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Let $G$ be a finitely generated abelian group. Then prove that it is not isomorphic to $\frac{G}{N}$, for every subgroup $N\neq\langle 1\rangle$.

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Off the top of my head, I'd look at the orders of the generators. It would seem one of them would have to drop when you move to the quotient group. –  Alex Becker Apr 11 '12 at 8:11
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Suppose $f:G\to G/N$ is an isomorphism and let $p:G\to G/N$ be the projection.

Let $N_1=N$, $N_2=p^{-1}(f(N_1))$, $N_3=p^{-1}(f(N_2))$ and so on. This defines a increasing sequence $$N_1\subseteq N_2\subseteq N_3\subseteq\cdots$$ of subgroups of $G$. You can check that if $N$ is non-trivial, then the sequence is strictly increasing.

This is absurd, since $G$ is a noetherian $\mathbb Z$-module.

Later. Alternatively, let $N$ be a subgroup of $G$. We have a short exact sequence $$0\to N\to G\to G/N\to 0$$ so for the ranks of the groups we have $\newcommand\rk{\operatorname{rk}}\rk G=\rk N+\rk G/N$. If $G\cong G/N$ we must have $\rk N=0$, so $N$ is torsion. But then $N$ is contained in the torsion subgroup $t(G)$ of $G$, and clearly the torsion subgroup $t(G/N)$ has less elements than $t(G)$ unless $N=0$. It follows that $G\not\cong G/N$ unless $N=0$.

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What can we say when $G$ is a permutation group instead? –  B. S. Apr 11 '12 at 8:38
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Well, that is a completely different question... You'd better ask it separately! –  Mariano Suárez-Alvarez Apr 11 '12 at 8:39
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