Can a group act on the empty set?What's the idea of an action of a group?Group Actions of $S_n$ and $O(n)$If groups $G$ and $H$ act on $X$, does $Gtimes H$ act on $X$?Group action with two normal subgroups which induce same block systemWhat can we say about the sets X on which a group G acts freely and transitively?Functions between G-set and H-setGroup action onto power setUnderstanding the definition of the Group action.How group action works.Set of Double Cosets

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Can a group act on the empty set?


What's the idea of an action of a group?Group Actions of $S_n$ and $O(n)$If groups $G$ and $H$ act on $X$, does $Gtimes H$ act on $X$?Group action with two normal subgroups which induce same block systemWhat can we say about the sets X on which a group G acts freely and transitively?Functions between G-set and H-setGroup action onto power setUnderstanding the definition of the Group action.How group action works.Set of Double Cosets













10












$begingroup$


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.










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Though it's kind of empty to have a group action on an empty set, isn't it? =)
    $endgroup$
    – user21820
    2 days ago






  • 3




    $begingroup$
    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
    2 days ago










  • $begingroup$
    @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
    2 days ago










  • $begingroup$
    @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
    2 days ago






  • 1




    $begingroup$
    @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
    2 days ago















10












$begingroup$


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.










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Though it's kind of empty to have a group action on an empty set, isn't it? =)
    $endgroup$
    – user21820
    2 days ago






  • 3




    $begingroup$
    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
    2 days ago










  • $begingroup$
    @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
    2 days ago










  • $begingroup$
    @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
    2 days ago






  • 1




    $begingroup$
    @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
    2 days ago













10












10








10


1



$begingroup$


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.










share|cite|improve this question











$endgroup$




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.







group-theory group-actions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









rabota

14.2k32782




14.2k32782










asked 2 days ago









andrewandrew

999




999







  • 2




    $begingroup$
    Though it's kind of empty to have a group action on an empty set, isn't it? =)
    $endgroup$
    – user21820
    2 days ago






  • 3




    $begingroup$
    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
    2 days ago










  • $begingroup$
    @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
    2 days ago










  • $begingroup$
    @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
    2 days ago






  • 1




    $begingroup$
    @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
    2 days ago












  • 2




    $begingroup$
    Though it's kind of empty to have a group action on an empty set, isn't it? =)
    $endgroup$
    – user21820
    2 days ago






  • 3




    $begingroup$
    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
    2 days ago










  • $begingroup$
    @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
    2 days ago










  • $begingroup$
    @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
    2 days ago






  • 1




    $begingroup$
    @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
    2 days ago







2




2




$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
2 days ago




$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
2 days ago




3




3




$begingroup$
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
2 days ago




$begingroup$
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
2 days ago












$begingroup$
@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
2 days ago




$begingroup$
@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
2 days ago












$begingroup$
@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
2 days ago




$begingroup$
@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
2 days ago




1




1




$begingroup$
@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
2 days ago




$begingroup$
@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
2 days ago










1 Answer
1






active

oldest

votes


















11












$begingroup$

yes you can define the trivial action.



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



That is:



For all $xin emptyset$ we have that $e.x=x$.



For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$



Both statements hold trivially.






share|cite|improve this answer









$endgroup$












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    active

    oldest

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    11












    $begingroup$

    yes you can define the trivial action.



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



    That is:



    For all $xin emptyset$ we have that $e.x=x$.



    For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$



    Both statements hold trivially.






    share|cite|improve this answer









    $endgroup$

















      11












      $begingroup$

      yes you can define the trivial action.



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



      That is:



      For all $xin emptyset$ we have that $e.x=x$.



      For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$



      Both statements hold trivially.






      share|cite|improve this answer









      $endgroup$















        11












        11








        11





        $begingroup$

        yes you can define the trivial action.



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



        That is:



        For all $xin emptyset$ we have that $e.x=x$.



        For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$



        Both statements hold trivially.






        share|cite|improve this answer









        $endgroup$



        yes you can define the trivial action.



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



        That is:



        For all $xin emptyset$ we have that $e.x=x$.



        For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$



        Both statements hold trivially.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        YankoYanko

        7,7001830




        7,7001830



























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