# Is there an operational isomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Q}^{+},\cdot)$?

Let $\left(\mathbb{Z},+\right)$ and $\left(\mathbb{Q}^{+},\cdot\right)$ be groups (let the integers be a group with respect to addition and the positive rationals be a group with respect to multiplication). Is there a function $\phi\colon\mathbb{Z}\mapsto\mathbb{Q}^{+}$ such that:

• $\phi(a)=\phi(b) \implies a=b$ (injection)
• $\forall p\in\mathbb{Q}^{+} : \exists a\in\mathbb{Z} : \phi(a)=p$ (surjection)
• $\phi(a+b) = \phi(a)\cdot\phi(b)$ (homomorphism)

? If so, provide an example. If not, disprove.

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Note that every homomorphism from $\mathbb Z$ is fully determined by $\phi(1)$. –  Asaf Karagila May 24 '12 at 17:30
There is no surjection, because as Asaf said the homomorphism is completely determined by $\phi (1)$ –  Theorem May 24 '12 at 17:38

If such an isomorphism existed it would of course be onto so there would exist some $n \in \mathbb{Z}$ such that $\phi(n) = \frac{1}{2}$ for example. But then, $$\phi(n) = \phi(1+\cdots + 1) = \phi(1)\cdots \phi(1) = \phi(1)^n = \frac{1}{2}.$$ This implies $n=1$ since otherwise $\left(\frac{1}{2}\right)^{1/n} \notin \mathbb{Q}$. So, $\phi(1) = \frac{1}{2}$. Thus, for any $n \in \mathbb{Z}$, $\phi(n) = \phi(1)^n = \frac{1}{2^n}$, so clearly $\phi$ is not onto since we only achieve powers of two in the image. So such an isomorphism cannot exist.

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Expand on why $\left(\frac{1}{2}\right)^{\frac{1}{n}}\not\in\mathbb{Q}$. I feel like it's true but it is not made explicit. –  chharvey Aug 19 '12 at 13:18
Since $2^{1/n}$ is irrational for every $n$ (see: math.stackexchange.com/questions/11872/…, for example), we know that $\left(\frac{1}{2}\right)^{1/n} = \frac{1}{2^{1/n}}$ must be irrational. If it was not, then there would exist some integers $a,b$ such that $\frac{1}{2^{1/n}} = \frac{a}{b} \Rightarrow \frac{b}{a} = 2^{1/n}$ which is a contradiction. –  Derek Allums Aug 19 '12 at 13:54

No, because $\mathbb Z$ is generated by $1$ while $\mathbb Q^+$ as a group under multiplication is not finitely generated. If such a homomorphism $\phi$ existed, then we could write any element of $\mathbb Q$ as $\phi (n)=\phi(1)^n$, so $\mathbb Q$ would be generated by $\phi(1)$. But $\phi(1)=a/b$ for some $a,b\in\mathbb Z$, and clearly raising $a/b$ to an integral exponent wont give us $1/p$ for primes $p$ not dividing $b$.

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The OP asks about $\mathbb{Q}^+$ under multiplication, not addition. But a similar argument works. –  Derek Allums May 24 '12 at 17:49
@unit3000-21 Yes, fixed. –  Alex Becker May 24 '12 at 17:54
Neat proof - it looks sort of like a generalization of what I did below. –  Derek Allums May 24 '12 at 18:02
What does "generated by" mean? –  chharvey May 24 '12 at 18:06
@TestSubject528491 A multiplicative group $G$ is (finitely) generated by $\lbrace g_1,\ldots,g_n \rbrace$ if every $g\in G$ can be written as some combination of the generators: $g_1^{k_1}\cdots g_n^{k_n}$ for integers $k_i \in \mathbb{Z}$. –  Derek Allums May 24 '12 at 18:21

Every nontrivial subgroup of $\mathbb{Z}$ is infinite cyclic. But, $\mathbb{Q}^+$ contains a rank 2 abelian subgroup, namely all rational numbers of the form $2^m 3^n$ for integer values of $m,n$.

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