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Let (A,+) and (B,+) be commutative groups and suppose that A is isomorphic to B. Prove that $A/dA$ is isomorphic to $B/dB$, where $d \in \mathbb{N}$.

Any thoughts on this one?

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    We can try $\phi(\operatorname{cl}_A x)=\operatorname{cl}_B(\psi(x))$, where $\psi$ is an isomorphism between $A$ and $B$.2012-10-17
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    Could you please tell me what is $cl_Ax$? I have never encountered this notation before.2012-10-17
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    It's the equivalence class of the element $x$, here modulo $dA$.2012-10-17
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    @DavideGiraudo: But I think the OP mentioned $dG=\{g\in G|\exists n\in\mathbb N, ng=0\}$ instead.2012-10-17

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Let $\phi: G\to H$ is our isomorphism. Define $\phi|_{dG}:dG\to H$ and show that it is an injective homomorphism and so is between $dG$ and $dH$. Moreover, $\phi|_{dG}$ is onto (Why?). For your problem, regarding to @Marc's answer, define $\psi:G\to \frac{H}{dH}$ with $\psi(g)=\phi(g)+dH$ and use the first isomorphism theorem for it.

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    $\ddot\smile\quad +1$2013-03-29
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    @amWhy: Thanks again Amy. I am not native and don't know how express my gratitude as it should be. :-)2013-03-29
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    You do just fine, native or not!2013-03-29
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Morphisms of commutative groups commute with multiplication by an integer. So if $\phi:A\to B$ is an isomorphism then $\phi(dA)=dB$, and so $\phi$ and its inverse pass to the quotient, giving isomorphisms there.