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I am trying to prove $\mathbb{Z}^{\oplus \mathbb{N}} \times \mathbb{Z}^{\oplus \mathbb{N}} \cong \mathbb{Z}^{\oplus \mathbb{N}}$ and will appreciate hints to approach the question.

Background:

This question is taken from Aluffi.

$\mathbb{Z}^{\oplus \mathbb{N}}$ is defined as

$$\{ \alpha \colon \mathbb{N} \to \mathbb{Z} \mid \alpha(n) \neq 0 \textrm{ for only finitely many elements } n \in \mathbb{N}\}$$

with the group operation $(\alpha + \beta) (n) = \alpha(n) + \beta(n)$, and it forms the free group of $\mathbb{N}$.

I have tried to show the isomorphism by defining group homomorphisms $\phi \colon \mathbb{Z}^{\oplus \mathbb{N}} \times \mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\oplus \mathbb{N}}$ by $\phi(\alpha,\beta) = \alpha + \beta$ and $\psi \colon \mathbb{Z}^{\oplus \mathbb{N}} \times \mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\oplus \mathbb{N}}$ by $\psi(\alpha) = (\alpha, \alpha)$. However, the first homomorphism is not injective while the second isn't surjective.

Brian M. Scott
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Kelvin Soh
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    Define a morphism by placing the elements from the first sequence in the even entries and the elements in the second sequence in the odd entry. – User0112358 Dec 01 '15 at 06:01
  • This notion is interesting to me; a group which is isomorphic to a "proper" subgroup. Is there a name for such groups? Is there a characterization of them? The same will work for rings too. Cool stuff!! – User0112358 Dec 01 '15 at 06:53
  • Your question is solved in this comment. – user26857 Dec 01 '15 at 09:25
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    @User0112358 Those groups are the non-cohopfian groups. A cohopfian group is a group which is not isomorphic to any proper subgroups. It's actually quite common for infinite groups to not be cohopfian. For example, $\mathbb{Z}$ is isomorphic to $2\mathbb{Z}$. Inside the free group $F_2$ of rank 2 sits free groups of every rank (even infinite!). In particular, for any $x\in F_2$, if $y\in F_2$ and $y\notin \langle x\rangle$, then $\langle x,y\rangle\subsetneq F_2$ is also a free group of rank 2. – oxeimon Dec 01 '15 at 18:42
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    @User0112358 Another example of a non-cohopfian group is $PSL_2(\mathbb{Z})$ (which is the free product $C_2*C_3$. – oxeimon Dec 01 '15 at 18:44
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    @User0112358 the dual property is hopfian. A group hopfian if it is not isomorphic to any of its proper quotients. This is far more common. For example, any finitely generated residually finite group is hopfian. (Just google hopfian or cohopfian for more information) – oxeimon Dec 01 '15 at 18:46
  • You can map your module isomorphically to a module of infinite matrices with finitely many non-zero entries in each row (or column). Can you think of way to show the isomorphism using this representation? (Note: this is the classical example of a module that does not satisfy the invariant basis property). – GiantTortoise1729 Dec 01 '15 at 18:59
  • @oxeimon thank you very much for your reply :) – User0112358 Dec 02 '15 at 02:09

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$\newcommand{\NN}{\mathbb{N}}$ $\newcommand{\ZZ}{\mathbb{Z}}$

The key is to notice that $\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$ is just $\NN$-many copies of $\ZZ$, and $\ZZ^{\oplus\NN}$ is also just $\NN$-many copies of $\ZZ$. Concretely, $\ZZ^{\oplus\NN}$ is indexed by $\NN$, whereas $\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$ is indexed by two copies of $\NN$, so it's indexed by $\NN\sqcup\NN$, where $\sqcup$ denotes disjoint union.

The fact that "$\NN$-many copies of $\ZZ$ is isomorphic to $\NN\sqcup\NN$-copies of $\ZZ$" relies on the fact that $\NN$ and $\NN\sqcup\NN$ have the same cardinality, which in turn relies on the fact that there is a bijection between $\NN$ and $\NN\sqcup\NN$. Let $i_1,i_2 : \NN\hookrightarrow\NN\sqcup\NN$ denote the two injections of $\NN$ into each copy of $\NN$ inside $\NN\sqcup\NN$. Then, for example, the map $f : \NN\rightarrow\NN\sqcup\NN$ given by $$f(n)\ = \left\{\begin{array}{ll} i_1\left(\frac{n}{2}\right) & \text{if $n$ is even} \\ i_2\left(\frac{n+1}{2}\right) & \text{if $n$ is odd} \end{array}\right.$$

is a bijection.

Then, you can define $\phi : \ZZ^{\oplus\NN}\rightarrow\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$ as follows:

$$\phi(a_1,a_2,a_3,\ldots) = \phi(a_{f(1)},a_{f(2)},a_{f(3)},\ldots)$$ where recall that $\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$ is indexed by $\NN\sqcup\NN$.

Here's another way to think about $\phi$. Let $j_1,j_2$ denote the natural injections $$j_k : \ZZ^{\oplus\NN}\hookrightarrow\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$$ (For any group $A$, there are two "obvious" natural injections $A\rightarrow A\times A$). Let $e_i\in\ZZ^{\oplus\NN}$ denote the usual basis vector with a 1 in the $i$th position and 0's everywhere else. Then

$$\phi(e_i)\ = \left\{\begin{array}{ll} j_1\left(e_{\frac{n}{2}}\right) & \text{if $n$ is even} \\ j_2\left(e_{\frac{n+1}{2}}\right) & \text{if $n$ is odd} \end{array}\right.$$

oxeimon
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  • I don't know what you mean by "$\mathbb N$-many copies", but I think this should refer to $\mathbb Z^{\mathbb N}$. – user26857 Dec 01 '15 at 21:17
  • you can think of either $\mathbb{Z}^\mathbb{N}$ or $\mathbb{Z}^{\oplus\mathbb{N}}$ both as "$\mathbb{N}$-many copies". While it is true that $\mathbb{Z}^\mathbb{N}$ is more accurately described by "$\mathbb{N}$-many copies of $\mathbb{Z}$", the point was to illustrate an idea, which I then make rigorous in the rest of my answer. – oxeimon Dec 01 '15 at 21:30