If $G_1, G_2, G_3$ are abelian groups and $0 \to G_1 \xrightarrow{\varphi_1} G_2 \xrightarrow{\varphi_2} G_3 \to 0$ is exact, then $G_2 \simeq \ker(\varphi_2) \oplus \text{im}(\varphi_2) \simeq G_1 \oplus G_3$

The above is the conclusion of Example 12.2 in Nonlinear Analysis and Semilinear Elliptic Problems, by Ambrosetti and Malchiodi.

I worked out as follows:

By exactness, ${0} = \ker(\varphi_1)$, so $G_1 \simeq \text{im}(\varphi_1) = \ker(\varphi_2) \lhd G_2$, so it makes sense to consider $G_2/G_1$. On the other hand, again by exactness, $\text{im}(\varphi_2) = G_3$. By the First Isomorphism Theorem, $$ G_2/ \ker(\varphi_2) \simeq G_2/G_1 \simeq G_3. $$

What one would like to do now is to "multiply both sides by $G_1$ and cancel out in the left-hand side". My question is, how to do it in a rigorous way?

I tried to write $G_2/G_1$ explicitly, but it was a dead end. I also tried to follow the hint by Najib Idrissi in this question, but failed.

Thanks in advace.

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    $\begingroup$ This doesn't work: $0\to\mathbb{Z}\xrightarrow{2}\mathbb{Z}\to\mathbb{Z}/2\to 0$. $\endgroup$ Feb 25, 2021 at 13:07
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    $\begingroup$ As stated, it is false. Counterexample: $0 \to n\mathbf Z\to\mathbf Z\to\mathbf Z/n\mathbf Z\to 0$ is exact, but your assertion would imply the ideal $n\mathbf Z$ is generated by an idempotent, which is impossible as $\mathbf Z$ is an integral domain. I guess you're for getting a hypothesis. $\endgroup$
    – Bernard
    Feb 25, 2021 at 13:08
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    $\begingroup$ The claim is false. See split exact sequence. $\endgroup$
    – user239203
    Feb 25, 2021 at 13:08
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    $\begingroup$ The usual counterexample with finite groups is $0\to\mathbb Z/2\mathbb Z\to \mathbb Z/4\mathbb Z\to\mathbb Z/2\mathbb Z\to 0$, where the second map is reduction mod $2$ and the first map send $0, 1\mapsto 0, 2$. $\endgroup$
    – Mathmo123
    Feb 25, 2021 at 13:54
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    $\begingroup$ The authors of that book are just wrong, there is no way around this. I checked the book, that's verbatim what they wrote, and it's simply false. $\endgroup$ Feb 25, 2021 at 14:28

1 Answer 1


This is unfortunately false. Consider the standard counterexample

$$0 \to \mathbb{Z} \xrightarrow{\varphi} \mathbb{Z} \xrightarrow{\pi} \mathbb{Z}/2\mathbb{Z} \to 0$$

where $\varphi(x)=2x$ and $\pi(x)=x+2\mathbb{Z}$ is the quotient map. The sequence is exact but clearly $\mathbb{Z}$ is not a direct sum $\mathbb{Z}\oplus(\mathbb{Z}/2\mathbb{Z})$.

For that to hold we need the $0 \to G_1 \xrightarrow{\varphi_1} G_2 \xrightarrow{\varphi_2} G_3 \to 0$ sequence to be a split exact sequence (see also: splitting lemma). This is for example true whenever $G_3$ is free abelian, i.e. $G_3\simeq \bigoplus\mathbb{Z}$.


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