Group of order $6p^m$ is solvable for prime $pgeq 7$












2












$begingroup$



Let $pgeq 7$ be a prime and $m$ be a positive integer. Prove that group of order $6p^m$ is solvable.




Attempt:



By Sylow's theorems we have that $n_p mid 6$ so $n_pin {1,2,3,6}$ where $n_p$ is the number of Sylow $p$ groups. Also we have that $n_p equiv 1 pmod p$ so $n_p=1$ and is thus normal. So we investigate:



$$Gtrianglerighteq H_p$$



Where $H_p$ denotes the Sylow $p$ group. We know that $|G/H_p|=6$ and there are two groups of order $6$: $Bbb Z_6$ and $S_3$. Both are solvable. We also know that Sylow groups are solvable. Since $G/H_p$ is solvable and $H_p$ is solvable, $G$ is solvable.



Is this correct?










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$endgroup$












  • $begingroup$
    Well, as you've noticed denoting as $;S_7;$ a Sylow subgroup is a very bad idea...so edit your question and change it!
    $endgroup$
    – DonAntonio
    Dec 7 '18 at 20:58










  • $begingroup$
    @DonAntonio there.
    $endgroup$
    – Zachary Selk
    Dec 7 '18 at 21:16










  • $begingroup$
    you are using $7$ but you really mean $p$
    $endgroup$
    – the_fox
    Dec 8 '18 at 3:52










  • $begingroup$
    @the_fox sorry, of course
    $endgroup$
    – Zachary Selk
    Dec 8 '18 at 4:09










  • $begingroup$
    See also this question, with similar arguments.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 9:03


















2












$begingroup$



Let $pgeq 7$ be a prime and $m$ be a positive integer. Prove that group of order $6p^m$ is solvable.




Attempt:



By Sylow's theorems we have that $n_p mid 6$ so $n_pin {1,2,3,6}$ where $n_p$ is the number of Sylow $p$ groups. Also we have that $n_p equiv 1 pmod p$ so $n_p=1$ and is thus normal. So we investigate:



$$Gtrianglerighteq H_p$$



Where $H_p$ denotes the Sylow $p$ group. We know that $|G/H_p|=6$ and there are two groups of order $6$: $Bbb Z_6$ and $S_3$. Both are solvable. We also know that Sylow groups are solvable. Since $G/H_p$ is solvable and $H_p$ is solvable, $G$ is solvable.



Is this correct?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, as you've noticed denoting as $;S_7;$ a Sylow subgroup is a very bad idea...so edit your question and change it!
    $endgroup$
    – DonAntonio
    Dec 7 '18 at 20:58










  • $begingroup$
    @DonAntonio there.
    $endgroup$
    – Zachary Selk
    Dec 7 '18 at 21:16










  • $begingroup$
    you are using $7$ but you really mean $p$
    $endgroup$
    – the_fox
    Dec 8 '18 at 3:52










  • $begingroup$
    @the_fox sorry, of course
    $endgroup$
    – Zachary Selk
    Dec 8 '18 at 4:09










  • $begingroup$
    See also this question, with similar arguments.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 9:03
















2












2








2


0



$begingroup$



Let $pgeq 7$ be a prime and $m$ be a positive integer. Prove that group of order $6p^m$ is solvable.




Attempt:



By Sylow's theorems we have that $n_p mid 6$ so $n_pin {1,2,3,6}$ where $n_p$ is the number of Sylow $p$ groups. Also we have that $n_p equiv 1 pmod p$ so $n_p=1$ and is thus normal. So we investigate:



$$Gtrianglerighteq H_p$$



Where $H_p$ denotes the Sylow $p$ group. We know that $|G/H_p|=6$ and there are two groups of order $6$: $Bbb Z_6$ and $S_3$. Both are solvable. We also know that Sylow groups are solvable. Since $G/H_p$ is solvable and $H_p$ is solvable, $G$ is solvable.



Is this correct?










share|cite|improve this question











$endgroup$





Let $pgeq 7$ be a prime and $m$ be a positive integer. Prove that group of order $6p^m$ is solvable.




Attempt:



By Sylow's theorems we have that $n_p mid 6$ so $n_pin {1,2,3,6}$ where $n_p$ is the number of Sylow $p$ groups. Also we have that $n_p equiv 1 pmod p$ so $n_p=1$ and is thus normal. So we investigate:



$$Gtrianglerighteq H_p$$



Where $H_p$ denotes the Sylow $p$ group. We know that $|G/H_p|=6$ and there are two groups of order $6$: $Bbb Z_6$ and $S_3$. Both are solvable. We also know that Sylow groups are solvable. Since $G/H_p$ is solvable and $H_p$ is solvable, $G$ is solvable.



Is this correct?







abstract-algebra group-theory sylow-theory






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share|cite|improve this question













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share|cite|improve this question








edited Dec 8 '18 at 4:09







Zachary Selk

















asked Dec 7 '18 at 20:24









Zachary SelkZachary Selk

593311




593311












  • $begingroup$
    Well, as you've noticed denoting as $;S_7;$ a Sylow subgroup is a very bad idea...so edit your question and change it!
    $endgroup$
    – DonAntonio
    Dec 7 '18 at 20:58










  • $begingroup$
    @DonAntonio there.
    $endgroup$
    – Zachary Selk
    Dec 7 '18 at 21:16










  • $begingroup$
    you are using $7$ but you really mean $p$
    $endgroup$
    – the_fox
    Dec 8 '18 at 3:52










  • $begingroup$
    @the_fox sorry, of course
    $endgroup$
    – Zachary Selk
    Dec 8 '18 at 4:09










  • $begingroup$
    See also this question, with similar arguments.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 9:03




















  • $begingroup$
    Well, as you've noticed denoting as $;S_7;$ a Sylow subgroup is a very bad idea...so edit your question and change it!
    $endgroup$
    – DonAntonio
    Dec 7 '18 at 20:58










  • $begingroup$
    @DonAntonio there.
    $endgroup$
    – Zachary Selk
    Dec 7 '18 at 21:16










  • $begingroup$
    you are using $7$ but you really mean $p$
    $endgroup$
    – the_fox
    Dec 8 '18 at 3:52










  • $begingroup$
    @the_fox sorry, of course
    $endgroup$
    – Zachary Selk
    Dec 8 '18 at 4:09










  • $begingroup$
    See also this question, with similar arguments.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 9:03


















$begingroup$
Well, as you've noticed denoting as $;S_7;$ a Sylow subgroup is a very bad idea...so edit your question and change it!
$endgroup$
– DonAntonio
Dec 7 '18 at 20:58




$begingroup$
Well, as you've noticed denoting as $;S_7;$ a Sylow subgroup is a very bad idea...so edit your question and change it!
$endgroup$
– DonAntonio
Dec 7 '18 at 20:58












$begingroup$
@DonAntonio there.
$endgroup$
– Zachary Selk
Dec 7 '18 at 21:16




$begingroup$
@DonAntonio there.
$endgroup$
– Zachary Selk
Dec 7 '18 at 21:16












$begingroup$
you are using $7$ but you really mean $p$
$endgroup$
– the_fox
Dec 8 '18 at 3:52




$begingroup$
you are using $7$ but you really mean $p$
$endgroup$
– the_fox
Dec 8 '18 at 3:52












$begingroup$
@the_fox sorry, of course
$endgroup$
– Zachary Selk
Dec 8 '18 at 4:09




$begingroup$
@the_fox sorry, of course
$endgroup$
– Zachary Selk
Dec 8 '18 at 4:09












$begingroup$
See also this question, with similar arguments.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 9:03






$begingroup$
See also this question, with similar arguments.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 9:03












1 Answer
1






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$begingroup$

By the way, there's no need to take $pge 7$. The result holds for all primes $p$. For $p=2,3$ this follows from Burnside's $p^aq^b$ theorem. For $p=5$ you additionally use the result that any group of order divisible by 2 but not 4 has an index 2 subgroup (of even permutations in the regular representation). And that proof also works for larger $p$ as well, and so is another solution to the original question






share|cite|improve this answer











$endgroup$













  • $begingroup$
    See here for the result about groups of order $equiv2pmod4$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 14:23











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

By the way, there's no need to take $pge 7$. The result holds for all primes $p$. For $p=2,3$ this follows from Burnside's $p^aq^b$ theorem. For $p=5$ you additionally use the result that any group of order divisible by 2 but not 4 has an index 2 subgroup (of even permutations in the regular representation). And that proof also works for larger $p$ as well, and so is another solution to the original question






share|cite|improve this answer











$endgroup$













  • $begingroup$
    See here for the result about groups of order $equiv2pmod4$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 14:23
















3












$begingroup$

By the way, there's no need to take $pge 7$. The result holds for all primes $p$. For $p=2,3$ this follows from Burnside's $p^aq^b$ theorem. For $p=5$ you additionally use the result that any group of order divisible by 2 but not 4 has an index 2 subgroup (of even permutations in the regular representation). And that proof also works for larger $p$ as well, and so is another solution to the original question






share|cite|improve this answer











$endgroup$













  • $begingroup$
    See here for the result about groups of order $equiv2pmod4$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 14:23














3












3








3





$begingroup$

By the way, there's no need to take $pge 7$. The result holds for all primes $p$. For $p=2,3$ this follows from Burnside's $p^aq^b$ theorem. For $p=5$ you additionally use the result that any group of order divisible by 2 but not 4 has an index 2 subgroup (of even permutations in the regular representation). And that proof also works for larger $p$ as well, and so is another solution to the original question






share|cite|improve this answer











$endgroup$



By the way, there's no need to take $pge 7$. The result holds for all primes $p$. For $p=2,3$ this follows from Burnside's $p^aq^b$ theorem. For $p=5$ you additionally use the result that any group of order divisible by 2 but not 4 has an index 2 subgroup (of even permutations in the regular representation). And that proof also works for larger $p$ as well, and so is another solution to the original question







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '18 at 22:57

























answered Dec 7 '18 at 22:49









C MonsourC Monsour

6,2191325




6,2191325












  • $begingroup$
    See here for the result about groups of order $equiv2pmod4$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 14:23


















  • $begingroup$
    See here for the result about groups of order $equiv2pmod4$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 14:23
















$begingroup$
See here for the result about groups of order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:23




$begingroup$
See here for the result about groups of order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:23


















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