Can a group act on the empty set?












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.






group-theory group-actions







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$endgroup$








  • 2




    $begingroup$
    Though it's kind of empty to have a group action on an empty set, isn't it? =)
    $endgroup$
    – user21820
    Mar 10 at 11:30






  • 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
    Mar 10 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 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 at 14:10






  • 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
    Mar 10 at 14:16
















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.






group-theory group-actions







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
    Mar 10 at 11:30






  • 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
    Mar 10 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 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 at 14:10






  • 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
    Mar 10 at 14:16














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.






group-theory group-actions







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




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 Mar 10 at 10:50









rabota

14.2k32782




14.2k32782










asked Mar 10 at 10:43









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
    Mar 10 at 11:30






  • 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
    Mar 10 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 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 at 14:10






  • 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
    Mar 10 at 14:16














  • 2




    $begingroup$
    Though it's kind of empty to have a group action on an empty set, isn't it? =)
    $endgroup$
    – user21820
    Mar 10 at 11:30






  • 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
    Mar 10 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 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 at 14:10






  • 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
    Mar 10 at 14:16








2




2




$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
Mar 10 at 11:30




$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
Mar 10 at 11:30




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
Mar 10 at 11:57




$begingroup$
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 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 at 13:41




$begingroup$
@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 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 at 14:10




$begingroup$
@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 at 14:10




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
Mar 10 at 14:16




$begingroup$
@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 at 14:16










1 Answer
1






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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









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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 Mar 10 at 10:46









        YankoYanko

        7,8801830




        7,8801830






























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