Classify all groups of order 4165

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Classify all groups of order $4165=5(7^2)17$




I've determined the following possibilities for each of the sylow subgroups



$r_5 = 1$



$r_7 = 1$ or $5(17)$



$r_17 = 1$ or $5(7)$



I'm trying to show either the sylow 7 subgroup or the sylow 17 subgroup is normal so that I can create a subgroup of index 5. Then I would use semi-direct product theorem. But maybe this is not necessary and maybe there is a simpler solution.




abstract-algebra group-theory finite-groups




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  • 2




    None of $7, 7^2$ or $7^2cdot17$ are $equiv1pmod5$, so it looks like you have a normal Sylow $5$
    – Jyrki Lahtonen
    17 mins ago











  • Only $7^2cdot 17$ is 1 mod 5
    – iYOA
    15 mins ago






  • 2




    $7cdot7cdot17equiv2cdot2cdot2=8equiv3pmod5$.
    – Jyrki Lahtonen
    14 mins ago






  • 1




    And the automorphism group of $C_5$ is $simeq C_4$, so that Sylow $5$-subgroup must be central.
    – Jyrki Lahtonen
    12 mins ago










  • wow...I can't do arithmetic
    – iYOA
    11 mins ago














up vote
1
down vote

favorite













Classify all groups of order $4165=5(7^2)17$




I've determined the following possibilities for each of the sylow subgroups



$r_5 = 1$



$r_7 = 1$ or $5(17)$



$r_17 = 1$ or $5(7)$



I'm trying to show either the sylow 7 subgroup or the sylow 17 subgroup is normal so that I can create a subgroup of index 5. Then I would use semi-direct product theorem. But maybe this is not necessary and maybe there is a simpler solution.




abstract-algebra group-theory finite-groups




share|cite|improve this question

















  • 2




    None of $7, 7^2$ or $7^2cdot17$ are $equiv1pmod5$, so it looks like you have a normal Sylow $5$
    – Jyrki Lahtonen
    17 mins ago











  • Only $7^2cdot 17$ is 1 mod 5
    – iYOA
    15 mins ago






  • 2




    $7cdot7cdot17equiv2cdot2cdot2=8equiv3pmod5$.
    – Jyrki Lahtonen
    14 mins ago






  • 1




    And the automorphism group of $C_5$ is $simeq C_4$, so that Sylow $5$-subgroup must be central.
    – Jyrki Lahtonen
    12 mins ago










  • wow...I can't do arithmetic
    – iYOA
    11 mins ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Classify all groups of order $4165=5(7^2)17$




I've determined the following possibilities for each of the sylow subgroups



$r_5 = 1$



$r_7 = 1$ or $5(17)$



$r_17 = 1$ or $5(7)$



I'm trying to show either the sylow 7 subgroup or the sylow 17 subgroup is normal so that I can create a subgroup of index 5. Then I would use semi-direct product theorem. But maybe this is not necessary and maybe there is a simpler solution.




abstract-algebra group-theory finite-groups




share|cite|improve this question














Classify all groups of order $4165=5(7^2)17$




I've determined the following possibilities for each of the sylow subgroups



$r_5 = 1$



$r_7 = 1$ or $5(17)$



$r_17 = 1$ or $5(7)$



I'm trying to show either the sylow 7 subgroup or the sylow 17 subgroup is normal so that I can create a subgroup of index 5. Then I would use semi-direct product theorem. But maybe this is not necessary and maybe there is a simpler solution.





abstract-algebra group-theory finite-groups





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 3 mins ago
























asked 24 mins ago









iYOA

40339




40339







  • 2




    None of $7, 7^2$ or $7^2cdot17$ are $equiv1pmod5$, so it looks like you have a normal Sylow $5$
    – Jyrki Lahtonen
    17 mins ago











  • Only $7^2cdot 17$ is 1 mod 5
    – iYOA
    15 mins ago






  • 2




    $7cdot7cdot17equiv2cdot2cdot2=8equiv3pmod5$.
    – Jyrki Lahtonen
    14 mins ago






  • 1




    And the automorphism group of $C_5$ is $simeq C_4$, so that Sylow $5$-subgroup must be central.
    – Jyrki Lahtonen
    12 mins ago










  • wow...I can't do arithmetic
    – iYOA
    11 mins ago












  • 2




    None of $7, 7^2$ or $7^2cdot17$ are $equiv1pmod5$, so it looks like you have a normal Sylow $5$
    – Jyrki Lahtonen
    17 mins ago











  • Only $7^2cdot 17$ is 1 mod 5
    – iYOA
    15 mins ago






  • 2




    $7cdot7cdot17equiv2cdot2cdot2=8equiv3pmod5$.
    – Jyrki Lahtonen
    14 mins ago






  • 1




    And the automorphism group of $C_5$ is $simeq C_4$, so that Sylow $5$-subgroup must be central.
    – Jyrki Lahtonen
    12 mins ago










  • wow...I can't do arithmetic
    – iYOA
    11 mins ago







2




2




None of $7, 7^2$ or $7^2cdot17$ are $equiv1pmod5$, so it looks like you have a normal Sylow $5$
– Jyrki Lahtonen
17 mins ago





None of $7, 7^2$ or $7^2cdot17$ are $equiv1pmod5$, so it looks like you have a normal Sylow $5$
– Jyrki Lahtonen
17 mins ago













Only $7^2cdot 17$ is 1 mod 5
– iYOA
15 mins ago




Only $7^2cdot 17$ is 1 mod 5
– iYOA
15 mins ago




2




2




$7cdot7cdot17equiv2cdot2cdot2=8equiv3pmod5$.
– Jyrki Lahtonen
14 mins ago




$7cdot7cdot17equiv2cdot2cdot2=8equiv3pmod5$.
– Jyrki Lahtonen
14 mins ago




1




1




And the automorphism group of $C_5$ is $simeq C_4$, so that Sylow $5$-subgroup must be central.
– Jyrki Lahtonen
12 mins ago




And the automorphism group of $C_5$ is $simeq C_4$, so that Sylow $5$-subgroup must be central.
– Jyrki Lahtonen
12 mins ago












wow...I can't do arithmetic
– iYOA
11 mins ago




wow...I can't do arithmetic
– iYOA
11 mins ago















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