 0 {} {} \end{CD} $$ (Original image of this diagram here.)

Where for example I let $a$ be given by $t \mapsto 4t$ and $b$ be given by $t \mapsto t/6$. I defined all the other functions in similar ways such that all rows and all columns became exact, however I missed the crucial part that the diagram must be commutative and everything failed. I need some hints on this one please.

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Okay I don't know how to make commutative diagrams on this site, but you know that the second row, third column must have $4$ elements since it is in between two sets of two elements, so it is either $(\mathbb{Z}/2\mathbb{Z})^2$ or $\mathbb{Z}/4\mathbb{Z}$. Since $\mathbb{Z}$ must surject onto it from the left, it can only be $\mathbb{Z}/4\mathbb{Z}$.

From this we know that the second row, first column must be $4\mathbb{Z}$, since it is the kernel of the map $\mathbb{Z} \rightarrow \mathbb{Z}/4\mathbb{Z}: x \rightarrow x \mod(4)$. The natural map $4\mathbb{Z} \rightarrow \mathbb{Z}/3\mathbb{Z}$ takes an element $x \rightarrow x \mod(12)$, so in the first column, first row we should have $12\mathbb{Z}$.

Since we need an exact sequence $0 \rightarrow 12\mathbb{Z} \rightarrow ? \rightarrow \mathbb{Z}/2\mathbb{Z}$, by the same logic we have been using we get $6\mathbb{Z}$ in the first row, second column. Finally we need the kernel of the inclusion map $6 \mathbb{Z} \rightarrow \mathbb{Z}$ in the third row, second column, which is $\mathbb{Z}/6\mathbb{Z}$.

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  • $\begingroup$ Thank you for your answer. I will read it through carefully, is the diagram you have constructed commutative? $\endgroup$ – Eric Sep 11 '15 at 21:30
  • $\begingroup$ Yes it is. all of the maps are either inclusion mappings or quotient maps that take integers modulo some number, so this should be easy to check. $\endgroup$ – JHalliday Sep 11 '15 at 21:34
  • $\begingroup$ OK its late over here, let me get some sleep and I will read through your answer. I will leave a comment or a question to you by tomorrow. $\endgroup$ – Eric Sep 11 '15 at 21:36
  • $\begingroup$ Okay, happy to help! $\endgroup$ – JHalliday Sep 11 '15 at 21:39
  • $\begingroup$ I have one problem, u said second row, third column should be $\mathbb{Z}/ 4\mathbb{Z}$. But, the inclusion map $\mathbb{Z}/2 \mathbb{Z} \to \mathbb{Z}/ 4\mathbb{Z}$ is not well-defined. Am I missing something? $\endgroup$ – Eric Sep 12 '15 at 0:16

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