0
$\begingroup$

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?

  • 1
    @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

1 Answers 1

4

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