Let $\Bbb R^*$ be the group of real numbers under multiplication (number $0$ is not in $ \Bbb R^*$) and $\Bbb R^+$ be the group of positive real numbers under multiplication. Define $f:\Bbb R^*\to\Bbb R^+: x\mapsto\lvert x\rvert$.
Is $f$ a group homomorphism?
For this, we need to show that $f(ab)=f(a)\,f(b)$.
$$\lvert xy\rvert=\lvert x\rvert\lvert y\rvert$$
This would lead to both $x$ and $y$ being positive numbers. Therefore they will be equal under multiplication. Therefore $f$ is a group homomorphism.
Find the kernel of $f$.
What can you conclude in this question by using the fundamental homomorphism theorem?
For $x, y\in(\mathbb R^*, \cdot)$, by definition of absolute value function, $f(x\cdot y)=|x\cdot y|=|x|\cdot|y|$ , hence $f$ is a homomorphism.
$\begin{align}\text{kernel }f &=\{x\in\mathbb R\colon f(x)=1\}\\&=\{x\in\mathbb R\colon |x|=1\}\\&=\{x=\pm 1\}\end{align}$
$f$ is surjective because for any positive real $x$, it is the absolute value of exactly two (nonzero) real numbers : $\pm x$.
Fundamental homomorphism theorem ensures $(\mathbb R^*, \cdot)/\{\pm 1\}\cong (\mathbb R^+, \cdot)$