# Proving every free action faithful.

I can see this as clearly true because - assuming the action free - if $g_1 \not=g_2$, then $g_1 \cdot x \not= g_2 \cdot x$ for all $x \in X$ implies it holds true for some $x \in X$, so the actions is faithful.

However, our teacher defined faithful actions interms of the action induced homomorphism $\phi : G \to S(X)$ and called the action faithful whenever $\phi$ is injective.

To prove the map injective, how should I proceed? If I define $\phi(g)=g \cdot x$, then $$g_1 \not= g_2 \implies \phi(g_1)\not= \phi(g_2)$$ proves $\phi$ is injective. I'm not sure if the mapping is defined correctly, however.

Also can someone show via an example how a faithful action isn't always free?

Note that a free action is precisely one with no fixed points, i.e. if $gx=x$ for some $x$, then $g=1$. A faithful action is just one where no $g$ acts trivially. I.e. if $gx=x$ for all $x$, then $g=1$.
Suppose $\phi$ is not injective, then for some $g_1\ne g_2$, $\phi(g_1)=\phi(g_2)$. Take an element $x\in X$. Then $g_1x=\phi(g_1)x=\phi(g_2)x=g_2x$, so since the action is free, $g_1=g_2$. Contradiction. $\phi$ is injective.
The action by $D_3$ on the vertices of a triangle is faithful, but the reflections all fix one vertex, so it is not free. Alternatively $S(X)$ acting on $X$ is clearly faithful, but if $X$ has more than two elements, a transposition will fix the other elements of $X$, so that the action is not free.