Torsion subgroup Prove that in a finitely generated abelian group $G$ the torsion subgroup is a direct summand (from Scott, Group Theory). Clearly, the torsion subgroup is normal because $G$ is abelian, so we have to prove that the non-torsion elements of $G$ constitute a subgroup.
 A: One way to prove this is via the following lemma:
Lemma 1. If $G$ is a finitely generated abelian group and $G$ has no torsion, then $G$ is free abelian.
There are several ways of proving this; one is to use the Smith Normal Form to show that given a subgroup $H$ of $\mathbb{Z}^n$, there is a basis $x_1,\ldots,x_n$ of $\mathbb{Z}^n$ and positive integers $d_1|d_2|\cdots |d_k$, $0\leq k\leq n$, such that $d_1x_1,d_2x_2,\ldots,d_kx_k$ are a basis for $H$. (This is what Hungerford does). Then let $g_1,\ldots,g_n$ generate $G$, and consider the map $\mathbb{Z}^n\to G$ mapping the standard basis to $g_1,\ldots,g_n$. If $H$ is the kernel, then the fact that $G$ is torsion-free shows that either $k=0$ or $d_1=\ldots=d_k=1$, so $G\cong \mathbb{Z}^m$ for some $m$.
Lang's Algebra has a slightly different argument. First he proves that a subgroup of a free abelian group of finite rank is free abelian by induction: let $A=\mathbb{Z}^{n}$, $B$ a subgroup. Let $B_1$ be the kernel of the projection of $B$ onto the first coordinate of $A$, which by induction is free; then to the same thing we will do below (define a retraction from $f(B)\subseteq \mathbb{Z}$ to $B$) to get that $B=B_1\oplus C$ with cyclic $C$ and conclude that $B$ is free abelian.
Having done that, let $S$ be a finite set of generators for $G$, and let $g_1,\ldots,g_n$ be a maximal subset of independent elements of $S$. Let $B$ be the subgroup generated by $g_1,\ldots,g_n$. Then $B$ is free; given $y\in A$, there are integers $m_1,\ldots,m_n,m$, not all zero and $m\neq 0$, with
$$my + m_1g_1+\cdots+m_ng_n = 0.$$
So $my\in B$; this holds for every element of $S$, and since there are finitely many, there exists $m_0$ such that $m_0s\in B$ for all $s\in S$. Hence $m_0A\subseteq B$. Since $a\mapsto m_0a$ is a one-to-one homomorphism of $A$ into itself, $A$ is isomorphic to a subgroup of $B$, and so $A$ is free abelian.
With this lemma in hand, we then show:
Lemma 2. If $G$ is abelian, and $N$ is the torsion subgroup of $G$, then $G/N$ is torsion-free.
I'll leave the easy exercise to you.
So suppose $G$ is finitely generated, and $N$ is the torsion subgroup of $G$. Then $G/N$ is finitely generated and free abelian, so $G/N\cong \mathbb{Z}^k$ for some $k$. Let $\varphi\colon G/N\to \mathbb{Z}^k$ be an isomorphism, and let $\mathbf{e}_1,\ldots,\mathbf{e}_k$ be the standard basis of $\mathbb{Z}^k$. For each $i$, $1\leq i\leq k$, let $g_i\in G$ be an element such that $\varphi(g_i+N) = \mathbf{e}_i$. Let $\psi\colon\mathbb{Z}^k\to G$ be the map that sends $\mathbf{e}_i$ to $g_i$. 
This is a retraction of the map $\varphi\circ\pi\colon G\to \mathbb{Z}^k$. Show that $G = N\oplus \psi(\mathbb{Z}^k)$. 
