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A homework problem asked to find a short exact sequence of abelian groups $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ such that $$B \cong A \oplus C$$ although the sequence does not split.

My solution: $$0 \rightarrow \mathbb{Z} \overset{i}{\rightarrow} \mathbb{Z} \oplus (\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \overset{p}{\rightarrow} (\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \rightarrow 0$$ with $i(x)=(2x,0,0,\dotsc)$ and $p(x,y_1,y_2,\dotsc)=(x+2\mathbb{Z},y_1,y_2,\dotsc)$.

My new questions:

  1. Is there an example with finite/finitely generated abelian groups?
  2. If the answer to $(1)$ is negative, will it help to pass to general $R$-modules for some ring $R$?
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FYI: A different example, still with infinite groups, is Example 6.35 of Rotman's Advanced Modern Algebra (2nd ed., AMS). –  KCd Apr 22 '12 at 21:55
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2 Answers 2

I'm sorry, but my answer uses Ext groups, which may not be in the scope of your course.

I don't know of a simple example with finitely generated abelian groups for the following reason: if $0 \rightarrow \mathbb Z \rightarrow E \rightarrow \mathbb Z / n \mathbb Z \rightarrow 0$ is an extension represented by $r+n \mathbb Z \in \operatorname{Ext}^1(\mathbb Z,\mathbb Z/n\mathbb Z)\cong \mathbb Z/n\mathbb Z$, then one can show that $E \cong \mathbb Z \oplus \mathbb Z/d\mathbb Z$ where $d$ is the highest common factor of $r$ and $n$. If you assume this sequence doesn't split, then $r$ is not a multiple of $n$ and $d<n$, so the middle term will never be isomorphic to the direct sum of the of the first and third.

However, you can get a simple example using modules in the following way. Let $G=\langle g\rangle$ be an infinite cyclic group and let $\mathbb Z G$ be the associated commutative group ring. Consider $0 \rightarrow \mathbb Z \rightarrow \mathbb Z \oplus \mathbb Z \rightarrow \mathbb Z \rightarrow 0$ a short exact sequence of $\mathbb Z G$-modules where the first map is inclusion into the first component and the second map is projection onto the second. Let $g$ act on $\mathbb Z \oplus \mathbb Z$ by $\bigl(\begin{smallmatrix} 1&1\\ 0&1 \end{smallmatrix} \bigr)$. Then $\operatorname{Ext}^1_{\mathbb Z G}(\mathbb Z, \mathbb Z) \cong \mathbb Z$ and this extension corresponds to $1$. If you let $g$ act by $\bigl(\begin{smallmatrix} 1&n\\ 0&1 \end{smallmatrix} \bigr)$ it corresponds to $n$. As long as $n>0$ this is a non-split short exact sequence where the direct sum of the outside terms are isomorphic to the middle one. However, this example is far from elementary, and I apologize for that.

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I should add that there is a 1-1 correspondence between elements of $Ext^1(C,A)$ and short exact sequences $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ and this sequence splits iff it corresponds to the $0$ element of the Ext group. –  Parsa Apr 22 '12 at 22:50
This looks great, but I don't yet have the tools to understand it. When I do, I will verify it's correct and (probably) accept it. Thank you very much! –  user3533 Apr 30 '12 at 20:34
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I believe that the answer to Question 1 is 'yes'.

Consider the following sequence of finitely generated abelian groups: \begin{equation} 0 \longrightarrow \mathbb{Z} \stackrel{f}{\longrightarrow} \mathbb{Z} \oplus \mathbb{Z} \stackrel{g}{\longrightarrow} \mathbb{Z} \longrightarrow 0, \end{equation} where $ f(n) := (2n,3n) $ and $ g(a,b) := 3a - 2b $ for all $ a,b,n \in \mathbb{Z} $.

Firstly, observe that $ f $ is injective. Secondly, as $ \text{gcd}(2,3) = 1 $, we see that $ g $ is surjective. Lastly, notice that \begin{equation} \ker(g) = \{ (a,b) \in \mathbb{Z}^{2} \,|\, 3a - 2b = 0 \} = \text{im}(f). \end{equation} With these three facts established, we conclude that the sequence above is short exact. However, the sequence does not split.

Erratum: I just realized that the last sentence is totally wrong. We have the following splitting map $ s $ from $ \mathbb{Z} \oplus \mathbb{Z} $ to the first $ \mathbb{Z} $: \begin{equation} \forall (a,b) \in \mathbb{Z}^{2}: \quad s(a,b) := 2a - b. \end{equation} A quick way of seeing that the sequence splits is to note that it ends in $ \mathbb{Z} $, which is free. I thank the users for their comments.

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Since $\mathbb{Z}$ is free, doesn't every short exact sequence ending in $\mathbb{Z}$ split? –  Jason DeVito Sep 3 '12 at 12:20
split by $k \mapsto (k,k)$ –  mt_ Sep 3 '12 at 12:38
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