Proving a mapping is a group action












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Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.



My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.



For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.



Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?






group-theory finite-groups group-actions







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    4












    $begingroup$


    Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.



    My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.



    For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.



    Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?






    group-theory finite-groups group-actions







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.



      My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.



      For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.



      Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?






      group-theory finite-groups group-actions







      share|cite|improve this question









      $endgroup$




      Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.



      My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.



      For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.



      Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?






      group-theory finite-groups group-actions




      group-theory finite-groups group-actions






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked 4 hours ago









      cb7cb7

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          $begingroup$

          Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.






          share|cite|improve this answer









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            $begingroup$

            Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.






                share|cite|improve this answer









                $endgroup$



                Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                MarkMark

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