# Prove that the group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$ implies 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?

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.

• Can you explain what $\otimes_{\mathbb{Z}}$ means? Is it just the direct product? I've not seen this notation before. – chubbycantorset Dec 9 '12 at 7:42
• It's the tensor product. Look at my edit, now that I see you don't know tensor products. – Alex Youcis Dec 9 '12 at 7:44
• Alternatively, couldn't you go from the group isomorphism to a module isomorphism, and then use the fact that the rank is invariant? – wj32 Dec 9 '12 at 7:46
• @Rankeya Sure, localization commutes with products. But really, localizing at, say $0$, is the same thing as tensoring with $\mathbb{Q}$. – Alex Youcis Dec 9 '12 at 8:11
• @BenjaLim Obviously this was true. But, tensoring with a finite quotient field allows one to bypass having to prove the result for fields. – Alex Youcis Dec 9 '12 at 16:30

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 $e_k\in\Bbb Z^m$.

• This makes perfect sense, but as I just said in reply to Alex's comment, we aren't allowed to use anything related to vector spaces in our exams. – chubbycantorset Dec 9 '12 at 7:55
• @chubbycantorset: Well $\mathbb{Z}^n$ isn't exactly a vector space here, and you're not using any existing results about vector spaces in this proof. – wj32 Dec 9 '12 at 7:57
• @chubbycantorset: Aargh! In my opinion it’s a poor instructor who can’t come up with good problems without introducing stupid artificial restrictions $-$ and if that prohibition covers this argument, it qualifies as stupid. (Sorry for the outburst, but that’s been a pet peeve of mine for decades.) – Brian M. Scott Dec 9 '12 at 7:57
• I certainly agree. We've been shown pretty much no applications of theorems, and hence I find it more difficult to manipulate n-cycles or computing orders of groups than proving results that follow theorems. Incidentally, would you know any online resources that show you how to apply group theoretical concepts? – chubbycantorset Dec 9 '12 at 8:02
• @chubbycantorset: One fun application is in combinatorics. Download this book and take a look at Chapter 6. Also, this book is good: Applied Abstract Algebra – wj32 Dec 9 '12 at 8:07

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 $2^n = 2^m$.

Highfalutin approach: Suppose we have $f : \Bbb{Z}^m \stackrel{\simeq}{\longrightarrow} \Bbb{Z}^n$ and isomorphism. 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 $2$ 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 $m\neq n$?

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 $m=n$.