# Faithful Group Actions

## Homework Statement

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.

## Homework Equations

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?

## Homework Statement

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.

## Homework Equations

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?

The definition of a faithful right group action on itself is

For any two distinct g_1 and g_2 in G (g_1 \neq g_2), there exists h in G such that h^{g_1} \neq h^{g_2}.

The contrapositive of the above is

If h^{g_1} = h^{g_2} for all h in G, then g_1 = g_2.

Your professor probably used this contrapositive method.