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Let $m$ and $n$ be two nonnegative integers. Assume that there is a group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$. Prove that $m = n$.

I tried using a contrapositive, ($m \neq n$ implies $\mathbb{Z}^m \ncong \mathbb{Z}^n$), and I think the problem is that there won't be a homomorphism, but did not get anywhere. Is there a better approach to this problem?

5 Answers 5

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Once again, I'll give you the outline, you fill in the details.

Suppose that $\mathbb{Z}^m\cong\mathbb{Z}^n$. Then, you see that $\mathbb{Z}^m\otimes_{\mathbb{Z}}\mathbb{Z}_2\cong\mathbb{Z}^n\otimes_{\mathbb{Z}}\mathbb{Z}_2$, which tells you that $\mathbb{Z}_2^m\cong\mathbb{Z}_2^n$, and so $2^m=2^n$--thus $m=n$.

EDIT: Since you don't know tensor products, perhaps this will be more understandable. If $\mathbb{Z}^m\cong\mathbb{Z}^n$, then $\text{Hom}_\mathbb{Z}(\mathbb{Z}^m,\mathbb{Q})\cong\text{Hom}_\mathbb{Z}(\mathbb{Z}^n,\mathbb{Q})$ as $\mathbb{Q}$-vector spaces. But, you can prove that as $\mathbb{Q}$-vector spaces one has that $\text{Hom}_\mathbb{Z}(\mathbb{Z}^k,\mathbb{Q})\cong\mathbb{Q}^k$, and so we have $\mathbb{Q}^m\cong\mathbb{Q}^n$, from where normal vector space theory tells us that $m=n$.

EDIT EDIT: You seemed to only take issue with the previous proof because of the vector spaces. In patticular,you seemed ok with the Hom manipulation. Here's a way to combine the first and the previous proof! Show first that $\text{Hom}(\mathbb{Z}^k,A)\cong A^k$ for any abelian group A. Then apply this fact to show that our problem implies $\mathbb{Z}_2^m\cong\mathbb{Z}_2^n$ again.

  • 0
    @BenjaLim It does not, it takes injections to injections precisely when the fixed module is flat, but that is not important here. Every single functor is isomorphism preserving, only special ones are exact. So, in particular while functors preserve properties that have to do with mappings and identities isomorphism (there exists an inverse), having a section or a retraction (having a left or right inverse), properties not phrased this way such as injectivity aren't in general preserved. If you have more questions you can email me, but we should stop this convo.2012-12-10
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You can prove it in the same way that you prove that the dimension of a vector space is well-defined.

Suppose that $m\le n$ and $h:\Bbb Z^m\to\Bbb Z^n$ is an isomorphism. For $k=1,\dots,m$ let $e_k$ be the element $\langle a_1,\dots,a_m\rangle$ such that $a_k=1$ and $a_i=0$ for $i\ne k$. Note for any $\langle a_1,\dots,a_m\rangle\in\Bbb Z^m$,

$\langle a_1,\dots,a_m\rangle=\sum_{k=1}^ma_ke_k\;,$

and this representation is unique: if

$\sum_{k=1}^ma_ke_k=\sum_{k=1}^mb_ke_k\;,$ then $\sum_{k=1}^m(a_k-b_k)e_k=\langle \underbrace{0,\dots,0}_m\rangle\;,$ and therefore $a_1=b_1,\dots,a_m=b_m$. In other words, $\{e_1,\dots,e_m\}$ behaves very much like a basis for a vector space.

Now show that $\{h(e_1),\dots,h(e_n)\}$ behaves like a basis for $\Bbb Z^n$: each $z\in\Bbb Z^n$ can be written uniquely in the form $z=\sum_{k=1}^ma_kh(e_k)$ for some integers $a_k$, $k=1,\dots,m$.

Now get a contradiction when $n>m$ by considering the elements of $\Bbb Z^n$ that are analogous to the $e_k\in\Bbb Z^m$.

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    @user26857: Pretty much the same way you’d prove it for $\Bbb Q^m$ and $\Bbb Q^n$. Either of the arguments [here](http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces#Case_2) works with minor modifications, since elements of $\Bbb Z^n$ are linearly independent over $\Bbb Z$ iff they’re lin. indep. over $\Bbb Q$.2015-02-08
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Let $G=\mathbb{Z}^n$ and $H=\mathbb{Z}^m$. Then $G/2G = (\mathbb{Z}/2\mathbb{Z})^n$. In particular, $|G/2G| = 2^n$. By a similar argument $|H/2H| = 2^m$. If they are isomorphic then $2^n = 2^m$.

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Highfalutin approach: Suppose we have an isomorphism $f : \Bbb{Z}^m \stackrel{\simeq}{\longrightarrow} \Bbb{Z}^n$. Then we recall that a module over a PID is flat iff it is torsion-free. Considering $\Bbb{Q}$ as an abelian group, it will now follow that the functor $- \otimes_{\Bbb{Z}} \Bbb{Q}$ is an exact functor and so

$f \otimes 1 : \Bbb{Z}^m \otimes_{\Bbb{Z}} \Bbb{Q} \to \Bbb{Z}^n \otimes_{\Bbb{Z}} \Bbb{Q}$

is injective. It is also clearly surjective and hence is an isomorphism. Using the fact that

  1. Tensor products commute with direct sums
  2. For any $R$ - module $M$ we have a canonical isomorphism $R \otimes_R M\cong M$ which on elementary tensors sends $r \otimes m \to rm$

we conclude that $f \otimes 1$ is an isomorphism between $\Bbb{Q}^m$ and $\Bbb{Q}^n$ and rank - nullity now gives that $m = n$.

More concrete approach:

Let us try to understand via elementary methods why there can be no isomorphism between say $\Bbb{Z}^3$ and $\Bbb{Z}^2$. Suppose there were some isomorphism $f : \Bbb{Z}^2 \to \Bbb{Z}^3$. Then if $x_1,x_2,x_3$ are the canonical basis generators for $\Bbb{Z}^3$ and $y_1,y_2$ that of $\Bbb{Z}^2$ we can find integers $a_{11},a_{12},a_{21},a_{22},a_{31},a_{32}$ such that

$\sum_{j=1}^2 a_{ij}y_j = x_i$

for $1 \leq j \leq 3$. More concretely, this means that given any triple $(a,b,c)$ we can find integers $d,e$ such that

$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} \begin{pmatrix} d \\ e \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}.$

Now from elementary row reduction when working over a field we know that this is not going to be possible simply because the number of pivots is only going to be $\leq 2$ and not $3$. But what if we work over the integers? I will now show using a concrete example that we can always get the last row to be zero. This will then give a contradiction because the matrix applied to no pair $(d,e)$ of integers will ever be equal to

$\begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}.$

Now for a concrete example. Suppose you take the matrix \begin{pmatrix} 7 & 6 \\ 5 & 12 \\ 4 & 1 \end{pmatrix}.

Let $R_1$ mean row $1$, $R_2$ row $2$ and so on. Then if we do $7R_3 - 4R_1$ and $7R_2 - 5R_1$ we get the matrix

$\begin{pmatrix} 7 & 6 \\ 0 & 54 \\ 0 & -17 \end{pmatrix}.$

But now if we do $54R_3 + 17R_2$ the last row will be zero and you get your desired contradiction.

From this concrete example do you see now why there can never be an isomorphism between $\Bbb{Z}^2$ and $\Bbb{Z}^3$. More generally do you see why there can never be an isomorphism between $\Bbb{Z}^m$ and $\Bbb{Z}^n$ for $m\neq n$?

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Embed $\Bbb Z$ in $\Bbb Q$ and thereby $\Bbb Z^n$ in $\Bbb Q^n$. Now clearly $\Bbb Z^n$ has an $n$-tuple of elements that are linearly independent over $\Bbb Z$ (the standard basis will do as an example), and it does not have any $(n+1)$-tuple that is linearly independent over $\Bbb Z$, since already $\Bbb Q^n$ doesn't admit any $(n+1)$-tuple that is linearly independent over $\Bbb Q$, let alone over $\Bbb Z$. Then you can recover the $n$ in $A\cong\Bbb Z^n$ as the maximum number of linearly independent elements over $\Bbb Z$ one can find in the free Abelian group $A$. Thus $\Bbb Z^m\cong\Bbb Z^n$ implies $m=n$.