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

A fact about preimages under group homomorphisms

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.2

Let $\varphi : G \rightarrow H$ be a group homomorphism with kernel $K$ and let $a,b \in \varphi[G]$. Let $X = \varphi^*(a)$ and $Y = \varphi^*(b)$. Fix $u \in X$. Prove that if $XY = Z$ in $G/K$ and $w \in Z$, then there exists $v \in Y$ such that $uv = w$. [Hint: Show that $u^{-1}v \in Y$.]

Solution: We saw in Exercise 3.1.1 that $K \trianglelefteq G$, so that $G/K$ is a group. Thus $Y = X^{-1}Z$ in $G/K$, where $$X^{-1} = \{ x^{-1} \ |\ x \in X \}.$$ In particular, we have $u^{-1}w \in Y$. So there exists an element $v \in Y$ with $u^{-1}w = v$, so that $uv = w$.


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.
Close Menu