Prove that $\mathbb{Q}^*/\langle-1\rangle$ is isomorphic to $\mathbb{Q}$

Question

Question is from An Inquiry-Based approach to abstract algebra. Problem $7.27$: Use the First Isomorphism Theorem to prove that $\mathbb{Q}^*/\langle-1\rangle$ is isomorphic to $\mathbb{Q}$.

I need help solving this question, I can't think of a solution. Below are some of my ideas, but I seem to have come up with an erroneous contradiction.

To apply the first isomorphism theorem to get the desired result, I need to find a surjective homomorphism $f$ from $\mathbb{Q}^*$ to $\mathbb{Q}$ st $\ker(f)=\langle-1\rangle$.

1. For any rational $r \in \mathbb{Q^*}$, $f(r)=f(-r)$ since they belong in the same coset $r\langle-1\rangle$.
2. In particular $f(1)=f(-1)=1$ since $\ker(f)=\langle-1\rangle$
3. Since $f$ is surjective, there exists $x \in \mathbb{Q^*}$ st $f(x)=-1$.
4. Since $f(x^{-1})=(-1)^{-1}$, $f(x^{-1})=1$(additive inverse). But then this means that $x^{-1}=1$ or $x^{-1}=1$ which further imply that $x \in \ker(f)$ but this is a contradiction.

Some extra information: The question also hints to use Problem $6.49$ which is "describe the quotient group $Q^*/\langle-1\rangle$." My ans to this one was These cosets look like distinct equivalence classes of the positive rationals. I.E. $\langle-1\rangle$ corresponds to $[1]$, $\frac{p}{q}\langle-1\rangle$ corresponds to $[|\frac{p}{q}|]$. which might also be wrong. I also thought that maybe $7.27$ was erroneous and meant $\mathbb{Q}^+$ instead of $\mathbb{Q}$ but that cannot be because "$\mathbb{Q}^+$ does not even form a group."(Edit: nvm this is not true)

Answer 1

The group $(\mathbb{Q}^*,\cdot)/\langle-1\rangle$ is canonically isomorphic to $(\mathbb{Q}^+,\cdot)$, hence not to $(\Bbb Q,+)$ because the latter is divisible, or simply because $\forall r\in\Bbb Q,\frac r2\in\Bbb Q$, whereas for instance $2$ has no square root in $\Bbb Q^+$.

Answer 2

The map $f: \Bbb{Q}^\ast \to \Bbb{Q}^+: q\mapsto |q|$ is a surjective group morphism (note that on both groups, multiplication is the group operation!), with kernel equal to $\langle -1 \rangle$.