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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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    There is no surjection, because as Asaf said the homomorphism is completely determined by $\phi (1)$2012-05-24

3 Answers 3

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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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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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    @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 $k_i \in \mathbb{Z}$.2012-05-24
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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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    Since $2^{1/n}$ is irrational for every $n$ (see: http://math.stackexchange.com/questions/11872/is-nth-root-of-2-an-irrational-number, 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.2012-08-19