If you find any mistakes, please make a comment! Thank you.

Absolute value is a group homomorphism on the multiplicative real numbers

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.8

Let $\varphi : \mathbb{R}^\times \rightarrow \mathbb{R}^\times$ be given by $x \mapsto |x|$. Prove that $\varphi$ is a homomorphism and find its image. Describe the kernel and fibers of $\varphi$.

Solution: $\varphi$ is a homomorphism since $$\varphi(ab) = |ab| = |a||b| = \varphi(a) \varphi(b);$$ by the definition of $|\cdot|$, the image of $\varphi$ is $\{ x \in \mathbb{R}^\times \ |\ x > 0 \}$. The kernel of $\varphi$ is $\{ 1, -1 \}$, and the fiber of $c$ is $\{ c, -c \}$.


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu