There is a definition from Group Theory about isomorphisms and homomorphisms. For example, if $G$ and $H$ are groups and $\varphi: G \rightarrow H$ is a homomorphism, it must satisfy that $\forall x,y \in G$, then $\varphi (xy) = \varphi(x) \varphi(y)$. Isomorphism is just a bijection and a homomorphism.
Now, when talking about normal subgroups, it is defined as follows:
Suppose $H$ is a subgroup of $G$. Then $H$ is a normal subgroup of $G$ iff for all $g \in G$, $gHg^{-1} = H$, and it is denoted as $H \trianglelefteq G$.
I got this definition from Normal Subgroup (paraphrased).
Now, my question is: Is there any construction of a homomorphism $𝜑: \mathbb{Q}^{\times}\rightarrow \mathbb{Q}^{+}$? Can you give an example?
I already proved that $\langle -1 \rangle = \{1, -1\}$ is a normal subgroup of $\mathbb{Q}^{\times}$ (group of rational numbers under multiplication). Also, I know there are many methods for proving these things, such as the Fundamental Homomorphism Theorem. Normally, to demonstrate that $G/H \cong K$, prove that $\varphi: G \rightarrow K$ is a homomorphism, it is surjective, and that the $\ker(\varphi) = H$, for some group $H$ (this is the one I want to use). However, I've been trying to work on this, and I couldn't construct any homomorphism between $\mathbb{Q}^{\times}$ and $\mathbb{Q}^{+}$. I have some knowledge about this topic, but not so much. Any hints will help.
Edit (11/6/2021): Making my question clearer
To make the question clear, I want to prove the isomorphism of $$\varphi: \mathbb{Q}^{\times} \rightarrow \mathbb{Q}^{+}$$
I needed a closed formula for homomorphism of $\varphi$, which was complicated to find, so I needed some help with this.
Edit 2 (11/6/2021): Finding my mistake
I finally found my mistake! I thought that $\mathbb{Q}^{+}$ and $(\mathbb{Q}, +)$ were the same group under the same operation! Thanks to Thomas Andrews, Arturo Margidin, and Greg Martin for helping out on this problem!
Best Answer
Assume $\phi:\mathbb Q^\times\to (\mathbb Q,+)$ is a homomorphism with kernel $\{1,-1\},$ then let $\frac pq=\phi(2)$ and $\frac rs=\phi(3),$ with $p,q,r,s\in \mathbb Z, q,s\neq 0.$
If $p=0$ or $r=0,$ then either $2$ or $3$ is in the kernel, so $\ker \phi\neq \{ 1,-1\}.$
So $p,q,r,s\neq 0.$
Let $x=2^{qr}3^{-sp}.$ Then $$\phi(x)=qr\cdot\frac pq+(-sp)\cdot\frac rs=0,$$ So $x\in\ker\phi.$
Show $2^{qr}3^{-sp}\neq \pm 1$ when $p,q,r,s$ are non-zero integers.