# Torsion subgroup quotient

Let G be an abelian group, $T$ the torsion subgroup of $G$. If $G/T$ is torsion-free, then $T$ and $G/T$ must be disjoint. $G=T \bigoplus G/T$ implies this as well. I don't understand why they are disjoint in the first place.

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This is a mess: $T$ and $G/T$ don't even live in the same group. And aren't you assuming that $G/T$ is finitely generated or something? –  Arturo Magidin Apr 26 '12 at 22:05

What you write does not make sense: $T$ "lives" in side $G$, but $G/T$ is a quotient of $G$, not a subgroup. It does not really make any sense to talk about the intersection of $T$ and $G/T$.

What is true is that if $G/T$ is finitely generated, then since $G/T$ is torsion free then it is isomorphic to $\mathbb{Z}^r$. Pick an isomorphism. Then for each of the elements $e_i \in \mathbb{Z}^r$ ($e_i$ has a $1$ in the $i$th coordinate and zeros elsewhere) we can pick an element $g_i\in G$ such that $g_i+T$ corresponds to $e_i$ in $G/T$ under the chosen isomorphism.

Then the $g_i$ generate a subgroup isomorphic to $\mathbb{Z}^r$: if $a_1g_1+\cdots+a_rg_r=0$, then projecting onto $G/T$ we get $a_1e_1+\cdots+a_re_r=0$, hence $a_i=0$ for all $i$; so $\langle g_1,\ldots,g_r\rangle = \langle g_1\rangle\oplus\langle g_2\rangle\oplus\cdots\oplus \langle g_r\rangle$.

And now we have the important point: $\langle g_1,\ldots,g_r\rangle\cap T = \{0\}$. For if $a_1g_1+\cdots+a_rg_r\in T$, then there exists $n\gt 0$ such that $n(a_1g_1+\cdots+a_rg_r) = 0$. This gives $$na_1g_1+\cdots+na_rg_r=0,$$ hence $na_i=0$ for all $i$, so $a_i=0$ for all $i$. That is, the intersection is contained in $\{0\}$.

Since $T+\langle g_1,\ldots,g_r\rangle = G$ (e.g., by the isomorphism theorems), and the intersection is trivial, we get $$G \cong T\oplus \langle g_1,\ldots,g_r\rangle \cong T\oplus \mathbb{Z}^r\cong T\oplus (G/T).$$

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What is not true. If $G=\prod_{p}\mathbb{F}_p=\mathbb{F}_2\times\mathbb{F}_3\times\cdots$, then $G/T$ can not inject to $G$. In particular, $G$ is not isomorphic to $T\oplus G/T$.

Proof It is clear that $\bigcap_ppG=0$. It suffices to show $\cap_pp(G/T)\neq 0$. The image of $x=(1,1,\ldots)$ is in $p(G/T)$ for every prime $p$.

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