# Does the short exact sequence $0\mapsto \mathbb{Z}^k\mapsto \Gamma\mapsto \mathbb{Z}^\ell\mapsto 0$ split?

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.

• Are you talking about abelian groups, or groups in general?
– jMdA
Feb 5, 2021 at 5:08
• A general group, not necessarily abelian Feb 5, 2021 at 5:09
• 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? Feb 5, 2021 at 10:06
• I was thinking in $\Gamma$ as a lattice of a Lie group, Feb 5, 2021 at 22:35

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.

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

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 $$\bf Z$$.