# Some element has trivial stabilizer or isotropy subgroup, i.e. $G$ and orbit $xG$ bijectively correspond. What's the term for this?

Let $$M$$ be a set. Let $$G$$ be a group. For convenience, let $$M$$ be non-empty and $$G$$ be not trivial. Let $$G$$ act right on $$M$$ by the action $$\mu: M \times G \to G$$. For each $$x \in M$$, let $$\text{Stab}(x):=\{g \in G | \mu(x,g) = x\}$$ denote a stabilizer subgroup of $$G$$. Let $$1_G$$ be the identity of $$G$$.

Consider the following conditions

• Condition 1: $$\mu$$ is free.

• Condition 2: For all $$x \in M$$, $$\text{Stab}(x)$$ is trivial, i.e. $$\text{Stab}(x) = \{1_G\}$$

• Condition 2.1: $$\bigcup_{x \in M} \text{Stab}(x)$$ is trivial, i.e. $$\bigcup_{x \in M} \text{Stab}(x) = \{1_G\}$$
• Condition 3: There exists $$x \in M$$ such that $$\text{Stab}(x)$$ is trivial, i.e. $$\text{Stab}(x) = \{1_G\}$$

• Condition 4: $$\mu$$ is faithful.

• Condition 5: $$\bigcap_{x \in M} \text{Stab}(x)$$ is trivial, i.e. $$\bigcap_{x \in M} \text{Stab}(x) = \{1_G\}$$

I understand that

• A. Conditions 1 and 2 (and I think 2.1) are equivalent (or identical, depending on your definitions),

• B. Conditions 4 and 5 are equivalent (or identical, depending on your definitions),

• C. Condition 2 implies Condition 3 (I think this is obvious) but not conversely , and

• D. Condition 3 implies Condition 5 (I think this is obvious) but not conversely .

Question: If one of A,B,C or D is wrong, then which, and why? If they are all correct, then what's the term, if any, for Condition 3?

• I was thinking something like $$\mu$$ is "free at element $$x$$" and then we can say an action $$\mu$$ in general is "free" if $$\mu$$ is "free at element $$x$$" for every $$x$$" (Again, this is under the assumption $$M$$ is non-empty)

Edit: I realized something.

Let $$x \in M$$. Observe that we can define a map $$\mu_x: G \to M, \mu_x(g) := \mu(x,g)$$. We have that image of $$\mu_x$$ is equal to $$\text{Orbit}(x)$$ $$:= xG :=$$ $$\{\mu(x,g) | g \in G\}$$. Then, we can define a surjective map $$\tilde{\mu_x}: G \to \text{Orbit}(x),$$ $$\tilde{\mu_x}(g) :=$$ $$\mu_x(g)$$ $$:= \mu(x,g)$$ , which is simply $$\mu_x$$ with range restricted to image.

I think the orbit-stabilizer theorem somehow implies that:

$$\text{Stab}(x)$$ is trivial if and only if $$\mu_x$$ is injective if and only if $$\tilde{\mu_x}$$ is bijective if and only if $$G$$ and $$\text{Orbit}(x)$$ are in bijective correspondence.

• @Shaun Thanks for the edit! Does your edit mean that A-D are all correct? – user636532 Oct 14 '19 at 16:04
• Not necessarily, no, @SeleneAukland. – Shaun Oct 14 '19 at 16:07
• What have you tried? – verret Oct 14 '19 at 18:32
• @verret Made a mistake. Edited post to make C and D obvious, so there's nothing to try (except possibly for the "but not conversely" I guess). Thanks – user636532 Oct 16 '19 at 5:30