I searched Wikipedia for definitions of free and faithful actions. As I understand them, the two concepts are the same thing!
If they are one concept, what is the point of introducing both or even of naming them in distinct ways?
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Free mean that if there is $x \in X$ and $g,h$ with $gx = hx$ then $g = h$. Faithful means that the morphism $G \to Sym(X)$ induced by the action is injective, i.e for all $g\ne h$ there is a $x \in X$ with $gx \neq hx$.
Of course, being free is stronger. It's not equivalent since the action of $\text{SO}(2)$ on $\Bbb R^2$ this is not free since there is a fixed point but it's faithful (take $x = (1,0)$ works for all $g,h \in \text{SO}(2)$).
Kenta S
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3@mja : you're welcome ! Don't hesitate to activate the little green mark next to the answer if this was useful to you :) – Jun 21 '17 at 08:29
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Is "being free" actually stronger than "being faithful"? The action $\mathbb{Z} \rightarrow \text{Sym}(\emptyset)$ is free but not faithful. – Lucas Sep 20 '22 at 21:08
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As Keith Conrad explains on his notes:
"It is worth comparing faithful and free actions. An action is faithful when ${g_1 \neq g_2 \Rightarrow g_1 x \neq g_2 x}$ for some ${x \in X}$ (different elements of $G$ act differently at some point) while an action is free when ${g_1 \neq g_2 \rightarrow g_1 x \neq g_2 x}$ for all ${x \in X}$ (different elements of $G$ act differently at every point). Since ${g_1 x = g_2 x}$ if and only if $g_2^{-1} g_1 x = x$ we can describe faithful and free actions in terms of fixed points: an action is faithful when each $g \neq e$ has Fix${_g}(X) \neq X$ while an action is free when each $g \neq e$ has Fix${_g}(X) = \emptyset$."
C.F.G
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Alberto Centelles
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Faithful means $$\forall g \in G ;[\forall x \in X, g x = x \implies g = 1]$$
– Watson Jun 23 '20 at 08:40