Is the Function λg:G→G Both Onto and One-to-One?

[gottfried]

Homework Statement

Let G be a group and define λ_g:G→G to be λ_g(x)=g.x , x$\in$G.

Show that λ_g is onto and one-to-one.

The Attempt at a Solution

Suppose g.x=g.x' g^-1.g.x=g^-1.g.x' which means x=x'.

How should I show that the function is onto?

 

[HallsofIvy]

If y is any member of G, then $\lambda(g^{-1}y)= y$