Throwing out an element of a field












2












$begingroup$


I was reading my book on Elementary Algebra and saw this theorem:




Suppose that $F$ is a finite field of order $q$, then the group $F^*$ is a cyclic group of order $q-1$.




I don't understand the transition from a field to a group. How come the theorem says $F^*$ is a group when the element $0$ is omitted?






abstract-algebra discrete-mathematics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Please clarify: are you asking why it is a (cyclic) group, or why $0$ had to be omitted?
    $endgroup$
    – Bill Dubuque
    46 mins ago


















2












$begingroup$


I was reading my book on Elementary Algebra and saw this theorem:




Suppose that $F$ is a finite field of order $q$, then the group $F^*$ is a cyclic group of order $q-1$.




I don't understand the transition from a field to a group. How come the theorem says $F^*$ is a group when the element $0$ is omitted?






abstract-algebra discrete-mathematics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Please clarify: are you asking why it is a (cyclic) group, or why $0$ had to be omitted?
    $endgroup$
    – Bill Dubuque
    46 mins ago
















2












2








2





$begingroup$


I was reading my book on Elementary Algebra and saw this theorem:




Suppose that $F$ is a finite field of order $q$, then the group $F^*$ is a cyclic group of order $q-1$.




I don't understand the transition from a field to a group. How come the theorem says $F^*$ is a group when the element $0$ is omitted?






abstract-algebra discrete-mathematics







share|cite|improve this question









$endgroup$




I was reading my book on Elementary Algebra and saw this theorem:




Suppose that $F$ is a finite field of order $q$, then the group $F^*$ is a cyclic group of order $q-1$.




I don't understand the transition from a field to a group. How come the theorem says $F^*$ is a group when the element $0$ is omitted?






abstract-algebra discrete-mathematics




abstract-algebra discrete-mathematics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 56 mins ago









ZacharyZachary

688




688












  • $begingroup$
    Please clarify: are you asking why it is a (cyclic) group, or why $0$ had to be omitted?
    $endgroup$
    – Bill Dubuque
    46 mins ago




















  • $begingroup$
    Please clarify: are you asking why it is a (cyclic) group, or why $0$ had to be omitted?
    $endgroup$
    – Bill Dubuque
    46 mins ago


















$begingroup$
Please clarify: are you asking why it is a (cyclic) group, or why $0$ had to be omitted?
$endgroup$
– Bill Dubuque
46 mins ago






$begingroup$
Please clarify: are you asking why it is a (cyclic) group, or why $0$ had to be omitted?
$endgroup$
– Bill Dubuque
46 mins ago












4 Answers
4






active

oldest

votes


















2












$begingroup$

Well, $mathbb{F}^*$ is the group $(mathbb{F}setminus {0},cdot)$ hence you are only considering multiplication as operation, in particular, you need to throw out $0$. In more general algebra (Rings) this still holds, but there you throw out every element which is not invertible. hence you can imagine $mathbb{F}^*$ as the multiplicative group of invertible elements.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If $F^{*}$ were to be a field, it would need an additive identity. But this would have to be the additive identity of the original field $F$, which as you pointed out, is omitted from $F^{*}$ since it isn't a unit. So $F^{*}$ is just a group.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      $F^ast$ is a set with a binary operation (the multiplication operation on $F$) which is associative (since it's associative on all of $F$) and in which all elements have inverses.



      That's a group.



      It wasn't a group before when you had all of $F$ since $0$ did not have an inverse. $F$ was a monoid though, with its multiplication operation.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Every field is a group if you remove the $0$ element and retstrict it to just the multiplication operation because:



        1) The multiplication operation is associative



        2) There is an element $1$ where $1*q = q*1 =q$ for all $qin Fsetminus {0} subset F$.



        3) For every $q in F; qne 0$ there is a $q^{-1} in F$ so that $q*q^{-1} = q^{-1}*q = 1$. It takes a minor prove to show that $q^{-1} ne 0$ (because $q*0 = 0ne 1$) but that means $q^{-1} in Fsetminus {0}$.



        that is the very definition of "group".



        What the theorem is saying is that if $F$ is finite then the group is cyclic. Which is not so trivial.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

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          4 Answers
          4






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Well, $mathbb{F}^*$ is the group $(mathbb{F}setminus {0},cdot)$ hence you are only considering multiplication as operation, in particular, you need to throw out $0$. In more general algebra (Rings) this still holds, but there you throw out every element which is not invertible. hence you can imagine $mathbb{F}^*$ as the multiplicative group of invertible elements.






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            Well, $mathbb{F}^*$ is the group $(mathbb{F}setminus {0},cdot)$ hence you are only considering multiplication as operation, in particular, you need to throw out $0$. In more general algebra (Rings) this still holds, but there you throw out every element which is not invertible. hence you can imagine $mathbb{F}^*$ as the multiplicative group of invertible elements.






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              Well, $mathbb{F}^*$ is the group $(mathbb{F}setminus {0},cdot)$ hence you are only considering multiplication as operation, in particular, you need to throw out $0$. In more general algebra (Rings) this still holds, but there you throw out every element which is not invertible. hence you can imagine $mathbb{F}^*$ as the multiplicative group of invertible elements.






              share|cite|improve this answer









              $endgroup$



              Well, $mathbb{F}^*$ is the group $(mathbb{F}setminus {0},cdot)$ hence you are only considering multiplication as operation, in particular, you need to throw out $0$. In more general algebra (Rings) this still holds, but there you throw out every element which is not invertible. hence you can imagine $mathbb{F}^*$ as the multiplicative group of invertible elements.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 47 mins ago









              EnkiduEnkidu

              1,30119




              1,30119























                  1












                  $begingroup$

                  If $F^{*}$ were to be a field, it would need an additive identity. But this would have to be the additive identity of the original field $F$, which as you pointed out, is omitted from $F^{*}$ since it isn't a unit. So $F^{*}$ is just a group.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    If $F^{*}$ were to be a field, it would need an additive identity. But this would have to be the additive identity of the original field $F$, which as you pointed out, is omitted from $F^{*}$ since it isn't a unit. So $F^{*}$ is just a group.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      If $F^{*}$ were to be a field, it would need an additive identity. But this would have to be the additive identity of the original field $F$, which as you pointed out, is omitted from $F^{*}$ since it isn't a unit. So $F^{*}$ is just a group.






                      share|cite|improve this answer









                      $endgroup$



                      If $F^{*}$ were to be a field, it would need an additive identity. But this would have to be the additive identity of the original field $F$, which as you pointed out, is omitted from $F^{*}$ since it isn't a unit. So $F^{*}$ is just a group.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 52 mins ago









                      pwerthpwerth

                      2,730414




                      2,730414























                          1












                          $begingroup$

                          $F^ast$ is a set with a binary operation (the multiplication operation on $F$) which is associative (since it's associative on all of $F$) and in which all elements have inverses.



                          That's a group.



                          It wasn't a group before when you had all of $F$ since $0$ did not have an inverse. $F$ was a monoid though, with its multiplication operation.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            $F^ast$ is a set with a binary operation (the multiplication operation on $F$) which is associative (since it's associative on all of $F$) and in which all elements have inverses.



                            That's a group.



                            It wasn't a group before when you had all of $F$ since $0$ did not have an inverse. $F$ was a monoid though, with its multiplication operation.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              $F^ast$ is a set with a binary operation (the multiplication operation on $F$) which is associative (since it's associative on all of $F$) and in which all elements have inverses.



                              That's a group.



                              It wasn't a group before when you had all of $F$ since $0$ did not have an inverse. $F$ was a monoid though, with its multiplication operation.






                              share|cite|improve this answer









                              $endgroup$



                              $F^ast$ is a set with a binary operation (the multiplication operation on $F$) which is associative (since it's associative on all of $F$) and in which all elements have inverses.



                              That's a group.



                              It wasn't a group before when you had all of $F$ since $0$ did not have an inverse. $F$ was a monoid though, with its multiplication operation.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 47 mins ago









                              rschwiebrschwieb

                              105k12101246




                              105k12101246























                                  0












                                  $begingroup$

                                  Every field is a group if you remove the $0$ element and retstrict it to just the multiplication operation because:



                                  1) The multiplication operation is associative



                                  2) There is an element $1$ where $1*q = q*1 =q$ for all $qin Fsetminus {0} subset F$.



                                  3) For every $q in F; qne 0$ there is a $q^{-1} in F$ so that $q*q^{-1} = q^{-1}*q = 1$. It takes a minor prove to show that $q^{-1} ne 0$ (because $q*0 = 0ne 1$) but that means $q^{-1} in Fsetminus {0}$.



                                  that is the very definition of "group".



                                  What the theorem is saying is that if $F$ is finite then the group is cyclic. Which is not so trivial.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Every field is a group if you remove the $0$ element and retstrict it to just the multiplication operation because:



                                    1) The multiplication operation is associative



                                    2) There is an element $1$ where $1*q = q*1 =q$ for all $qin Fsetminus {0} subset F$.



                                    3) For every $q in F; qne 0$ there is a $q^{-1} in F$ so that $q*q^{-1} = q^{-1}*q = 1$. It takes a minor prove to show that $q^{-1} ne 0$ (because $q*0 = 0ne 1$) but that means $q^{-1} in Fsetminus {0}$.



                                    that is the very definition of "group".



                                    What the theorem is saying is that if $F$ is finite then the group is cyclic. Which is not so trivial.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Every field is a group if you remove the $0$ element and retstrict it to just the multiplication operation because:



                                      1) The multiplication operation is associative



                                      2) There is an element $1$ where $1*q = q*1 =q$ for all $qin Fsetminus {0} subset F$.



                                      3) For every $q in F; qne 0$ there is a $q^{-1} in F$ so that $q*q^{-1} = q^{-1}*q = 1$. It takes a minor prove to show that $q^{-1} ne 0$ (because $q*0 = 0ne 1$) but that means $q^{-1} in Fsetminus {0}$.



                                      that is the very definition of "group".



                                      What the theorem is saying is that if $F$ is finite then the group is cyclic. Which is not so trivial.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Every field is a group if you remove the $0$ element and retstrict it to just the multiplication operation because:



                                      1) The multiplication operation is associative



                                      2) There is an element $1$ where $1*q = q*1 =q$ for all $qin Fsetminus {0} subset F$.



                                      3) For every $q in F; qne 0$ there is a $q^{-1} in F$ so that $q*q^{-1} = q^{-1}*q = 1$. It takes a minor prove to show that $q^{-1} ne 0$ (because $q*0 = 0ne 1$) but that means $q^{-1} in Fsetminus {0}$.



                                      that is the very definition of "group".



                                      What the theorem is saying is that if $F$ is finite then the group is cyclic. Which is not so trivial.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 45 mins ago









                                      fleabloodfleablood

                                      69.2k22685




                                      69.2k22685






























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