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I've read the following argument (on Wikipedia, IIRC) :

Suppose that $\Bbb Z^n \cong \Bbb Z^m$ as abelian groups. Then quotienting both sides by $(2\Bbb Z)^n$ and $(2\Bbb Z)^m$ respectively, we get the group isomorphism $\Bbb F_2^n \cong \Bbb F_2^m$, which implies $2^n=2^m$ i.e. $n=m$ [NB : we just compare cardinalities, we can't conclude $n=m$ because of some vector space isomorphism].

I have the following question : Why can we ensure that the image of $(2\Bbb Z)^n$ under the given isomorphism is $(2\Bbb Z)^m$ ? I want to use $A/I \cong B/f(I)$, where $f : A \to B$ is an injective group morphism. But I don't know how.

Thanks!

Alphonse
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  • By the way, if you have other proofs of $\Bbb Z^n \cong \Bbb Z^m \implies n=m$, I would be interested :-) – Alphonse Aug 17 '16 at 09:15
  • Interestingly, this doesn't hold for modules over any ring $R$. See invariant basis number https://en.wikipedia.org/wiki/Invariant_basis_number – Ulrik Aug 17 '16 at 09:33
  • Z^0 is not isomorphic to Z^(0+k) for all k (by cardinality difference). Assume Z^n is not isomorphic to Z^(n+k). Z^nZ is not isomorphic to Z^(n+k)Z because otherwise the isomorphism truncated to Z_n would be an isomorphism to Z^(n+k). – Jacob Wakem Aug 17 '16 at 09:45
  • I found http://math.stackexchange.com/questions/989944 – Alphonse Feb 10 '17 at 10:14

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If $\phi:\mathbb Z^n\to\mathbb Z^m$ is an isomorphism, then $\forall x\in\mathbb Z^n$, $\phi(2x) = \phi(x+x) = \phi(x)+\phi(x) = 2\phi(x)$ so the restriction of $\phi$ to $2(\mathbb Z^n)=(2\mathbb Z)^n$ is an isomorphism onto $2(\mathbb Z^m)=(2\mathbb Z)^m$.

George Law
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  • What did you prove?! – A_Sh Aug 17 '16 at 09:34
  • @user26857 I don't see $n=m$ in his answer! – A_Sh Aug 17 '16 at 09:44
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    @Ab_Sh, why didn't you just read the question: "I have the following question : Why can we ensure that the image of $(2\mathbb Z)^n$ under the given isomorphism is $(2\mathbb Z)^m$ ? " – Ennar Aug 17 '16 at 09:46
  • @Ennar Thanks, so I should read the question and not the answer! (see the comment of user26857) – A_Sh Aug 17 '16 at 09:49