# A group action is regular if and only if free and transitive. I have questions on the non-empty part. [duplicate]

Regular can of course be defined free and transitive. For this post, I choose the definition of regular to be second definition of regular given in Wikipedia.

Let $$X$$ be a set, possibly empty. Let $$G$$ be a group, possibly a singleton. Suppose there exists a right group action $$\mu: M \times G \to M$$.

Unless I misunderstand the meaning of "for every two" or "for each pair", I think the definitions are as follows:

$$\mu$$ is defined regular if for all $$x,y \in M$$, there exists a unique $$g \in G$$ such that $$\mu(x,g)=y$$

$$\mu$$ is defined free if for all $$g \in G$$, if there exists $$x \in M$$ such that $$\mu(x,g)=x$$, then we have that $$g=1_G$$

$$\mu$$ is defined transitive if $$M$$ is non-empty and for all $$x,y \in M$$, there exists a $$g \in G$$ such that $$\mu(x,g)=y$$

1. In proving regular implies free and transitive, how do we prove $$M$$ is non-empty?

2. In proving free and transitive implies regular, where do we use $$M$$ non-empty?

My guess for both: Every $$g \in G$$ satisfies the equation $$\mu(x,g)=y$$ if $$x,y \in M = \emptyset$$.

• With the definitions that you have there, $\mu$ regular does not imply that $M$ is nonempty, because if $M$ is empty then the condition for regular is satisfied vacuously. – Derek Holt Oct 14 at 15:55
• This definition of "regular" does not imply non-empty. (If it were up to me, I'd include "nonempty" (or something that implies it) in the definition of "regular". Specifically, I'd want "regular" to mean isomorphic to the multiplication action of $G$ on itself.) – Andreas Blass Oct 14 at 15:56
• @AndreasBlass Oh thanks. So for (1), Wikipedia is wrong? (assuming I did not misunderstand the meaning of "for every two" or "for each pair") (Well it is Wikipedia, but so far I notice Wikipedia rarely makes mistakes in math) So for (2), $M$ non-empty is not used? – user636532 Oct 14 at 15:57
• The issue in this situation is not whether wikipedia (or any other source) is right or wrong, although keep in mind that wikipedia is written by humans and, well, to err is .... human. Instead, the issue is that definitions can easily vary between different sources. These variations are actually rather common at the "extreme" cases, such as the case where $M$ is empty. This is why some sources will outlaw the empty set (or will outlaw other extreme cases) when specifying the scope of their definition. – Lee Mosher Oct 14 at 16:21
• @SeleneAuckland For the record, I think asking 5 questions (some including sub-questions) in a few hours about the connection between free vs faithful, with an extra focus on the empty set, is not a very good use of this site. It would have been better if you thought a bit about what the question you really wanted to ask was, and ask that. I'm voting to close all of them (except the first) as duplicate. – verret Oct 14 at 18:38

You are right. If $$M=\emptyset$$, the action is regular and free but by definition it can't be transitive. Just impose the condition that $$M\neq\emptyset$$ and everything will work.
• Oh thanks. So for (1), Wikipedia is wrong? (assuming I did not misunderstand the meaning of "for every two" or "for each pair") (Well it is Wikipedia, but so far I notice Wikipedia rarely makes mistakes in math) So for (2), $M$ non-empty is not used? – user636532 Oct 14 at 15:57
• I wouldn't call it a mistake. Empty set is pathological in action theory. Letting a group act on the empty set will never get you any useful information, so you might as well ask that the set on which the groups acts is non-empty in the definition imho. Wikipedia defines regular = transitive + free. If you impose the condition that the set on which the group acts is non-empty, then this is equivalent with what you wrote. For your question (2), you strictly don't need that $M \neq \emptyset$ because if $M= \emptyset$ you get regular and free automatically (vacuously). – EpsilonDelta Oct 14 at 16:07