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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? –  Chris Eagle Apr 11 '12 at 9:03
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Then $G$ is finite, and the result is obvious by counting: $|G/N|=|G|/|N|<|G|$. –  Chris Eagle Apr 11 '12 at 9:09
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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. –  Arjang Apr 11 '12 at 10:28
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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. –  Nicky Hekster Apr 11 '12 at 10:42
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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. –  Babak S. Apr 11 '12 at 12:04

1 Answer 1

up vote 4 down vote accepted

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

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