Hw7 Problem 11
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Problem 11
Let $G$ be a group and let $a\in G$ be a fixed element of $G$. Show that the map $\lambda_{a}: G\rightarrow G$, given by $\lambda_{a}(g)=ag$ for all $g\in G$, is a permutation of the set $G$.
Solution
A permutation on a set $G$ is a function $\phi : G \rightarrow G$ that is 1-1 and onto.
Let $g,h \in G$. Suppose $\lambda_{a}(g)=\lambda_{a}(h)$. Now, $ag=ah$ and by cancellation, $g=h$. Hence, $\phi$ is 1-1.
Let $h \in G$. Note $\lambda_{a}(a^{-1}h)=h$. Now,
(1)\begin{align} \lambda_{a}(a^{-1}h)=a(a^{-1}h)=(aa^{-1})h=eh=h. \end{align}
We know $G$ is a group, therefore $a^{-1} \in G$ and the function $\phi$ is onto.
Therefore, the map $\lambda_{a}:G \rightarrow G$ is a permutation on $G$.