Exercise 3.1.2 (Fibers of $\varphi:G o H$)

Question

Let $\varphi : G 	o H$ be a homomorphism of groups with kernel $K$ and let $a,b \in \varphi(G)$. Let $X \in G/K$ be the fiber above $a$ and let $Y$ be the fiber above $b$, i.e., $X = \varphi^{-1}(a)$, $Y =\varphi^{-1}(b)$. Fix an element $u$ of $X$ (so $\varphi(u) =a$). Prove that if $XY = Z$ in the quotient group $G/K$ and $w$ is any member of $Z$, then there is some $v \in Y$ such that $uv = w$. [Show $u^{-1} w \in Y$].

Richard Ganaye · Accepted Answer

\begin{proof} 
By definition of the product in $G/K$, $Z$ is the fiber above $ab$, i.e., $Z = \varphi^{-1}(ab)$, so $Z = \varphi^{-1}(c)$, where $c =ab$.

Since $w \in Z$,  then  $w \in \varphi^{-1}(c)$, therefore $\varphi(w) =c = ab$ (and $\varphi(u) =a$, since $u \in X = \varphi^{-1}(a)$).

Put $v = u^{-1} w$. Then $uv = w$ and
$$\varphi(u^{-1} w) = \varphi(u)^{-1} \varphi(w) = a^{-1} ab = b.$$
Hence $v = u^{-1} w \in \varphi^{-1} (b) = Y$. So there is some $v \in Y$  such that $uv = w$.

\end{proof}

Exercise 3.1.2 (Fibers of $\varphi:G \to H$)

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Exercise 3.1.2 (Fibers of $\varphi:G \to H$)

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