# Two split exact sequence implies another one is split?

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

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 $\mathbb{Z}_4 \oplus \mathbb{Z}_2$?

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I don't think xymatrix is supported by MathJax. –  Arturo Magidin Mar 3 '12 at 5:39
xymatrix doesn't work here at present. Some alternatives that do work are discussed in this meta thread. –  Dylan Moreland Mar 3 '12 at 5:39
I think this is what the diagram was meant to be; if it is not, please correct it appropriately. –  Arturo Magidin Mar 3 '12 at 5:46
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 3 '12 at 5:53
Right, that's the point; both are possible. Your diagram doesn't determine $G$. –  Hurkyl Mar 3 '12 at 7:32

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 $H$.

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