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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.

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    $\begingroup$ @Shaun Thanks for the edit! Does your edit mean that A-D are all correct? $\endgroup$ – user636532 Oct 14 at 16:04
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    $\begingroup$ Not necessarily, no, @SeleneAukland. $\endgroup$ – Shaun Oct 14 at 16:07
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    $\begingroup$ What have you tried? $\endgroup$ – verret Oct 14 at 18:32
  • $\begingroup$ @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 $\endgroup$ – user636532 Oct 16 at 5:30

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