I am trying to show that a short exact sequence of abelian groups splits. I have a short exact sequence,

$$0\rightarrow \mathbb{Z} \rightarrow G \rightarrow \mathbb{Z}_2 \rightarrow 0$$

and I know that $\mathbb Z \oplus \mathbb Z_2$ is a subgroup of $G$. Do I have a splitting by sending the generator of $\mathbb Z_2$ to the generator of $\mathbb Z_2$ in $G$?


1 Answer 1


If you have a sequence $$0 \to F \to G \to H \to 0$$ then $H$ is a quotient of $G$. In general there will be no map $H \to G$ giving a section of $G\to H$. However, if such a section exists then the sequence is indeed split on the right (by definition), and therefore split. The important thing is not only the existence of an injection $H\to G$, but the existence of one which is a section to the given map $G\to H$.

In your case, since $G$ contains an element $\epsilon$ of order $2$, there is a unique injection $H\to G$ taking the generator of $H$ to $\epsilon$. So the question is: is this injection necessarily a section of $G \to H$? In other words, does $G\to H$ necessarily map $\epsilon$ to the generator of $H$? Well, if it doesn't, then $\epsilon$ is in the kernel, hence in the image of $\mathbf Z$. But $\mathbf Z$ has no element of finite order. Therefore, the map you have described is necessarily a section, and the sequence splits.

  • $\begingroup$ Thank you for your response. To extend my question further, what can I say if $\bigoplus_{j=2}^{k} \mathbb Z_j$ was a subgroup of $G$ and $H=\mathbb Z_k$? $\endgroup$
    – Felix Y.
    Commented Mar 4, 2014 at 2:35

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