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There is a well-known theorem that:

If $S\le G$ and $\frac{G}{S}$ is torsion-free, so $S$ is pure in $G$.

Please hint me about the reverse. If $S$ is pure in $G$, then will $\frac{G}{S}$ is necessarily torsion-free group? Thanks.

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This link could be useful: – Giovanni De Gaetano Oct 15 '12 at 9:51
Am I right if I say that your question is equivalent to ask for a pure subgroup $S$ which does not contain the whole torsion of the original group $G$? If this is the case a negative answer could be deduced from this answer:… . Please let me know if I'm just saying a pack of rubbish or not! – Giovanni De Gaetano Oct 15 '12 at 10:00
up vote 1 down vote accepted

Pure subgroups are limits of direct summands. In particular every direct summand is pure. Take $G$ of order 2, and $S$ of order 1 for a minimal counter example.

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Thanks for the counter example. What about $\mathbb Q/\mathbb Z$? – Babak S. Oct 15 '12 at 13:26
For $\mathbb{Q}/\mathbb{Z}$ the only pure subgroups are the direct summands, such as $\mathbb{Z}[\tfrac12]/\mathbb{Z}$, but they all have torsion quotients. – Jack Schmidt Oct 15 '12 at 16:04

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