Why does this cyclic subgroup have only 4 subgroups?












2












$begingroup$


Let the cyclic group have 6 elements and be denoted as $G = {1, a, a^2, a^3, a^4, a^5}$ where $a^6 = 1$.



Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, ${1, a^2, a^4}$ and ${1, a^3}$.



Why isnt ${1, a^5}$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?




There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.



If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^{-5}$ or $a^{10}$, where it is explicitly stated that $a^{10} = 1$ as well.



Is my thought process correct?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
    $endgroup$
    – Minus One-Twelfth
    Apr 3 at 19:58












  • $begingroup$
    why? Could you help me understand how you got to that conclusion?
    $endgroup$
    – Evan Kim
    Apr 3 at 20:00






  • 2




    $begingroup$
    ${1,a^5}$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^{10}=a^4$
    $endgroup$
    – J. W. Tanner
    Apr 3 at 20:00












  • $begingroup$
    The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
    $endgroup$
    – Minus One-Twelfth
    Apr 3 at 21:44
















2












$begingroup$


Let the cyclic group have 6 elements and be denoted as $G = {1, a, a^2, a^3, a^4, a^5}$ where $a^6 = 1$.



Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, ${1, a^2, a^4}$ and ${1, a^3}$.



Why isnt ${1, a^5}$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?




There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.



If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^{-5}$ or $a^{10}$, where it is explicitly stated that $a^{10} = 1$ as well.



Is my thought process correct?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
    $endgroup$
    – Minus One-Twelfth
    Apr 3 at 19:58












  • $begingroup$
    why? Could you help me understand how you got to that conclusion?
    $endgroup$
    – Evan Kim
    Apr 3 at 20:00






  • 2




    $begingroup$
    ${1,a^5}$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^{10}=a^4$
    $endgroup$
    – J. W. Tanner
    Apr 3 at 20:00












  • $begingroup$
    The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
    $endgroup$
    – Minus One-Twelfth
    Apr 3 at 21:44














2












2








2





$begingroup$


Let the cyclic group have 6 elements and be denoted as $G = {1, a, a^2, a^3, a^4, a^5}$ where $a^6 = 1$.



Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, ${1, a^2, a^4}$ and ${1, a^3}$.



Why isnt ${1, a^5}$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?




There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.



If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^{-5}$ or $a^{10}$, where it is explicitly stated that $a^{10} = 1$ as well.



Is my thought process correct?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




Let the cyclic group have 6 elements and be denoted as $G = {1, a, a^2, a^3, a^4, a^5}$ where $a^6 = 1$.



Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, ${1, a^2, a^4}$ and ${1, a^3}$.



Why isnt ${1, a^5}$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?




There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.



If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^{-5}$ or $a^{10}$, where it is explicitly stated that $a^{10} = 1$ as well.



Is my thought process correct?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 3 at 20:02









J. W. Tanner

4,7471420




4,7471420










asked Apr 3 at 19:57









Evan KimEvan Kim

69519




69519












  • $begingroup$
    The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
    $endgroup$
    – Minus One-Twelfth
    Apr 3 at 19:58












  • $begingroup$
    why? Could you help me understand how you got to that conclusion?
    $endgroup$
    – Evan Kim
    Apr 3 at 20:00






  • 2




    $begingroup$
    ${1,a^5}$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^{10}=a^4$
    $endgroup$
    – J. W. Tanner
    Apr 3 at 20:00












  • $begingroup$
    The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
    $endgroup$
    – Minus One-Twelfth
    Apr 3 at 21:44


















  • $begingroup$
    The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
    $endgroup$
    – Minus One-Twelfth
    Apr 3 at 19:58












  • $begingroup$
    why? Could you help me understand how you got to that conclusion?
    $endgroup$
    – Evan Kim
    Apr 3 at 20:00






  • 2




    $begingroup$
    ${1,a^5}$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^{10}=a^4$
    $endgroup$
    – J. W. Tanner
    Apr 3 at 20:00












  • $begingroup$
    The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
    $endgroup$
    – Minus One-Twelfth
    Apr 3 at 21:44
















$begingroup$
The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
$endgroup$
– Minus One-Twelfth
Apr 3 at 19:58






$begingroup$
The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
$endgroup$
– Minus One-Twelfth
Apr 3 at 19:58














$begingroup$
why? Could you help me understand how you got to that conclusion?
$endgroup$
– Evan Kim
Apr 3 at 20:00




$begingroup$
why? Could you help me understand how you got to that conclusion?
$endgroup$
– Evan Kim
Apr 3 at 20:00




2




2




$begingroup$
${1,a^5}$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^{10}=a^4$
$endgroup$
– J. W. Tanner
Apr 3 at 20:00






$begingroup$
${1,a^5}$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^{10}=a^4$
$endgroup$
– J. W. Tanner
Apr 3 at 20:00














$begingroup$
The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
$endgroup$
– Minus One-Twelfth
Apr 3 at 21:44




$begingroup$
The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
$endgroup$
– Minus One-Twelfth
Apr 3 at 21:44










4 Answers
4






active

oldest

votes


















4












$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$



But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    $lbrace 1, a^5 rbrace$ is not a subgroup because
    $$a^5 . a^5 = a^4$$
    is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.



      It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^{gcd(n,k)} rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.




        To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          Why the downvote?
          $endgroup$
          – Shaun
          Apr 3 at 20:46












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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        $[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$



        But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          $[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$



          But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            $[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$



            But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.






            share|cite|improve this answer









            $endgroup$



            $[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$



            But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 3 at 20:01









            PeterPeter

            49.1k1240138




            49.1k1240138























                3












                $begingroup$

                $lbrace 1, a^5 rbrace$ is not a subgroup because
                $$a^5 . a^5 = a^4$$
                is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  $lbrace 1, a^5 rbrace$ is not a subgroup because
                  $$a^5 . a^5 = a^4$$
                  is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    $lbrace 1, a^5 rbrace$ is not a subgroup because
                    $$a^5 . a^5 = a^4$$
                    is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.






                    share|cite|improve this answer









                    $endgroup$



                    $lbrace 1, a^5 rbrace$ is not a subgroup because
                    $$a^5 . a^5 = a^4$$
                    is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 3 at 20:00









                    TheSilverDoeTheSilverDoe

                    5,427216




                    5,427216























                        2












                        $begingroup$

                        Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.



                        It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^{gcd(n,k)} rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.



                          It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^{gcd(n,k)} rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.



                            It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^{gcd(n,k)} rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.






                            share|cite|improve this answer









                            $endgroup$



                            Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.



                            It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^{gcd(n,k)} rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Apr 3 at 21:06









                            Jack PfaffingerJack Pfaffinger

                            3961212




                            3961212























                                0












                                $begingroup$

                                Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.




                                To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.






                                share|cite|improve this answer









                                $endgroup$









                                • 1




                                  $begingroup$
                                  Why the downvote?
                                  $endgroup$
                                  – Shaun
                                  Apr 3 at 20:46
















                                0












                                $begingroup$

                                Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.




                                To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.






                                share|cite|improve this answer









                                $endgroup$









                                • 1




                                  $begingroup$
                                  Why the downvote?
                                  $endgroup$
                                  – Shaun
                                  Apr 3 at 20:46














                                0












                                0








                                0





                                $begingroup$

                                Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.




                                To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.






                                share|cite|improve this answer









                                $endgroup$



                                Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.




                                To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Apr 3 at 20:27









                                ShaunShaun

                                10.5k113687




                                10.5k113687








                                • 1




                                  $begingroup$
                                  Why the downvote?
                                  $endgroup$
                                  – Shaun
                                  Apr 3 at 20:46














                                • 1




                                  $begingroup$
                                  Why the downvote?
                                  $endgroup$
                                  – Shaun
                                  Apr 3 at 20:46








                                1




                                1




                                $begingroup$
                                Why the downvote?
                                $endgroup$
                                – Shaun
                                Apr 3 at 20:46




                                $begingroup$
                                Why the downvote?
                                $endgroup$
                                – Shaun
                                Apr 3 at 20:46


















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                                Musa

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                                Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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