Question on a proof of $\Bbb Z^n \cong \Bbb Z^m \implies n=m$

Question

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!

Answer 1

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 $2(\mathbb Z^m)=(2\mathbb Z)^m$.