# G additive group isomorphic to $\mathbf{Z}^{n}, \mathbf{Z}^{m}$

I have been stuck with this problem for quite a while now:

G is an additive group with $n,m \in \mathbf{N}$ ; $\mathbf{Z}^{n}, \mathbf{Z}^{m}$ isomorphic. Then it holds that: $\#(G/2G) = 2^{n} = 2^{m}$. Show also that m=n without using Theorem 2.1 (or a later one).

I wanted to show the cardinality first by arguing with equivalence classes: $\mathbf{Z}^{n} = \mathbf{Z}\times \cdots \mathbf{Z}$. Let $G= \mathbf{Z}x_{1} \times \mathbf{Z}x_{2}\times\cdots \times \mathbf{Z}x_{n}$ , $2G= 2\mathbf{Z}x_{1} \times 2\mathbf{Z}x_{2}\times \cdots\times 2\mathbf{Z}x_{n}$

So then by taking: $G/2G= \mathbf{Z}x_{1}/2\mathbf{Z}x_{1}\times\cdots\times \mathbf{Z}x_{n} /2\mathbf{Z}x_{n}$

I think that every "factor" adds 2 equivalence classes (because I know that $\mathbf{Z}/2\mathbf{Z} = \{\tilde{0},\tilde{1}\}$, but I have no clue how to show that there aren't any duplicates in the product. Theorem 2.1 states that if R is an integral domain, then if there are $R^{n}, R^{m}$ which are isomorphic, $m=n$ holds.

Does anybody know how to show that there aren't any duplicates while showing the cardinality? Also how to show that $m=n$ without using any theorem? Merci

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Note that $G/2G$ is actually a vector space over the field $\mathbf{Z}/2\mathbf{Z}$. Vector spaces have dimensions that play by very nice rules. –  Dylan Moreland Dec 1 '11 at 19:53
What exactly does "Theorem 2.1" say? And what happened to $z$? –  Arturo Magidin Dec 1 '11 at 19:56
@Arturo Magidin: Theorem 2.1: If R is an integral domain, where $n.m \in \mathbf{Z}^{n}, \mathbf{Z}^{m}$ isomorphic, then $m=n$. –  VVV Dec 1 '11 at 20:04
What is $R$, in that statement? –  Dylan Moreland Dec 1 '11 at 20:10
It's not explicitly mentioned, only that it is an integral domain… But I assume it to be a ring. –  VVV Dec 1 '11 at 20:16

In short, you want to show that if $\mathbf{Z}^n\cong\mathbf{Z}^m$, as additive groups (with $\mathbf{Z}$ being the additive group of integers), then $n=m$.

One route is the one given in the statement. There are a couple of ways of showing it.

First:

Claim 1. If $G$ is abelian, and $G=H\times K$, then for any positive integer $n$ we have $nG=nH\times nK$.

Proof. If $(nh,nk)\in nH\times nK$, then since $(nh,nk) = n(h,k)\in nG$ we have that $(nh,nk)\in nG$. Conversely, if $(x,y)\in nG$, then $(x,y) = n(h,k) = (nh,nk)$ for some $h\in H$, $k\in K$, so $(x,y)\in nH\times nK$.

Claim 2. If $G = H\times K$, $M\triangleleft H$, and $N\triangleleft K$, then $M\times N\triangleleft G$, and $G/(M\times N) \cong (H/M)\times (K/N)$.

Sketch. Consider the map $G\to (H/M)\times (K/N)$ given by $\varphi(x,y) = (xM,yN)$. Prove that this is an onto group homomorphism. By the Isomorphism Theorems, $$\frac{G}{\mathrm{ker}(\varphi)} \cong \frac{H}{M}\times \frac{K}{N}.$$ Prove that $\mathrm{ker}(\varphi) = M\times N$ to complete the proof.

Now show that the above claims hold for any number of finite factors (in fact, they hold for any number of factors, finite or infinite). Together they will show that $G/2G\cong (\mathbf{Z}/2\mathbf{Z})^n$ and $G/2G\cong (\mathbf{Z}/2\mathbf{Z})^m$.

So, if $G\cong\mathbf{Z}^n$ and $G\cong\mathbf{Z}^m$, then applying the ideas above it follows that on the one hand you have $G/2G$ has $2^n$ elements, and on the other that it has $2^m$ elements. But $G/2G$ does not depend on how you write $G$, it just depends on $G$. So we must have $2^n=2^m$.

The following is just basic combinatorics:

If $A$ has $k$ elements and $B$ has $\ell$ elements, then $A\times B$ has $k\ell$ elements:

Proof. The number of elements in $A\times B$ is the number of ordered pairs $(a,b)$ with $a\in A$ and $b\in B$. There are $k$ possibilities for the first entry, $\ell$ possibilities for the second entry, so by the Multiplication Rule, the total number of possibilities is $k\times \ell = k\ell$.

By induction, if $A_1,\ldots,A_n$ are all finite, then $$|A_1\times\cdots\times A_n| = |A_1|\times\cdots\times |A_n|.$$

(In fact, for possibly infinite sets, one defines the product of the cardinalities to be the cardinality of the cartesian product, $|A|\cdot|B|=|A\times B|$)

So if $A_i = \mathbb{Z}/2\mathbb{Z}$ for each $i$, then since $|\mathbb{Z}/2\mathbb{Z}|=2$...

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These claims appear later in a similar form, I don't think they are allowed to be used…. Thank you, though. –  VVV Dec 1 '11 at 20:30
@VVV: You could prove them, instead of just using them. In any case, you are invoking/using/proving that $G/2G=(\mathbb{Z}/2\mathbb{Z})\times\cdots\times(\mathbb{Z}/2\mathbb{Z})$, and your only problem is the cardinality of this product, that's a set theory issue: $|A\times B| = |A|\times|B|$: just count the ordered pairs. –  Arturo Magidin Dec 1 '11 at 20:34
Merci!……………………………………. –  VVV Dec 1 '11 at 21:07