I have to find out if these groups are isomorphic:

  1. $( { \mathbb {Q} };+) $ and $(\mathbb{Z}\times\mathbb{Z};+)$ - (this exercise is the most important for me)

  2. $( { \mathbb {Q} };+)$ and $( { \mathbb {Q} \times \mathbb {Q} };+)$

  3. $(ℝ;+)$and $(\mathbb{R}\times \mathbb{R};+)$

I have tried to solve 1. and first of all I have tried to find some property which could show that these two groups can not be isomorphic. But I have not find such property so I think that $( { \mathbb {Q} };+) $ and $(\mathbb{Z}\times\mathbb{Z};+)$ are isomorphic, but I can not to prove it, I can not find some isomorphism.

Thank you very much for any help.

  • 3
    $\begingroup$ $\Bbb Q$ is divisible. $\endgroup$ Nov 22, 2017 at 20:25
  • $\begingroup$ Try to think about the difference between $\mathbb{Q}$ and $\mathbb{Z}\times\mathbb{Z}$ as sets. $\endgroup$
    – Alex
    Nov 22, 2017 at 20:25
  • $\begingroup$ If $x\in\Bbb Q$ there is $y\in\Bbb Q$ with $x=y+y$. $\endgroup$ Nov 22, 2017 at 20:28
  • 3
    $\begingroup$ @Alex How would thinking just of the sets help? $\endgroup$ Nov 22, 2017 at 20:29
  • $\begingroup$ Is "$+$" in $(\mathbb{Z}\times\mathbb{Z};+)$ defined as $(p_1,q_1)+(p_2,q_2)=(p_1q_2+p_2q_1,p_1q_2)$ (i.e. like normal fractions) ? $\endgroup$
    – Jepsilon
    Nov 22, 2017 at 20:29

2 Answers 2

  1. Every homomorphism from $\mathbb Q$ to $C_2$ is trivial. Indeed, let $x \in \mathbb Q$. Write $x=2y$. If $y \mapsto g$, then $x \mapsto g^2 = e$. On the other hand, there is a nontrivial homomorphism from $\mathbb Z \times \mathbb Z$ to $C_2$ given by $(m,n) \mapsto g^m$.
    A simpler reason is that every element in $\mathbb Q$ is a square, that is, $x \mapsto 2x$ is surjective. But not every element of $\mathbb Z \times \mathbb Z$ is a square, that is, of the form $(2m,2n)$.

  2. Consider $\mathbb Q$ and $\mathbb Q \times \mathbb Q$ as vector spaces over $\mathbb Q$. Every additive map $\mathbb Q \to \mathbb Q \times \mathbb Q$ is actually a linear transformation. But $\mathbb Q$ and $\mathbb Q \times \mathbb Q$ have different dimensions.

  3. The argument above does not apply to $\mathbb R$ and $\mathbb R \times \mathbb R$ as vector spaces over $\mathbb R$ because not every additive map is a linear transformation. Nevertheless, $\mathbb R$ and $\mathbb R \times \mathbb R$ have the same dimension as vector spaces over $\mathbb Q$ and so are isomorphic.

  • $\begingroup$ Plus one. But I would say that #2 needs more talk. The groups $\Bbb Q$ and $\Bbb Q\times\Bbb Q$ have one and only one structure as $\Bbb Z$-modules. As they are uniquely divisible, each has one and only one structure as $\Bbb Q$-module. Then I would continue as you do. $\endgroup$
    – Lubin
    Nov 23, 2017 at 1:42

We know that $\mathbb{Q}$ is not a free-abelian group, see here:

How can we show that $\mathbb Q$ is not free?

On the other hand $\mathbb{Z}^2$ is free of rank $2$.


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