Hi I am working on following hw problem and I want to make sure that I am doing this correctly? I think I am going about this in the right way but I still need some reassurance.
Let $G$ be a group and for a fixed element $a \in G$ define a map $\lambda_a:G \rightarrow G$ by $\lambda_a(x) = ax.$
To prove that $\lambda_a$ is a permutation of $G$ we must show that $\lambda_a$ is a bijective mapping from $G \rightarrow G$.
To show that $\lambda_a$ is injective let $x_1,x_2 \in G$ and let $a \in G$ be fixed. Suppose that $\lambda_a(x_1) = \lambda_a(x_2)$ since $a \in G$ and since $G$ is a group then $a^{-1} \in G$. Then $ax_1 = ax_2$ and,
$$ x_1 = a^{-1}ax_1 = a^{-1}ax_2 = x_2$$
Therefore $\lambda_a$ is injective. Next to show that $\lambda_a$ is surjective let $x \in G$, then $\lambda_a$ is surjective if for all $x \in G$, there exists an $x \in G$ such that $\lambda_a(x) = ax$. Thus $\lambda_a$ is both injective and surjective and is therefore a bijective mapping from $G \rightarrow G$.
Best Answer
Injectivity is fine, but you just state the definition of surjectivity without proving anything. To show that $\lambda_a$ is surjective, you should fix an $x \in G$ and find an element $g \in G$ such that $\lambda_a(g) = x$; manipulating the definition, this strongly suggests that you look at
$$g = a^{-1} x$$