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I've seen in Nonsplit extension of $\mathbb{Z}$ by itself that every short exact sequence of the form $0\mapsto \mathbb{Z}\mapsto G\mapsto \mathbb{Z}\mapsto 0$ splits.

I wonder if this is still valid if we change the exact sequence by $$0\mapsto \mathbb{Z}^k\mapsto \Gamma\mapsto \mathbb{Z}^\ell\mapsto0$$ where $\Gamma$ is a discrete group.

Thanks in advance

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    $\begingroup$ Are you talking about abelian groups, or groups in general? $\endgroup$
    – jMdA
    Feb 5, 2021 at 5:08
  • $\begingroup$ A general group, not necessarily abelian $\endgroup$ Feb 5, 2021 at 5:09
  • $\begingroup$ Any free finitely generated nilpotent group of class 2 or 3 has this property. But there are also many examples that are not nilpotent. By the way, what's the difference between a discrete group and a group? $\endgroup$
    – Derek Holt
    Feb 5, 2021 at 10:06
  • $\begingroup$ I was thinking in $\Gamma$ as a lattice of a Lie group, $\endgroup$ Feb 5, 2021 at 22:35

2 Answers 2

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A counterexample is the following short exact sequence: $0 \rightarrow \mathbb{Z} \rightarrow T \rightarrow \mathbb{Z}^2 \rightarrow 0$. Where $T$ is the group of upper triangular matrices in $\text{GL}_3(\mathbb{R})$ with unit diagonals.

The injection is defined by the formula $z\mapsto \begin{pmatrix} 1 &0&z\\ 0 &1 &0 \\ 0 &0&1\end{pmatrix}$. The surjective morphism is defined by the formula $\begin{pmatrix} 1 &a&c\\0&1&b\\0&0&1\end{pmatrix} \mapsto (a, b)$.

The image of the injection and the kernel of the surjection is the subgroup generated by $\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$. This shows that this is a short exact sequence.

However the sequence is clearly not split because $T$ is not isomorphic to the product of $\mathbb{Z}$ and $\mathbb{Z}^2$ because $T$ is not abelian.

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  • $\begingroup$ The previous answer was incorrect, since the sequence I described was not exact, but this should be correct. $\endgroup$
    – jMdA
    Feb 5, 2021 at 5:43
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    $\begingroup$ The kernel is not $\bf Z$, but a a non finitely free group. $\endgroup$
    – Thomas
    Feb 5, 2021 at 5:51
  • $\begingroup$ I added another fix, I was much more careful about the example I chose this time. $\endgroup$
    – jMdA
    Feb 5, 2021 at 7:28
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    $\begingroup$ Now it is Ok, the group T is the Heisenberg group (see my answer below) $\endgroup$
    – Thomas
    Feb 5, 2021 at 7:32
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For abelian groups, The proof is the same than for vector spaces.

Choose a $\bf Z$ base of the kernel $e_1,..,e_k$ and a family of vectors $f_1,...,f_l$ which projects onto a base $f'_1,...,f'_l$of $\bf Z ^l$. We wil prove that $(e_1,..e_k, f_1,...f_l)$ is a $\bf Z$ base of $\Gamma$, ie the obvious map $\bf Z^{k+l}$ $\to \Gamma$ is an isomorphism.

Let $x\in \Gamma$; then we write $\pi (x)= \sum y_i f'_i$, so that $x-\sum y_i f_i$ is in the kernel, and there exists $x_1,..x_k$ so that $x= \sum x_i e_i +\sum y_i f_i$.

To check that we have a $\bf Z$ base, assume $x=0$. By projecting and using that $f'_i$, we see that $y_i=0$, and using that $e_i$ is a base of the kernel, we see that $x_i=0$.

But for the Heisenberg group,(see https://en.wikipedia.org/wiki/Heisenberg_group), the sequence do not split. This groupe is the group of upper triangular $(3,3)$ matrices with 1 on the diagonal and coefficients in $\bf Z$.

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