Prove or disprove: $\mathbb{Q}$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$. I mean the groups $(\mathbb Q, +)$ and $(\mathbb Z \times \mathbb Z,+).$ Is there an isomorphism?
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Yet another way to see the two cannot be isomorphic as additive groups: if $a,b\in\mathbb{Q}$, and neither $a$ nor $b$ are equal to $0$, then $\langle a\rangle\cap\langle b\rangle\neq\{0\}$; that is, any two nontrivial subgroups intersect nontrivially. To see this, write $a=\frac{r}{s}$, $b=\frac{u}{v}$, with $r,s,u,v\in\mathbb{Z}$, $\gcd(r,s)=\gcd(u,v)=1$. Then $(su)a = (rv)b\neq 0$ lies in the intersection, so the intersection is nontrivial. However, in $\mathbb{Z}\times\mathbb{Z}$, the elements $(1,0)$ and $(0,1)$ are both nontrivial, but $\langle (1,0)\rangle\cap\langle (0,1)\rangle = \{(0,0)\}$. |
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I assume that you are asking whether we have an isomorphism of additive groups. In that case, assume that $\phi: \mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ is such an isomorphism. So we have for example that $\phi(0) = (0,0)$. Let $a\in \mathbb{Q}$ be such that $\phi(a) = (1,0)$ and $b$ be such that $\phi(b) = (0,1)$. Then we see that $\mathbb{Q}$ is equal to $\{na + mb \lvert n,m \in \mathbb{Z}\}$. This is a contradiction... (Hence the argument is that $\mathbb{Q}$ is not finitely generated while $\mathbb{Z}\times \mathbb{Z}$ is.) (I will leave the details to you.) |
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Another argument that you can construct with the following (be sure you can prove/answer every section): 1) An abelian additive group $\,A\,$ is said to be divisible if $\,\forall a\in A\,\,n\in\mathbb{N}\,\,\exists b\in A\,\,s.t.\,\,nb=a\,$ . To be sure, $n\neq 0$ 2) $\,\mathbb{Q}\,$ is a divisible group 3) Any homomorphic image of a divisible group is a divisible group 4) Is $\,\mathbb{Z}\times\mathbb{Z}\,$ divisible? |
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Let $ \phi: \mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ be a homomorphism. Fix $u/v\in\mathbb{Q}$ and let $(a_n,b_n)=\phi(u/v^n)$. Since $\phi(u/v)=v^{n-1}\phi(u/v^n)$, we get $a_1=v^{n-1}a_n$ and $b_1=v^{n-1}b_n$ for all $n\in\mathbb N$, which is clearly impossible unless $\phi(u/v)=(a_1,b_1)=(0,0)$. So, the only homomorphism $\mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ is the zero map, and there is no chance of an isomorphism. |
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Another idea: suppose there is an isomorphism $\mathbb Q \to \mathbb Z \oplus \mathbb Z$, then tensor both sides $\otimes_\mathbb{Z} \mathbb Q$, and get a $\mathbb Q$-module isomorphism $\mathbb Q = \mathbb Q \otimes_\mathbb{Z} \mathbb Q \to (\mathbb Z \oplus \mathbb Z)\otimes_\mathbb{Z} \mathbb Q = \mathbb Q\oplus \mathbb Q$. |
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You can show that the group $(\mathbb{Q}, +)$ has no proper subgroup of finite index, but for example $\mathbb{Z} \times 2\mathbb{Z}$ has finite index in ($\mathbb{Z} \times \mathbb{Z}, +)$. |
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Another way of seeing this (yes, there are many ways!) is to notice that two isomorphic groups have the same quotients. That is, if $G\cong H$ and $G\twoheadrightarrow K$ then $H \twoheadrightarrow K$. Now, by this question (which was only asked the other day, which is why I am posting this answer!), we know that every proper quotient of $\mathbb{Q}$ is torsion (that is, every element has finite order). On the other hand, $\mathbb{Z}\times\mathbb{Z}$ has a torsion-free proper quotient, $\mathbb{Z}\times\mathbb{Z}\twoheadrightarrow \mathbb{Z}$. Thus, they cannot be isomorphic. (Indeed, this actually proves that there cannot be a homomorphism from $\mathbb{Q}$ to $\mathbb{Z}\times\mathbb{Z}$, as lhf has already shown - the result we use tells us that the map cannot have non-trivial kernel, while this proves that the kernel cannot be trivial either.) |
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In ($\mathbb{Q}$, +) every element has a "square root": for any $q \in \mathbb{Q}$ there is an $x \in \mathbb{Q}$ such that $x + x = q$. But in $(\mathbb{Z} \times \mathbb{Z}, +)$ for any element $(a,b) \in \mathbb{Z} \times \mathbb{Z}$ there is an $x \in \mathbb{Z} \times \mathbb{Z}$ such that $x + x = (a,b)$ if and only if both $a$ and $b$ are even. This motivates the following proof. If $\phi: \mathbb{Q} \rightarrow \mathbb{Z} \times \mathbb{Z}$ were an isomorphism, then $\phi(q) = (1,1)$ for some $q \in \mathbb{Q}$. Then $(1,1) = \phi(q/2 + q/2) = \phi(q/2) + \phi(q/2)$, but this is a contradiction since $1$ is not even. |
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