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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.

Jendrik Stelzner
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Eric
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1 Answers1

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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 $\mathbb{Z}/6\mathbb{Z}$.

JHalliday
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  • Thank you for your answer. I will read it through carefully, is the diagram you have constructed commutative? – Eric Sep 11 '15 at 21:30
  • 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. – JHalliday Sep 11 '15 at 21:34
  • 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. – Eric Sep 11 '15 at 21:36
  • Okay, happy to help! – JHalliday Sep 11 '15 at 21:39
  • 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? – Eric Sep 12 '15 at 00:16
  • I was defining it as taking 0 to 0 and 1 to 2. Does that make sense? – JHalliday Sep 12 '15 at 00:23
  • Yes that makes sense, but the first row u have the map $6 \mathbb{Z} \to \mathbb{Z} / 2 \mathbb{Z}$. This function must be surjective for the first row to be exact, I dont see how this is surjective? – Eric Sep 12 '15 at 21:43
  • that map would naturally be mod 12, since you have $12\mathbb{Z}$ in the slot before it. – JHalliday Sep 12 '15 at 23:04
  • If its mod 12, then what element of $6 \mathbb{Z}$ is mapped to 1? I defined the function as $6,-6 \mapsto 1$ and $12 \mathbb{Z} \mapsto 0$. – Eric Sep 12 '15 at 23:51
  • Uh every element not divisible by 12 is mapped to 1. The coset $12 \mathbb{Z}+6$ – JHalliday Sep 13 '15 at 00:07
  • The fact that you're having this much confusion about these things probably means you don't understand the ideas of quotient groups as well as you think you do. Consider looking or re-looking through a section on them in an abstract algebra book if you're studying exact sequence stuff which requires fluidity in these things. – JHalliday Sep 13 '15 at 00:09