Group of order $6p^m$ is solvable for prime $pgeq 7$
$begingroup$
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?
abstract-algebra group-theory sylow-theory
asked Dec 7 '18 at 20:24
Zachary SelkZachary Selk
593311
$endgroup$
add a comment |
$begingroup$
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?
abstract-algebra group-theory sylow-theory
asked Dec 7 '18 at 20:24
Zachary SelkZachary Selk
593311
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
abstract-algebra group-theory sylow-theory
asked Dec 7 '18 at 20:24
Zachary SelkZachary Selk
593311
$endgroup$
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
abstract-algebra group-theory sylow-theory
asked Dec 7 '18 at 20:24
Zachary SelkZachary Selk
593311
asked Dec 7 '18 at 20:24
Zachary SelkZachary Selk
593311
edited Dec 8 '18 at 4:09
Zachary Selk
asked Dec 7 '18 at 20:24
Zachary SelkZachary Selk
593311
asked Dec 7 '18 at 20:24
Zachary SelkZachary Selk
593311
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
answered Dec 7 '18 at 22:49
C MonsourC Monsour
6,2191325
$endgroup$
$begingroup$
See here for the result about groups of order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:23
add a comment |
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1 Answer
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1 Answer
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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
answered Dec 7 '18 at 22:49
C MonsourC Monsour
6,2191325
$endgroup$
$begingroup$
See here for the result about groups of order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:23
add a comment |
$begingroup$
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
answered Dec 7 '18 at 22:49
C MonsourC Monsour
6,2191325
$endgroup$
$begingroup$
See here for the result about groups of order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:23
add a comment |
$begingroup$
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
answered Dec 7 '18 at 22:49
C MonsourC Monsour
6,2191325
$endgroup$
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
answered Dec 7 '18 at 22:49
C MonsourC Monsour
6,2191325
edited Dec 7 '18 at 22:57
answered Dec 7 '18 at 22:49
C MonsourC Monsour
6,2191325
answered Dec 7 '18 at 22:49
C MonsourC Monsour
6,2191325
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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