Does exactness of $0 \to \Bbb Z^n \to G \to \Bbb Z^m \to 0$ imply $G$ is isomorphic to $\Bbb Z^{m+n}$?

Question

Suppose we have an exact sequence of abelian groups $0 \to \Bbb Z^n \to G \to \Bbb Z^m \to 0$. By exactness, we know that $G/\Bbb Z^n$ is isomorphic to $\Bbb Z^m$ (here we are identifying $\Bbb Z^n $ with its image, since $\Bbb Z^n \to G$ is injective).

From this, can I conclude that $G$ is isomorphic to $\Bbb Z^{m+n}$?

Intuitively this seems true, but how do I have to prove this?

Answer 1

Yes, consider a map $h:\mathbb{Z}^m\rightarrow G$ such that $p\circ h=Id$ where $p:G\rightarrow\mathbb{Z}^m$ is the projection map. Let $e_1,...,e_m$ generators of $\mathbb{Z}^m$. $H$ the subgroup generated by $h(e_1),...,h(e_n)$. $H$ is isomorphic to $\mathbb{Z}^m$, and $G$ is isomorphic to $i(\mathbb{Z}^n)\oplus H$ where $i:\mathbb{Z}^n\rightarrow G$.

Answer 2

This exact sequence splits. If $\pi:G\to\Bbb Z^m$ is the surjection in the exact sequence, then there is a homomorphism $\phi:\Bbb Z^m\to G$ with $\pi\circ\phi$ the identity on $\Bbb Z^m$. To do this, let $e_1,\ldots,e_m$ be the standard basis of $\Bbb Z^m$ and pick some $g_i$ with $\pi(g_i)=e_i$. Define $\phi(a_1e_1+\cdots+a_me_m)=a_1g_1+\cdots+a_me_m$.

Then $G$ is the direct sum of the $\Bbb Z^n$ inside $G$ with $\pi(\Bbb Z^m) \cong\Bbb Z^m$.

Answer 3

It's true because free groups have the splitting property:

Theorem. Let $F$ be a free group. Then any surjective group homomorphism $\pi:G\rightarrow F$ has a right inverse homomorphism $\varphi:F\rightarrow G$, i.e. $\pi\circ\varphi=\mathrm{id}$.

(Even if $\pi$ isn't surjective, its image is still free by the Nielsen--Schreier Theorem. So $\varphi$ is defined on the image of $\pi$.)

The theorem isn't that hard to prove either. Suppose $F$ is free on the set of symbols $\{x_i\}$. Since $\pi$ is surjective, select any $g_i\in G$ so that $\pi(g_i)=x_i$. Now simply define $\varphi$ on the $x_i$'s by $\varphi(x_i):=g_i$; by the universal property of free groups, $\varphi$ extends to a homomorphism $\varphi:F\rightarrow G$. QED!

When a homomorphism $\pi:G\rightarrow H$ splits (i.e. has a right inverse), you always reconstruct $G$ as a semidirect product of $H$ with the kernel: $G\simeq \ker(\pi)\rtimes H$.

Basically the same thing holds in the abelian category:

• if $F$ is free abelian (i.e. $F\simeq \mathbb{Z}^I$), then any short exact sequence $0\rightarrow A\rightarrow B\rightarrow F\rightarrow 0$ splits; and,
• if a short exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ splits, then $B\simeq A\oplus C$.

Note that if things are abelian, you don't need a semidirect product; a direct sum suffices.