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There was a question here on the site that was asked:

Let $G$ be a finitely generated abelian group. Then prove that it is not isomorphic to $G/N$, for every subgroup $N≠⟨1⟩$.

Would the answer for above question , as Mariano did, be valid if we assumed $G$ was a permutation group instead?

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    What does "permutation group" mean?2012-04-11
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    $G$ acts on a set of $n$ letters faithfully.2012-04-11
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    Then $G$ is finite, and the result is obvious by counting: $|G/N|=|G|/|N|<|G|$.2012-04-11
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    @ChrisEagle: Thanks. The answer was very easy.2012-04-11
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    @Babak : The correct thing to say to Chris is that "Your answer is very simple to understand" as oppose to "answer is easy", the latter does not have nice tone to it, although I know you meant to say "The answer Chris gave was simple and clear" Regards.2012-04-11
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    This also holds for infinite groups, use the fundamental theorem of finitely generated abelian groups $G$, which says that $G$ is a direct sum of a finite number of copies of $\mathbb{Z}$ and a torsion part.2012-04-11
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    @NickyHekster: While I don't really doubt the conclusion for infinite groups, the argument you suggest is not entirely convincing without additional detail. The subgroup $N$ could be badly positioned with respect to the direct sum decomposition.2012-04-11
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    @Arjang: Yes. I meant exactly the way in which my question has an answer is very simple and clear as you said.2012-04-11

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Permutation groups are finite, so we have $|G/N|=|G|/|N|<|G|$.