Show that $\Bbb Q^+ $ under multiplication is isomorphic to a proper subgroup of itself

Question

Show that $\Bbb Q^+ $ under multiplication is isomorphic to a proper subgroup of itself

Proof:

Let $H= \{{2^p 3^q \; | \; p,q \in Z}\}\leq (subgroup)\;\Bbb Q^+$

define, $f: \Bbb Q^+ \to H$ by

$\frac{p}{q} \mapsto 2^p 3^q \; \forall p, q \in \Bbb N\cup\{0\} \ni \gcd(p,q) = 1$ which is an isomorphism.

Is this correct?

Answer 1

Let $\mathbb{P}$ be the set of primes. By the fundamental theorem of arithmetic every positive rational number $r$ can be uniquely factorized as $$r = \prod_{p\in \mathbb{P}}p^{\alpha(p)}$$ where $\alpha(p)\in \mathbb{Z}$ is zero for almost all $p$. This implies that the group $\mathbb{Q}^+$ under the multiplication is isomorphic to a direct sum $\mathbb{Z}^{\oplus \mathbb{P}}$ of $\mathbb{P}$-many copies of $\mathbb{Z}$ ($\mathbb{Z}$ is a group with respect to integer addition). The isomorphism is given by $$\mathbb{Q}^+\ni r = \prod_{p\in \mathbb{P}}p^{\alpha(p)}\mapsto \left(\alpha(p)\right)_{p\in \mathbb{P}}\in \mathbb{Z}^{\oplus \mathbb{P}}$$ The bottom line is that $\mathbb{Q}^+$ is a free abelian group on countably many generators.

Can you show that a free abelian group on countably many generators is isomorphic with its proper subgroup?

Hint. Shift generators.

Answer 2

I'm afraid your mapping $f:p/q\mapsto 2^p\cdot3^q$ is not a homomorphism of multiplicative groups. After all, that definition tells $$ f(1)=f(1/1)=2\cdot3=6, $$ so $f(1)\neq1$. On the other hand, the related mapping $$g:\Bbb{Z}\times\Bbb{Z}\to \Bbb{Q}^+, (m,n)\mapsto 2^m3^n$$ is a homomorphism from the additive group $\Bbb{Z}\times\Bbb{Z}$ to the multiplicative group $\Bbb{Q}^+$. But that does not help you here.

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Assuming $\Bbb{Q}^+$ stands for positive rationals, we can ignore signs, so the following simple answer suggests itself.

Consider the mapping $f:\Bbb{Q}^+\to\Bbb{Q}^+$ defined by $$ f(q)=q^2. $$ Clearly $f$ is a homomorphism of multiplicative groups. Its kernel is trivial, so the first isomorphism theorem says that the image of $f$ is isomorphic to $\Bbb{Q}^+$. As $2$ is not the square of a rational number, the image is a proper subgroup.