Why does this cyclic subgroup have only 4 subgroups?












1












$begingroup$


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
    2 hours ago












  • $begingroup$
    why? Could you help me understand how you got to that conclusion?
    $endgroup$
    – Evan Kim
    2 hours ago






  • 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
    2 hours ago












  • $begingroup$
    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
    1 hour ago
















1












$begingroup$


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
    2 hours ago












  • $begingroup$
    why? Could you help me understand how you got to that conclusion?
    $endgroup$
    – Evan Kim
    2 hours ago






  • 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
    2 hours ago












  • $begingroup$
    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
    1 hour ago














1












1








1





$begingroup$


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 2 hours ago









J. W. Tanner

4,3651320




4,3651320










asked 2 hours ago









Evan KimEvan Kim

66319




66319












  • $begingroup$
    The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
    $endgroup$
    – Minus One-Twelfth
    2 hours ago












  • $begingroup$
    why? Could you help me understand how you got to that conclusion?
    $endgroup$
    – Evan Kim
    2 hours ago






  • 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
    2 hours ago












  • $begingroup$
    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
    1 hour ago


















  • $begingroup$
    The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
    $endgroup$
    – Minus One-Twelfth
    2 hours ago












  • $begingroup$
    why? Could you help me understand how you got to that conclusion?
    $endgroup$
    – Evan Kim
    2 hours ago






  • 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
    2 hours ago












  • $begingroup$
    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
    1 hour ago
















$begingroup$
The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
$endgroup$
– Minus One-Twelfth
2 hours ago






$begingroup$
The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
$endgroup$
– Minus One-Twelfth
2 hours ago














$begingroup$
why? Could you help me understand how you got to that conclusion?
$endgroup$
– Evan Kim
2 hours ago




$begingroup$
why? Could you help me understand how you got to that conclusion?
$endgroup$
– Evan Kim
2 hours ago




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
2 hours ago






$begingroup$
${1,a^5}$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^{10}=a^4$
$endgroup$
– J. W. Tanner
2 hours ago














$begingroup$
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
1 hour ago




$begingroup$
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
1 hour ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

$[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









$endgroup$





















    2












    $begingroup$

    $lbrace 1, a^5 rbrace$ 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









    $endgroup$





















      1












      $begingroup$

      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









        $endgroup$













        • $begingroup$
          Why the downvote?
          $endgroup$
          – Shaun
          2 hours ago












        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        $[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









        $endgroup$


















          2












          $begingroup$

          $[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









          $endgroup$
















            2












            2








            2





            $begingroup$

            $[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 2 hours ago









            PeterPeter

            49k1240137




            49k1240137























                2












                $begingroup$

                $lbrace 1, a^5 rbrace$ 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












                  $begingroup$

                  $lbrace 1, a^5 rbrace$ 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












                    2








                    2





                    $begingroup$

                    $lbrace 1, a^5 rbrace$ 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 2 hours ago









                    TheSilverDoeTheSilverDoe

                    5,157215




                    5,157215























                        1












                        $begingroup$

                        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









                        $endgroup$


















                          1












                          $begingroup$

                          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









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            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 1 hour ago









                            Jack PfaffingerJack Pfaffinger

                            3841112




                            3841112























                                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









                                $endgroup$













                                • $begingroup$
                                  Why the downvote?
                                  $endgroup$
                                  – Shaun
                                  2 hours ago
















                                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









                                $endgroup$













                                • $begingroup$
                                  Why the downvote?
                                  $endgroup$
                                  – Shaun
                                  2 hours ago














                                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 2 hours ago









                                ShaunShaun

                                10.1k113685




                                10.1k113685












                                • $begingroup$
                                  Why the downvote?
                                  $endgroup$
                                  – Shaun
                                  2 hours ago


















                                • $begingroup$
                                  Why the downvote?
                                  $endgroup$
                                  – Shaun
                                  2 hours ago
















                                $begingroup$
                                Why the downvote?
                                $endgroup$
                                – Shaun
                                2 hours ago




                                $begingroup$
                                Why the downvote?
                                $endgroup$
                                – Shaun
                                2 hours ago


















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