Does the multiplicative group of GF(81) have a subgroup of order 9?

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$$|GF(81)^*| = |GF(81)setminus 0| = 81 - 1 = 80.$$
Since $GF(81)^*$ is cyclic, $|GF(81)^*|=n$ : for any divisor $d$ of $n,$ there's exactly one unique subgroup of $G$ of order $d.$
$9$ is not a divisor of $80,$ therefore, there is no subgroups of order $9.$



Would someone please help me with this and tell me whether my argumentation is correct?
Thank you!




group-theory




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    Yes, $9nmid 80$.
    – Lord Shark the Unknown
    Jul 20 at 17:05







  • 1




    Worse, of course, this is a field of characteristic three, in which the polynomial $X^9-1$ has only the single root $1$ (of multiplicity nine).
    – Lubin
    Jul 20 at 18:54














up vote
2
down vote

favorite












$$|GF(81)^*| = |GF(81)setminus 0| = 81 - 1 = 80.$$
Since $GF(81)^*$ is cyclic, $|GF(81)^*|=n$ : for any divisor $d$ of $n,$ there's exactly one unique subgroup of $G$ of order $d.$
$9$ is not a divisor of $80,$ therefore, there is no subgroups of order $9.$



Would someone please help me with this and tell me whether my argumentation is correct?
Thank you!




group-theory




share|cite|improve this question

















  • 2




    Yes, $9nmid 80$.
    – Lord Shark the Unknown
    Jul 20 at 17:05







  • 1




    Worse, of course, this is a field of characteristic three, in which the polynomial $X^9-1$ has only the single root $1$ (of multiplicity nine).
    – Lubin
    Jul 20 at 18:54












up vote
2
down vote

favorite









up vote
2
down vote

favorite











$$|GF(81)^*| = |GF(81)setminus 0| = 81 - 1 = 80.$$
Since $GF(81)^*$ is cyclic, $|GF(81)^*|=n$ : for any divisor $d$ of $n,$ there's exactly one unique subgroup of $G$ of order $d.$
$9$ is not a divisor of $80,$ therefore, there is no subgroups of order $9.$



Would someone please help me with this and tell me whether my argumentation is correct?
Thank you!




group-theory




share|cite|improve this question













$$|GF(81)^*| = |GF(81)setminus 0| = 81 - 1 = 80.$$
Since $GF(81)^*$ is cyclic, $|GF(81)^*|=n$ : for any divisor $d$ of $n,$ there's exactly one unique subgroup of $G$ of order $d.$
$9$ is not a divisor of $80,$ therefore, there is no subgroups of order $9.$



Would someone please help me with this and tell me whether my argumentation is correct?
Thank you!





group-theory





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edited Jul 20 at 18:58









Michael Hardy

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204k23186462









asked Jul 20 at 17:04









pramort

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303







  • 2




    Yes, $9nmid 80$.
    – Lord Shark the Unknown
    Jul 20 at 17:05







  • 1




    Worse, of course, this is a field of characteristic three, in which the polynomial $X^9-1$ has only the single root $1$ (of multiplicity nine).
    – Lubin
    Jul 20 at 18:54












  • 2




    Yes, $9nmid 80$.
    – Lord Shark the Unknown
    Jul 20 at 17:05







  • 1




    Worse, of course, this is a field of characteristic three, in which the polynomial $X^9-1$ has only the single root $1$ (of multiplicity nine).
    – Lubin
    Jul 20 at 18:54







2




2




Yes, $9nmid 80$.
– Lord Shark the Unknown
Jul 20 at 17:05





Yes, $9nmid 80$.
– Lord Shark the Unknown
Jul 20 at 17:05





1




1




Worse, of course, this is a field of characteristic three, in which the polynomial $X^9-1$ has only the single root $1$ (of multiplicity nine).
– Lubin
Jul 20 at 18:54




Worse, of course, this is a field of characteristic three, in which the polynomial $X^9-1$ has only the single root $1$ (of multiplicity nine).
– Lubin
Jul 20 at 18:54










1 Answer
1






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up vote
3
down vote



accepted










That's fine, by the Fundamental Theorem of cyclic groups, all subgroups have to divide the order of the cyclic group. Since you point out $9$ does not divide $80$ ,no such subgroup can exist.






share|cite|improve this answer

















  • 1




    It's just Lagrange's theorem.
    – Lord Shark the Unknown
    Jul 20 at 17:07






  • 1




    @LordSharktheUnknown It's also the Fundamental Theorem of cyclic subgroups. I find the FT more basic than Lagrange. But you are right, it is just Lagrange.
    – Sorfosh
    Jul 20 at 17:10










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










That's fine, by the Fundamental Theorem of cyclic groups, all subgroups have to divide the order of the cyclic group. Since you point out $9$ does not divide $80$ ,no such subgroup can exist.






share|cite|improve this answer

















  • 1




    It's just Lagrange's theorem.
    – Lord Shark the Unknown
    Jul 20 at 17:07






  • 1




    @LordSharktheUnknown It's also the Fundamental Theorem of cyclic subgroups. I find the FT more basic than Lagrange. But you are right, it is just Lagrange.
    – Sorfosh
    Jul 20 at 17:10














up vote
3
down vote



accepted










That's fine, by the Fundamental Theorem of cyclic groups, all subgroups have to divide the order of the cyclic group. Since you point out $9$ does not divide $80$ ,no such subgroup can exist.






share|cite|improve this answer

















  • 1




    It's just Lagrange's theorem.
    – Lord Shark the Unknown
    Jul 20 at 17:07






  • 1




    @LordSharktheUnknown It's also the Fundamental Theorem of cyclic subgroups. I find the FT more basic than Lagrange. But you are right, it is just Lagrange.
    – Sorfosh
    Jul 20 at 17:10












up vote
3
down vote



accepted







up vote
3
down vote



accepted






That's fine, by the Fundamental Theorem of cyclic groups, all subgroups have to divide the order of the cyclic group. Since you point out $9$ does not divide $80$ ,no such subgroup can exist.






share|cite|improve this answer













That's fine, by the Fundamental Theorem of cyclic groups, all subgroups have to divide the order of the cyclic group. Since you point out $9$ does not divide $80$ ,no such subgroup can exist.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 20 at 17:06









Sorfosh

910616




910616







  • 1




    It's just Lagrange's theorem.
    – Lord Shark the Unknown
    Jul 20 at 17:07






  • 1




    @LordSharktheUnknown It's also the Fundamental Theorem of cyclic subgroups. I find the FT more basic than Lagrange. But you are right, it is just Lagrange.
    – Sorfosh
    Jul 20 at 17:10












  • 1




    It's just Lagrange's theorem.
    – Lord Shark the Unknown
    Jul 20 at 17:07






  • 1




    @LordSharktheUnknown It's also the Fundamental Theorem of cyclic subgroups. I find the FT more basic than Lagrange. But you are right, it is just Lagrange.
    – Sorfosh
    Jul 20 at 17:10







1




1




It's just Lagrange's theorem.
– Lord Shark the Unknown
Jul 20 at 17:07




It's just Lagrange's theorem.
– Lord Shark the Unknown
Jul 20 at 17:07




1




1




@LordSharktheUnknown It's also the Fundamental Theorem of cyclic subgroups. I find the FT more basic than Lagrange. But you are right, it is just Lagrange.
– Sorfosh
Jul 20 at 17:10




@LordSharktheUnknown It's also the Fundamental Theorem of cyclic subgroups. I find the FT more basic than Lagrange. But you are right, it is just Lagrange.
– Sorfosh
Jul 20 at 17:10












 

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