when $\Bbb{Z}^n$ isomorphic to $\Bbb{Z}^m$

Question

I would like to know when $H_1:=\Bbb{Z}^n$ is isomorphic (as a group) to $H_2:=\Bbb{Z}^m$, saying like that I would say only when $n=m$ but it's not a proof.

My idea : I can assume that they are isomorphic when $n\ne m$ and take $$G_1:=\{x\in H_1: y\in H_2, x=2y\}$$ and $$G_2:=\{x\in H_2: y\in H_1, x=2y\}.$$ I was thinking that it's a copy of $\Bbb{Z}/2\Bbb{Z}, n$ times and the other one $m$ times, and so it's a contradiction. Is that correct ? I am not sure because it's more an intuitive explanation than a proof.

Answer 1

Try this:

If $\mathbb Z\times \mathbb Z\times\cdot\cdot\cdot \mathbb Z (n)$ times $\cong \mathbb Z\times \mathbb Z\times\cdot\cdot\cdot \mathbb Z(m)$ times

Then $2\mathbb Z\times 2\mathbb Z\times\cdot\cdot\cdot 2\mathbb Z (n)$ times $\cong 2\mathbb Z\times 2\mathbb Z\times\cdot\cdot\cdot 2\mathbb Z(m)$ times then we get

$\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\cdot\cdot \mathbb Z/2\mathbb Z(n)$times $\cong\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\cdot\cdot \mathbb Z/2\mathbb Z(m)$times

$\implies 2^n=2^m$ which holds $\iff n=m$

NOTE: Definitely it is not true that two groups are isomorphic implies that their quotient groups are isomorphic too.Consider $\mathbb Z\cong\mathbb Z,2\mathbb Z\cong3\mathbb Z$ but $\mathbb Z_2\ncong \mathbb Z_3$

But the above result holds in my case because the isomorphism between $\mathbb Z\times \mathbb Z\times\cdot\cdot\cdot \mathbb Z (n)$ times $\cong \mathbb Z\times \mathbb Z\times\cdot\cdot\cdot \mathbb Z(m)$ times is the one which determines all the other isomorphisms involved in the answer which was not the case in the example provided above.In other words all the isomorphisms are completely determined by the first isomorphism mapping.

Hope this brings some clarity.

Answer 2

The result follows from the following theorem:

Let $M\le \mathbb Z^n$. Then $M\cong\mathbb Z^{m}$ for some $m\le n$,

since if $\mathbb Z^n\cong\mathbb Z^m$, then each can be viewed as a subgroup of the other.

The proof of the theorem is by induction on $n$. (The result is easy to prove when $n=1$)

Let $\{e_1,\ldots,e_n\}$ be a set of generators for $\mathbb Z^n$, and let $G=\langle e_1,\ldots,e_{n-1}\rangle\cong\mathbb Z^{n-1}$. Then by the inductive hypothesis, $G\cap M\cong\mathbb Z^m$ for some $m\le n-1$.

Now $\mathbb Z^n/G\cong\mathbb Z$ and the map$$M/(G\cap M)\overbrace{\cong}^{\text{2nd isomorphism theorem}} MG/G\hookrightarrow\mathbb Z^n/G\cong \mathbb Z$$is injective, so $M/(G\cap M)$ can be seen as a subgroup of $\mathbb Z$. Hence $$M/(G\cap M) \cong \mathbb Z^i$$where $i=0$ or $1$. Since $M, G\cap M$ are free groups and $G\cap M \cong \mathbb Z^{m}$, it follows (using some non-trivial results about free groups) that$$M\cong \mathbb Z^{i+m}$$and since $m\le n-1$, we have $m+i\le n$.

Answer 3

I would like to formulate your reasoning using category language, to give an another point of view. It follows the general idea that one uses functors to get "simpler" objects to refute isomorphy.

Consider the category $\def\Ab{\mathsf{Ab}}\Ab$ of abelian groups and the functor $F \colon \Ab \to \Ab$ given on objects by $F(A) = A/2A$, an on morphisms $f \colon A \to B$ by $F(f)(a + 2A) = f(a) + 2B$ (note that $f[2A] \subseteq 2B$), hence $F(f)$ is well-defined. As $F$ is a functor, it maps isomorphisms to isomorphims. If $\mathbf Z^n$ and $\mathbf Z^m$ are isomorphic, $F(\mathbf Z^n) = (\mathbf Z/2\mathbf Z)^n$ and $F(\mathbf Z^m) = (\mathbf Z/2\mathbf Z)^m$ are also. Hence $n = m$ (for example by counting elements).

Answer 4

The easiest way would be to tensor over $\mathbb{Z}$ by $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$, which might be what you were trying to do. Then, you would have two $\mathbb{F}_2$-vector spaces $\mathbb{F}_2^m$ and $\mathbb{F}_2^n$, which are isomorphic iff $m=n$, either by counting the number of elements or by an argument involving bases.