A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{>0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} , \times )$, not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here?

  • $\begingroup$ This is Exercise II.4.6 in Aluffi Chapter 0. I was also struggling with it. $\endgroup$ – user125763 Sep 6 '14 at 13:50

The isomorphism would have to map some element of $(\mathbb{Q},+)$ to $2$. There is no element of $(\mathbb{Q}_{>0},\times)$ whose square is $2$, but whatever number is mapped to $2$ has a half in $(\mathbb{Q},+)$. More generally speaking, you can divide by any natural number $n$ in $(\mathbb{Q},+)$, but you can't generally draw $n$-th roots in $(\mathbb{Q}_{>0},\times)$. More abstractly speaking, you can introduce an invertible multiplication operation on $(\mathbb{Q},+)$ to turn it into a field (in fact that in a sense is the point of the construction of $\mathbb{Q}$) but you can't define a corresponding exponentiation operation within $(\mathbb{Q}_{>0},\times)$.

The isomorphism that you expected to exist exists not between $(\mathbb{Q},+)$ and $(\mathbb{Q}_{>0},\times)$ but between $(\mathbb{Q},+)$ and $(b^\mathbb{Q},\times)$ for any $b\in\mathbb{R}_{>0} \setminus\{1\}$. Since $b^\mathbb{Q}$ always contains irrational elements, this is never a subgroup of $(\mathbb{Q}_{>0},\times)$.

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    $\begingroup$ More-or-less, joriki's answer can be rephrased like this. If we find a property, which is preserved by isomorphisms and which is fulfilled only for one of the groups, then the groups are not isomorphic. In this case, the property of the group $(G,\circ)$ is $(\forall x\in G)(\exists y\in G) y\circ y=x$. $\endgroup$ – Martin Sleziak Apr 18 '11 at 9:11
  • $\begingroup$ Martin's comment referred to an earlier version of the answer that only had the first couple of sentences. $\endgroup$ – joriki Apr 18 '11 at 9:28
  • $\begingroup$ It might be added that the property mentioned here (being avle to divide by any natural number n, which would then correspond to being able to take n'th roots in the multiplicative case) is called being divisible. $\endgroup$ – Tobias Kildetoft Apr 18 '11 at 9:45
  • $\begingroup$ To clarify that slightly (since "being able to divide" is a property of humans and computers, not of groups :-): A group is called divisible if for each element $x$ and each natural number $n$ there is an element $y$ such that $ny$, defined as the $n$-fold sum of $y$, is $x$. Then $(\mathbb{Q},+)$ is divisible and $(\mathbb{Q}_{>0},\times)$ isn't. $\endgroup$ – joriki Apr 18 '11 at 10:19
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    $\begingroup$ The generalization of the result mentioned by Martin Sleziak can be found in Kuratowski's and Mostowski's "Set theory" (p.85 of the 2-nd edition). Given an isomorphism f between two algebraic systems (not necessarily groups), the result allows to "build" the whole spectrum of formulas preserved by f. $\endgroup$ – Elmar Guseinov Mar 11 '20 at 10:18

The fundamental theorem of arithmetic exactly says that $(\mathbb{Q}_{>0}, \times)$ is an abelian free group with the set of primes as a basis. Therefore, if $(\mathbb{Q}_{>0}, \times)$ and $(\mathbb{Q},+)$ were isomorphic, $(\mathbb{Q},+)$ would be an abelian free group.

Suppose by contradiction that $X$ is a free basis of $(\mathbb{Q},+)$ and let $x \in X$. Then there exist $x_1,\dots,x_m \in X$ and $a_1, \dots,a_m \in \mathbb{Z}$ (uniquely determined) so that $$ \frac{x}{n} = a_1x_1+ \dots + a_mx_m,$$ hence $$x= na_1x_1+ \dots+ na_mx_m.$$ Consequently, there exists $1 \leq i \leq m$ such that $x=na_ix_i=na_ix$ ie. $na_i=1$; a contradiction with $a_i \in \mathbb{Z}$ when $n>1$.


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