Hw7 Problem 11

### 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 $G$.