Group with order $30$ is not a simple group.












0












$begingroup$


Could someone prove the sentence given above in the title?



I know that Sylow theorems should be used here.



Let $N_{p}$ stands for number of $p$-Sylow subgroups in group $G$.



I tried to use following sentences:




  • $lvert G rvert = 30 =2 cdot 3 cdot 5$

  • $N_{p} equiv 1 pmod p$

  • $N_{p} mid 30$


But i can't figure out.



Regards.






group-theory sylow-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    While it doesn't really impact your result in this case. One important thing to note is Sylow's theorem actually guarantees us that $N_p|30/p$. This reduces the number of cases you have to check, for example, 6 has 4 divisors, while 30 has 8.
    $endgroup$
    – Melody
    Jan 14 at 10:33










  • $begingroup$
    You could also assume that there is more than one Sylow $p$-subgroup for each prime and count elements of order $p$ for each of the primes. This leads to a contradiction.
    $endgroup$
    – Tobias Kildetoft
    Jan 14 at 13:36
















0












$begingroup$


Could someone prove the sentence given above in the title?



I know that Sylow theorems should be used here.



Let $N_{p}$ stands for number of $p$-Sylow subgroups in group $G$.



I tried to use following sentences:




  • $lvert G rvert = 30 =2 cdot 3 cdot 5$

  • $N_{p} equiv 1 pmod p$

  • $N_{p} mid 30$


But i can't figure out.



Regards.






group-theory sylow-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    While it doesn't really impact your result in this case. One important thing to note is Sylow's theorem actually guarantees us that $N_p|30/p$. This reduces the number of cases you have to check, for example, 6 has 4 divisors, while 30 has 8.
    $endgroup$
    – Melody
    Jan 14 at 10:33










  • $begingroup$
    You could also assume that there is more than one Sylow $p$-subgroup for each prime and count elements of order $p$ for each of the primes. This leads to a contradiction.
    $endgroup$
    – Tobias Kildetoft
    Jan 14 at 13:36














0












0








0





$begingroup$


Could someone prove the sentence given above in the title?



I know that Sylow theorems should be used here.



Let $N_{p}$ stands for number of $p$-Sylow subgroups in group $G$.



I tried to use following sentences:




  • $lvert G rvert = 30 =2 cdot 3 cdot 5$

  • $N_{p} equiv 1 pmod p$

  • $N_{p} mid 30$


But i can't figure out.



Regards.






group-theory sylow-theory







share|cite|improve this question











$endgroup$




Could someone prove the sentence given above in the title?



I know that Sylow theorems should be used here.



Let $N_{p}$ stands for number of $p$-Sylow subgroups in group $G$.



I tried to use following sentences:




  • $lvert G rvert = 30 =2 cdot 3 cdot 5$

  • $N_{p} equiv 1 pmod p$

  • $N_{p} mid 30$


But i can't figure out.



Regards.






group-theory sylow-theory




group-theory sylow-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 10:21









Andreas Caranti

57.4k34498




57.4k34498










asked Jan 14 at 10:11









mkultramkultra

1048




1048












  • $begingroup$
    While it doesn't really impact your result in this case. One important thing to note is Sylow's theorem actually guarantees us that $N_p|30/p$. This reduces the number of cases you have to check, for example, 6 has 4 divisors, while 30 has 8.
    $endgroup$
    – Melody
    Jan 14 at 10:33










  • $begingroup$
    You could also assume that there is more than one Sylow $p$-subgroup for each prime and count elements of order $p$ for each of the primes. This leads to a contradiction.
    $endgroup$
    – Tobias Kildetoft
    Jan 14 at 13:36


















  • $begingroup$
    While it doesn't really impact your result in this case. One important thing to note is Sylow's theorem actually guarantees us that $N_p|30/p$. This reduces the number of cases you have to check, for example, 6 has 4 divisors, while 30 has 8.
    $endgroup$
    – Melody
    Jan 14 at 10:33










  • $begingroup$
    You could also assume that there is more than one Sylow $p$-subgroup for each prime and count elements of order $p$ for each of the primes. This leads to a contradiction.
    $endgroup$
    – Tobias Kildetoft
    Jan 14 at 13:36
















$begingroup$
While it doesn't really impact your result in this case. One important thing to note is Sylow's theorem actually guarantees us that $N_p|30/p$. This reduces the number of cases you have to check, for example, 6 has 4 divisors, while 30 has 8.
$endgroup$
– Melody
Jan 14 at 10:33




$begingroup$
While it doesn't really impact your result in this case. One important thing to note is Sylow's theorem actually guarantees us that $N_p|30/p$. This reduces the number of cases you have to check, for example, 6 has 4 divisors, while 30 has 8.
$endgroup$
– Melody
Jan 14 at 10:33












$begingroup$
You could also assume that there is more than one Sylow $p$-subgroup for each prime and count elements of order $p$ for each of the primes. This leads to a contradiction.
$endgroup$
– Tobias Kildetoft
Jan 14 at 13:36




$begingroup$
You could also assume that there is more than one Sylow $p$-subgroup for each prime and count elements of order $p$ for each of the primes. This leads to a contradiction.
$endgroup$
– Tobias Kildetoft
Jan 14 at 13:36










1 Answer
1






active

oldest

votes


















4












$begingroup$

$newcommand{Size}[1]{leftlvert #1 rightrvert}$If you know the result




if the finite group $G$ has order $Size{G} = 2 d$, with $d$ odd, then $G$ has a subgroup of order $d$,




then you are immediately in business.



Otherwise, check the possibilities for the number of Sylow $5$-subgroups, and consequently for the number of elements of order $5$. If $N_{5} > 1$, you will find that there won't be too many elements left.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    As regards to the given result- I am not sure whether this going to help. A group is not simple when there exists non-trivial normal subgroup not just ordinary subgroup.
    $endgroup$
    – mkultra
    Jan 14 at 11:14










  • $begingroup$
    @mkultra Yes, but the result tells you that in a group $G$ of order $2d, d$ odd, you have a subgroup of order $d$, so $[G:H] = (2d)/d = 2$ and every subgroup of index $2$ is normal, you can see a proof here math.stackexchange.com/questions/84632/…
    $endgroup$
    – Cosmin
    Jan 14 at 11:23












  • $begingroup$
    I want to only ask for proof of the result mentioned by @Andreas Caranti.
    $endgroup$
    – mkultra
    Jan 14 at 13:33












  • $begingroup$
    @mkultra there is a proof here.
    $endgroup$
    – Andreas Caranti
    Jan 14 at 14:28












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

$newcommand{Size}[1]{leftlvert #1 rightrvert}$If you know the result




if the finite group $G$ has order $Size{G} = 2 d$, with $d$ odd, then $G$ has a subgroup of order $d$,




then you are immediately in business.



Otherwise, check the possibilities for the number of Sylow $5$-subgroups, and consequently for the number of elements of order $5$. If $N_{5} > 1$, you will find that there won't be too many elements left.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    As regards to the given result- I am not sure whether this going to help. A group is not simple when there exists non-trivial normal subgroup not just ordinary subgroup.
    $endgroup$
    – mkultra
    Jan 14 at 11:14










  • $begingroup$
    @mkultra Yes, but the result tells you that in a group $G$ of order $2d, d$ odd, you have a subgroup of order $d$, so $[G:H] = (2d)/d = 2$ and every subgroup of index $2$ is normal, you can see a proof here math.stackexchange.com/questions/84632/…
    $endgroup$
    – Cosmin
    Jan 14 at 11:23












  • $begingroup$
    I want to only ask for proof of the result mentioned by @Andreas Caranti.
    $endgroup$
    – mkultra
    Jan 14 at 13:33












  • $begingroup$
    @mkultra there is a proof here.
    $endgroup$
    – Andreas Caranti
    Jan 14 at 14:28
















4












$begingroup$

$newcommand{Size}[1]{leftlvert #1 rightrvert}$If you know the result




if the finite group $G$ has order $Size{G} = 2 d$, with $d$ odd, then $G$ has a subgroup of order $d$,




then you are immediately in business.



Otherwise, check the possibilities for the number of Sylow $5$-subgroups, and consequently for the number of elements of order $5$. If $N_{5} > 1$, you will find that there won't be too many elements left.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    As regards to the given result- I am not sure whether this going to help. A group is not simple when there exists non-trivial normal subgroup not just ordinary subgroup.
    $endgroup$
    – mkultra
    Jan 14 at 11:14










  • $begingroup$
    @mkultra Yes, but the result tells you that in a group $G$ of order $2d, d$ odd, you have a subgroup of order $d$, so $[G:H] = (2d)/d = 2$ and every subgroup of index $2$ is normal, you can see a proof here math.stackexchange.com/questions/84632/…
    $endgroup$
    – Cosmin
    Jan 14 at 11:23












  • $begingroup$
    I want to only ask for proof of the result mentioned by @Andreas Caranti.
    $endgroup$
    – mkultra
    Jan 14 at 13:33












  • $begingroup$
    @mkultra there is a proof here.
    $endgroup$
    – Andreas Caranti
    Jan 14 at 14:28














4












4








4





$begingroup$

$newcommand{Size}[1]{leftlvert #1 rightrvert}$If you know the result




if the finite group $G$ has order $Size{G} = 2 d$, with $d$ odd, then $G$ has a subgroup of order $d$,




then you are immediately in business.



Otherwise, check the possibilities for the number of Sylow $5$-subgroups, and consequently for the number of elements of order $5$. If $N_{5} > 1$, you will find that there won't be too many elements left.






share|cite|improve this answer









$endgroup$



$newcommand{Size}[1]{leftlvert #1 rightrvert}$If you know the result




if the finite group $G$ has order $Size{G} = 2 d$, with $d$ odd, then $G$ has a subgroup of order $d$,




then you are immediately in business.



Otherwise, check the possibilities for the number of Sylow $5$-subgroups, and consequently for the number of elements of order $5$. If $N_{5} > 1$, you will find that there won't be too many elements left.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 14 at 10:19









Andreas CarantiAndreas Caranti

57.4k34498




57.4k34498












  • $begingroup$
    As regards to the given result- I am not sure whether this going to help. A group is not simple when there exists non-trivial normal subgroup not just ordinary subgroup.
    $endgroup$
    – mkultra
    Jan 14 at 11:14










  • $begingroup$
    @mkultra Yes, but the result tells you that in a group $G$ of order $2d, d$ odd, you have a subgroup of order $d$, so $[G:H] = (2d)/d = 2$ and every subgroup of index $2$ is normal, you can see a proof here math.stackexchange.com/questions/84632/…
    $endgroup$
    – Cosmin
    Jan 14 at 11:23












  • $begingroup$
    I want to only ask for proof of the result mentioned by @Andreas Caranti.
    $endgroup$
    – mkultra
    Jan 14 at 13:33












  • $begingroup$
    @mkultra there is a proof here.
    $endgroup$
    – Andreas Caranti
    Jan 14 at 14:28


















  • $begingroup$
    As regards to the given result- I am not sure whether this going to help. A group is not simple when there exists non-trivial normal subgroup not just ordinary subgroup.
    $endgroup$
    – mkultra
    Jan 14 at 11:14










  • $begingroup$
    @mkultra Yes, but the result tells you that in a group $G$ of order $2d, d$ odd, you have a subgroup of order $d$, so $[G:H] = (2d)/d = 2$ and every subgroup of index $2$ is normal, you can see a proof here math.stackexchange.com/questions/84632/…
    $endgroup$
    – Cosmin
    Jan 14 at 11:23












  • $begingroup$
    I want to only ask for proof of the result mentioned by @Andreas Caranti.
    $endgroup$
    – mkultra
    Jan 14 at 13:33












  • $begingroup$
    @mkultra there is a proof here.
    $endgroup$
    – Andreas Caranti
    Jan 14 at 14:28
















$begingroup$
As regards to the given result- I am not sure whether this going to help. A group is not simple when there exists non-trivial normal subgroup not just ordinary subgroup.
$endgroup$
– mkultra
Jan 14 at 11:14




$begingroup$
As regards to the given result- I am not sure whether this going to help. A group is not simple when there exists non-trivial normal subgroup not just ordinary subgroup.
$endgroup$
– mkultra
Jan 14 at 11:14












$begingroup$
@mkultra Yes, but the result tells you that in a group $G$ of order $2d, d$ odd, you have a subgroup of order $d$, so $[G:H] = (2d)/d = 2$ and every subgroup of index $2$ is normal, you can see a proof here math.stackexchange.com/questions/84632/…
$endgroup$
– Cosmin
Jan 14 at 11:23






$begingroup$
@mkultra Yes, but the result tells you that in a group $G$ of order $2d, d$ odd, you have a subgroup of order $d$, so $[G:H] = (2d)/d = 2$ and every subgroup of index $2$ is normal, you can see a proof here math.stackexchange.com/questions/84632/…
$endgroup$
– Cosmin
Jan 14 at 11:23














$begingroup$
I want to only ask for proof of the result mentioned by @Andreas Caranti.
$endgroup$
– mkultra
Jan 14 at 13:33






$begingroup$
I want to only ask for proof of the result mentioned by @Andreas Caranti.
$endgroup$
– mkultra
Jan 14 at 13:33














$begingroup$
@mkultra there is a proof here.
$endgroup$
– Andreas Caranti
Jan 14 at 14:28




$begingroup$
@mkultra there is a proof here.
$endgroup$
– Andreas Caranti
Jan 14 at 14:28


















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