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$$\begin{array}{ccccccccc} &&0&&0&&0\\ &&\downarrow &&\downarrow && \downarrow\\ 0 & \to & \mathbb{Z}_2\{a\} & \to & \mathbb{Z}_2\{a\} & \to & 0 & \to & 0\\ & &\downarrow & & \downarrow &&\downarrow\\ 0&\to&\mathbb{Z}_2\{a\}\oplus\mathbb{Z}_2\{b\} & \xrightarrow{f} & G & \xrightarrow{g} & \mathbb{Z}_2\{c\} & \to & 0\\ &&\downarrow & &\downarrow & & \downarrow\\ 0 & \to & \mathbb{Z}_2\{b\} & \xrightarrow{h} & \mathbb{Z}_2\{y\}\oplus\mathbb{Z}_2\{z\} & \xrightarrow{i} & \mathbb{Z}_2\{c\} & \to &0\\ &&\downarrow && \downarrow &&\downarrow\\ &&0 && 0 && 0 \end{array}$$

where $\mathbb{Z}_2\{a\}$ means that $a$ generates $\mathbb{Z}_2$ and $y=h(b)$ and $i(z)=c$.

In the diagram, first and third rows are split and first and third columns are split.

Then second row($0 \to \mathbb{Z}_2 \oplus \mathbb{Z}_2 \to G \to \mathbb{Z}_2 \to 0$) or second column($0 \to \mathbb{Z}_2 \to G\to \mathbb{Z}_2 \oplus \mathbb{Z}_2 \to 0$) is split? Or $G$ is isomorphic to $\mathbb{Z}_4 \oplus \mathbb{Z}_2$?

user26170
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  • I don't think xymatrix is supported by MathJax. – Arturo Magidin Mar 03 '12 at 05:39
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    xymatrix doesn't work here at present. Some alternatives that do work are discussed in this meta thread. – Dylan Moreland Mar 03 '12 at 05:39
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    I think this is what the diagram was meant to be; if it is not, please correct it appropriately. – Arturo Magidin Mar 03 '12 at 05:46
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    I think you can have $G$ be $\mathbb{Z}_2\oplus\mathbb{Z}_4$: let the maps in the first column be the embedding into the second component followed by the corresponding projection; the nontrivial map in the first row is the identity. The embedding in the middle column maps $\mathbb{Z}_2$ to the subgroup $\langle (0,2)\rangle$ of $\mathbb{Z}_2\oplus\mathbb{Z}_4$; the embedding in the middle row maps $(1,0)$ to $(1,0)$ and $(0,1)$ to $(0,2)$. – Arturo Magidin Mar 03 '12 at 05:53
  • @user: Have you tried assuming $G \cong \mathbb{Z}_4 \oplus \mathbb{Z}_2$ directly solving for the maps? (And similarly for $G \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2?$) In the first case, start by choosing a map $G \to \mathbb{Z_2}$ (there are only two possibilities), and the rest of the diagram pretty much fills itself in. I'm sure the latter is consistent too. (in particular, that this diagram doesn't uniquely determine $G$) –  Mar 03 '12 at 06:03
  • @Margidin: If $G\simeq \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$, is there a contradiction?? – user26170 Mar 03 '12 at 06:23
  • @Hurkyl: Yes. But I don't have a contradiction. – user26170 Mar 03 '12 at 06:24
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    Right, that's the point; both are possible. Your diagram doesn't determine $G$. –  Mar 03 '12 at 07:32
  • @user26170: Yes, both are possible; the direct sum is certainly possible, the question was whether the non-split case was possible. It is. And please don't e-mail me with questions that you are asking here. – Arturo Magidin Mar 03 '12 at 21:26

1 Answers1

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Although the problem has been answered in the comments, I thought I'd make an observation that makes things more obvious. It is possible to produce such diagrams as a direct sum of two subdiagrams:

$$ \begin{array}{ccccc} \mathbb{Z}_2 &\to& \mathbb{Z}_2 &\to& 0 \\ \downarrow & & \downarrow & & \downarrow \\ \mathbb{Z}_2 &\to& H &\to& \mathbb{Z}_2 \\ \downarrow & & \downarrow & & \downarrow \\ 0 &\to& \mathbb{Z}_2 &\to& \mathbb{Z}_2 \end{array} $$ and

$$ \begin{array}{ccccc} 0 &\to& 0 &\to& 0 \\ \downarrow & & \downarrow & & \downarrow \\ \mathbb{Z}_2 &\to& \mathbb{Z}_2 &\to& 0 \\ \downarrow & & \downarrow & & \downarrow \\ \mathbb{Z}_2 &\to& \mathbb{Z}_2 &\to& 0 \end{array} $$

Off the cuff, I believe every way to fill in your diagram has such a direct sum decomposition. This simplification makes it easier to see that there are two possibilities for $H$.