show for any group G, the function G x G --> G defined by (h)^g = hg is a faithful right action of the group G on itself. In this case, G is said to act on itself by right multiplication.
Definition of faithful: a right action of a group g on a set X is said to be faithful if the only element of G that acts trivially on the entire set is the identity element. So an action is faithful if x^g = x for all x in X i.e g = e, the identity element
The Attempt at a Solution
My professor did this in class and I am a bit confused.
Suppose (h)^g1 = (h)^g2 for all h in G. Then by definition hg1 = hg2.
Then we left multiply by h^(-1) and we have g1=g2
I do not understand how this shows faithful right action?
Can someone explain?