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Are the groups $(\Bbb Q^+,*)$ where $\Bbb Q^+$ denotes set of positive rationals isomorphic to $(A,*)$ where $A=\{\frac{p}{q}\in \Bbb Q^+:p,q\text{are odd}\}$ ?

I tried $f:(\Bbb Q^+,*)\to (A,*) $ by $f(\frac{a}{b})=\frac{2a+1}{2b+1}$ but alas it's not even a homomorphism.

How should I do it?

1 Answers 1

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Yes.

Map $$\prod_{k=1}^\infty p_k^{m_k}\mapsto \prod_{k=1}^\infty p_{k+1}^{m_k}$$ where the $p_k$ are the primes (and $p_1=2$).

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    I am not getting this at all.If $\frac{a}{b}$ is any element of $\Bbb Q^+$ how are you mapping this to $A$ – 2017-03-01
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    Are you taking $a=p_1^{m_1}p_2^{m_2}\dots p_k^{m_k}$ – 2017-03-01
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    Why is the product in your answer infinite then? – 2017-03-01
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    @Mathematics : Note. $p_{1},p_{2},\ldots$ are the enumeration of the prime numbers. Any homomorphism on $\mathbb{Q}$ is determined by the value it takes at prime numbers since $f$ is multiplicative. So assume $p_{1}=2,p_{2}=3,p_{3}=5$ and so on. Take a say $16 \in \mathbb{Q}^{+}$. So $2^{4} \mapsto 3^{4}$ under the above map which Hagen has mentioned. Once you know the value of homomorphism at integers you can easily figure out what value it will take at rationals. – 2017-04-07