- #1
gottfried
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Homework Statement
Let G be a group and define λg:G→G to be λg(x)=g.x , x[itex]\in[/itex]G.
Show that λg 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?