There isn't much more to add to this question. Can we define an action between some group and the null set?

I would have thought that there being no elements to act on it trivially satisfies the requirements for something to be an action but I'm not sure.

  • 2
    $\begingroup$ Though it's kind of empty to have a group action on an empty set, isn't it? =) $\endgroup$
    – user21820
    Mar 10, 2019 at 11:30
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    $\begingroup$ In particular, the symmetric group $S_0$, which has order $1$, acts naturally on the empty set. There is unique bijection between the empty set and itself. $\endgroup$
    – Derek Holt
    Mar 10, 2019 at 11:57
  • $\begingroup$ @user21820 the interest of a mathematical formalism is to avoid such philosophical considerations. In the same spirit, there were mathematicians fighting against the existence of infinite sets in the late XIX... $\endgroup$
    – YCor
    Mar 10, 2019 at 13:41
  • $\begingroup$ @YCor: Erm... I was just joking in my first comment, but I disagree with your comment, because anyone who claims they use ZFC as their foundational system necessarily has made some very weird philosophical assumptions whether or not they know it. $\endgroup$
    – user21820
    Mar 10, 2019 at 14:10
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    $\begingroup$ @YCor: But that's only if you think "truth within set theory" is meaningful. To refrain from prolonging this thread with our off-topic discussion, do you want to come to the logic chat-room? $\endgroup$
    – user21820
    Mar 10, 2019 at 14:16

1 Answer 1


yes you can define the trivial action.

Note that the axioms for group action begins with "for all"

That is:

For all $x\in \emptyset$ we have that $e.x=x$.

For all $x\in\emptyset$ and all $g,h\in G$ we have $(gh)x=g.(h.x)$

Both statements hold trivially.


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