Consider the family of short exact sequences $$ 0 \to \mathbb{Z}^m \to G \to \mathbb{Z}/n \mathbb{Z} \to 0 $$ where $G$ is a finitely generated abelian group. This is known to not always split, as per the classic example of $$ 0 \to \mathbb{Z} \to^{\times 2} \mathbb{Z} \to \mathbb{Z}/2 \mathbb{Z} \to 0.$$ But what about when the order is swapped, i.e. $$ 0 \to \mathbb{Z}/n\mathbb{Z} \to G \to \mathbb{Z}^m \to 0 ?$$ My intuition tells me this should always split as a consequence of the classification of finitely generated abelian groups. The torsion part is a subgroup, and taking the quotient gives the free part. But I know this isn't rigorous, and I'm struggling to actually come up with a proof, or even a counterexample. It feels like if it is true, the proof should be quite simple. What am I missing?

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    $\begingroup$ $\mathbb{Z}^m$ is free abelian, hence projective, hence every surjection onto $\mathbb{Z}^m$ always has a right inverse. So yes, this always splits. $\endgroup$ Apr 12 at 18:21
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    $\begingroup$ As @ArturoMagidin says, it is true under your assumption that $G$ is abelian, but it is not true without that assumption. For eaxample , in the nilpotent group of class $2$ with presentation $\langle x,y,z \mid z^2=1, [x,y]=z, [x,z]=[y,z]=1 \rangle$, the extension is non-split. $\endgroup$
    – Derek Holt
    Apr 12 at 18:29
  • $\begingroup$ @DerekHolt Beat me to it... $\endgroup$ Apr 12 at 18:34
  • $\begingroup$ @ArturoMagidin this is great, thanks! I'll accept this if you write it as an answer. $\endgroup$ Apr 12 at 18:34
  • $\begingroup$ And then as a result $G$ can be written as a direct product. $\endgroup$ Apr 12 at 19:34

1 Answer 1


Recall that an $R$-module $P$ is projective if and only if whenever we have a surjective homomorphism $f\colon M\to N$ and a homomorphism $g\colon P\to N$, there exists a homomorphism $h\colon P\to M$ such that $g=f\circ h$. That is: $$\begin{array}{rl} &P\\ &\downarrow g\\ M\stackrel{f}{\to}&N\to 0 \end{array} \implies \begin{array}{rl} &P\\ h\swarrow&\downarrow g\\ M\stackrel{f}{\to}&N\to 0 \end{array}$$

In particular, if $P$ is projective and $f\colon M\to P$ is surjective, then taking $g=\mathrm{id}_P$ in the above diagram we obtain that there exists $h\colon P\to M$ such that $fh=\mathrm{id}_P$.

It is clear that free modules are projective: given a basis $X$ of $P$, for each $x\in X$ let $m_x\in M$ be an element such that $f(m_x) = g(x)$. Then the assignment $i\colon X\to M$ given by $i(x)=m_x$ defines a homomorphism $h\colon P\to M$, and since $hf(x)=g(x)$ for all $x\in X$, we have $hf=g$. (More generally, one can show that an $R$-module $P$ is projective if and only if there exists an $R$-module $Q$ such that $P\oplus Q$ is free; in particular, this holds for free modules.)

With $R=\mathbb{Z}$, you are just talking about abelian groups and group homomorphisms between them. So every free abelian group is projective. In particular, any short exact sequence $$ 0 \stackrel{f}{\longrightarrow} A \to B \stackrel{g}{\longrightarrow} \mathbb{Z}^m \to 0$$ with $m\geq 0$ always splits, since $g$ must have a right inverse.

(Since subgroups of free abelian groups are free abelian, in this case projective is equivalent to free.)

If you leave the realm of abelian groups and consider general groups (so that $G$ is now known to be metabelian but not necessarily abelian) then the sequence need not split. For example, the group of all $3\times 3$ unitriangular matrices of the form $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right),\qquad a,b\in\mathbb{Z}, c\in\mathbb{Z}/n\mathbb{Z}$$ with $n\gt 1$ and the obvious matrix multiplication forms a group in which the subgroup with $a=b=0$ is central and isomorphic to $\mathbb{Z}/n\mathbb{Z}$, and the quotient is isomorphic to $\mathbb{Z}^2$, mapping the matrix to the pair $(a,b)$. But the sequence does not split: the preimages of the first generator correspond to elements with $a=1$, $b=0$; and the preimages of the second generator to elements with $a=0$ and $b=1$; but any matrix of the first type does not commute with any matrix of the second type, so we cannot define a group homomorphism $\mathbb{Z}^2\to G$ that splits the surjection.

  • $\begingroup$ So, the tune changes when you switch categories. $\Bbb Z^m$ is projective in the abelian category $\bf Ab,$ but not in $\bf Grp.$ Is that right? $\endgroup$ Apr 12 at 23:10
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    $\begingroup$ @calcll correct. The free groups are projective for groups (as is any group $G$ for which there exists $H$ with $G*H$ free... but by Nielsen-Schreier that again just gives you free groups). Same argument, since it has the relevant universal property relative to all groups, which $\mathbb{Z}^m$ does not when $m\gt 1$. But note that over arbitrary groups, a short exac sequence being right-split is not equivalent to being left-split. If $1\to N to G \to K\to 1$ splits on the right, then that tells you thT $G\cong N\rtimes K$. $\endgroup$ Apr 13 at 0:57

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