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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@Lucas Does $Sym(\emptyset)$ refers to the trivial group? Either way I believe the claim that this action is free is false. The cardinality of $\mathbb{Z}$ is strictly greater than that of either the empty set or the singleton (or any finite set), and therefore can not have a free group action on a finite set. (Assume by means of contradiction that it did -- then purely for cardinality reasons there would have to exist (at least) two distinct integers $n_1 \not= n_2$ such that $n_1 \cdot x = n_2 \cdot x$ for a fixed $x$, contradicting freeness. – hasManyStupidQuestions Oct 19 '25 at 00:47
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@hasManyStupidQuestions: I claim the action \mathbb{Z} \rightarrow \text{Sym}(\emptyset) is free because of the following reason: given m, n \in \mathbb{Z}, if there exists x\in \emptyset such that g\cdot x = h \cdot x then g = h. – Lucas Oct 20 '25 at 01:13
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@hasManyStupidQuestions: if there exists x \in \emptyset then I may deduce whatever I want(https://en.wikipedia.org/wiki/Principle_of_explosion) – Lucas Oct 20 '25 at 01:16
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@lucas So you mean $\mathbb{Z}$ is acting on the empty set, so a group homomorphism $\mathbb{Z} \to Sym(\emptyset)$? That's a group homomorphism into the trivial group, which is not injective, so not faithful, and therefore not free (because free implies faithful). Considered as a function $\mathbb{Z} \times \emptyset = \emptyset \to \emptyset$, it is just the empty function. So there are no $m,n \in \mathbb{Z}$ to quantify over, so I guess all properties defined using universal (rather than existential quantifiers) are vacuously satisfied https://en.wikipedia.org/wiki/Vacuous_truth . – hasManyStupidQuestions Oct 20 '25 at 01:49
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@Lucas So I would concede that you are probably right that, depending on one's formulation of "group action", one should technically specify that $X \not= \emptyset$ when defining certain properties (such as freeness or transitiveness) that use universal quantifiers to get the intended definitions. Related: (1) https://qchu.wordpress.com/2010/12/03/empty-sets/ (2) https://math.stackexchange.com/questions/3393427/whether-or-not-a-group-action-on-a-possibly-empty-set-is-transitive-if-for-all-e For almost all group actions however there is no ambiguity that free implies faithful. – hasManyStupidQuestions Oct 20 '25 at 01:51
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@hasManyStupidQuestions: freeness implies faithfullness when the set being acted is non-empty. The only possible counter-example uses the empty set. – Lucas Oct 20 '25 at 02:09
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@Lucas My point is that any correct definition of freeness is one such that the implication is 100%, no counterexamples. 99.9% of people, when formulating a definition with $\forall x \in X P(x)$, intend it such that $\forall x \in X P(x)$ implies $\exists x \in X P(x)$. Technically they should write $\exists x \in X \land \forall x \in X P(x)$ or $X \not= \emptyset \land \forall x \in X P(x)$, but generally no one cares about the empty set, and literally infinitely many sets of infinitely many cardinalities are not the empty set. 99.9% of the time people are working implicitly in the category – hasManyStupidQuestions Oct 22 '25 at 16:33
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of all sets excluding the empty set, and their definitions reflect that. Whether that's sloppiness, lack of rigor, imprecise, etc. is a subjective matter. Regardless, if all of the intended examples for a definition are non-empty, and the only counterexample for an implication involving that definition is the empty set, then the definition was clearly intended to exclude the empty set. You can define a new notion of "freeness" that includes the pathological "free" action on the empty set, but it's not a useful notion and makes things needlessly complicated without creating any new insights. – hasManyStupidQuestions Oct 22 '25 at 16:36
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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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Faithful means $$\forall g \in G ;[\forall x \in X, g x = x \implies g = 1]$$
– Watson Jun 23 '20 at 08:40