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This question is a little bit different from Group of positive rationals under multiplication not isomorphic to group of rationals since I was wondering if logarithmic function could solve this or not, thank you.

Consider two groups $(\mathbb Q^+,\cdot)$ and $(\mathbb Q,+)$, does an isomorphism exist between them?

My attempt: Let $\varphi:\mathbb Q^+\rightarrow\mathbb Q$ be the isomorphic function, then the below statement must hold true for all $a,b\in\mathbb Q^+$: $$\varphi(a\cdot b)=\varphi(a)+\varphi(b)$$ So I guess maybe a logarithmic function would be fine here, since it's bijective, too.

But the problem is I can not show for a specific base like $10$ for example, $\log(\mathbb Q^+)=\mathbb Q$.

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  • $\begingroup$ Had the question been about $\mathbb{R}^+$ and $\mathbb{R}$, the answer would have been different (and easier), because here $\log : \mathbb{R}^+ \to \mathbb{R}$ is truly a bijection, and in fact a group isomorphism. However, when restricted to $\mathbb{Q}^+$, most values of $\log$ are still irrational (in fact, transcendental, this is a transcendental function), so it will not work here. $\endgroup$
    – Jeppe Stig Nielsen
    Commented Feb 25, 2016 at 9:30
  • $\begingroup$ @Najib thank you, I explained how it has some difference in the top of the problem text. if it's still not acceptable, tell me to close my question. $\endgroup$
    – Amir Naseri
    Commented Feb 25, 2016 at 11:22
  • $\begingroup$ Since the answers to the other question already prove the two groups are not isomorphic, there can be no isomorphism. $\endgroup$
    – egreg
    Commented Feb 25, 2016 at 11:22
  • $\begingroup$ @egreg Yes sir, but at first I was not aware of that.. $\endgroup$
    – Amir Naseri
    Commented Feb 25, 2016 at 11:45
  • $\begingroup$ The answer that you accepted is the same (up to rewording) as the accepted answer of the duplicate thread. How are the two questions not duplicate? $\endgroup$
    – Najib Idrissi
    Commented Feb 25, 2016 at 12:29

2 Answers 2

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HINT: For each $q\in\Bbb Q$ the equation $x+x=q$ has a solution; is it true that $x\cdot x=q$ has a solution for each $q\in\Bbb Q^+$?

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  • $\begingroup$ Thanks, got it! there is simply no isomorphism between them.. $\endgroup$
    – Amir Naseri
    Commented Feb 24, 2016 at 22:32
  • $\begingroup$ @Amir: That’s right. You’re welcome! $\endgroup$
    – Brian M. Scott
    Commented Feb 24, 2016 at 22:33
  • $\begingroup$ But generally, is it true that $\log_{10}(\mathbb Q^+)=\mathbb Q$? $\endgroup$
    – Amir Naseri
    Commented Feb 24, 2016 at 22:36
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    $\begingroup$ @Amir: No, because $\log_{10}2$, for instance, is irrational. $\endgroup$
    – Brian M. Scott
    Commented Feb 24, 2016 at 22:37
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    $\begingroup$ Oh yes, and if it was, then also the above problem had an answer :].. $\endgroup$
    – Amir Naseri
    Commented Feb 24, 2016 at 22:43
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So now you know they're not isomorphic, but you can go a bit further with what you already know. For example, $\mathbb{Q}^+$ has a subgroup whose elements are the powers $2^n$ of $2$, including $1$ and $1/2$, etc. Because the map $n \mapsto 2^n$ from $(\mathbb Z,+)$ is injective, this is a cyclic subgroup. You can do the same thing with the powers $3^n$ and see these two subgroups intersect only in $\{1\}$.

Now, you know every natural number has a unique factorization in terms of primes. Can you use that to say more about $(\mathbb Q^+,\cdot)$?

By comparison, you can check for any two elements $g$ and $h$ of $(\mathbb Q,+)$,there are integers $m$ and $n$ such that $mg = nh$. This means that, in contrast to the case of $(\mathbb Q^+,\cdot)$, any two cyclic subgroups have a nontrivial intersection. One says that $(\mathbb Q^+,\cdot)$ has rank equal to 1, meaning there's a $\mathbb Z$ subgroup, but only one independent such $\mathbb Z$ subgroup, in that all other isomorphic copies of $\mathbb Z$ meet it nontrivially.

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