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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.

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The elements of $G$ without torsion do not constitute a subgroup of $G$ for $0$ is a torsion element and thus does not lie in the subgroup in question. In fact, the difficulty is more serious. For example, consider the group $G=\mathbb{Z}\oplus \mathbb{Z}/2$ and the elements $g=(1,1)$ and $h=(1,0)$ of $G$. You can check that the elements $g$ and $h$ of $G$ do not have torsion but that their difference $g-h=(0,1)$ is a torsion element of $G$. –  Amitesh Datta Dec 16 '11 at 7:06
    
If you have studied some form of the fundamental theorem for finitely generated abelian groups, then this exercise should be a straightforward application of the relevant form of this theorem. Could you please confirm whether or not you are familiar with the fundamental theorem for finitely generated abelian groups and, if so, state the form of the theorem with which you are familiar? –  Amitesh Datta Dec 16 '11 at 7:09
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Interestingly, you say that the torsion subgroup is normal because $G$ is abelian. However, if the torsion elements for a subgroup then the subgroup will always be normal. This is because conjugation preserves order. –  user1729 Dec 16 '11 at 14:41
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In fact, if $G$ is any group (abelian or not), and $T$ is the set of torsion elements of $G$, then the subgroup generated by $T$ is normal in $G$: because for any set $X$, the subgroup $\langle X\rangle$ is normal in $G$ if and only if for every $g\in G$ and every $x\in X$, $gxg^{-1}\in\langle X\rangle$, which holds when $X$ is the set of torsion elements. –  Arturo Magidin Dec 16 '11 at 17:42

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up vote 3 down vote accepted

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 $G = N\oplus \psi(\mathbb{Z}^k)$.

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Thanks Arturo ! –  WLOG Dec 16 '11 at 23:53

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