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.
Torsion subgroup quotient
0
$\begingroup$
abstract-algebra
abelian-groups
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2This 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? – 2012-04-26