Sylow subgroups and simple groups [closed]
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I recently began learning the Sylow Theorems. However, I don't understand why if you have a finite group $G$ with $|G|> 1$, and if $G$ has a unique Sylow p-subgroup, then $G$ cannot be simple.
Thanks
abstract-algebra
asked Dec 2 '18 at 21:43
jclayjclay
121
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closed as off-topic by Morgan Rodgers, Shailesh, Saad, KReiser, Jyrki Lahtonen Dec 3 '18 at 3:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Shailesh, Saad, KReiser, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I recently began learning the Sylow Theorems. However, I don't understand why if you have a finite group $G$ with $|G|> 1$, and if $G$ has a unique Sylow p-subgroup, then $G$ cannot be simple.
Thanks
abstract-algebra
asked Dec 2 '18 at 21:43
jclayjclay
121
$endgroup$
closed as off-topic by Morgan Rodgers, Shailesh, Saad, KReiser, Jyrki Lahtonen Dec 3 '18 at 3:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Shailesh, Saad, KReiser, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Do you know what simple means?
$endgroup$
– Fakemistake
Dec 2 '18 at 21:46
$begingroup$
If you have a unique Sylow $p$-subgroup then the $p$-subgroup is normal. So if this $p$-subgroup is proper, then $G$ won't be simple.
$endgroup$
– Morgan Rodgers
Dec 2 '18 at 21:46
add a comment |
$begingroup$
I recently began learning the Sylow Theorems. However, I don't understand why if you have a finite group $G$ with $|G|> 1$, and if $G$ has a unique Sylow p-subgroup, then $G$ cannot be simple.
Thanks
abstract-algebra
asked Dec 2 '18 at 21:43
jclayjclay
121
$endgroup$
I recently began learning the Sylow Theorems. However, I don't understand why if you have a finite group $G$ with $|G|> 1$, and if $G$ has a unique Sylow p-subgroup, then $G$ cannot be simple.
Thanks
abstract-algebra
abstract-algebra
asked Dec 2 '18 at 21:43
jclayjclay
121
asked Dec 2 '18 at 21:43
jclayjclay
121
asked Dec 2 '18 at 21:43
jclayjclay
121
asked Dec 2 '18 at 21:43
jclayjclay
121
asked Dec 2 '18 at 21:43
jclayjclay
121
121
closed as off-topic by Morgan Rodgers, Shailesh, Saad, KReiser, Jyrki Lahtonen Dec 3 '18 at 3:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Shailesh, Saad, KReiser, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Morgan Rodgers, Shailesh, Saad, KReiser, Jyrki Lahtonen Dec 3 '18 at 3:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Shailesh, Saad, KReiser, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Do you know what simple means?
$endgroup$
– Fakemistake
Dec 2 '18 at 21:46
$begingroup$
If you have a unique Sylow $p$-subgroup then the $p$-subgroup is normal. So if this $p$-subgroup is proper, then $G$ won't be simple.
$endgroup$
– Morgan Rodgers
Dec 2 '18 at 21:46
add a comment |
$begingroup$
Do you know what simple means?
$endgroup$
– Fakemistake
Dec 2 '18 at 21:46
$begingroup$
If you have a unique Sylow $p$-subgroup then the $p$-subgroup is normal. So if this $p$-subgroup is proper, then $G$ won't be simple.
$endgroup$
– Morgan Rodgers
Dec 2 '18 at 21:46
$begingroup$
Do you know what simple means?
$endgroup$
– Fakemistake
Dec 2 '18 at 21:46
$begingroup$
Do you know what simple means?
$endgroup$
– Fakemistake
Dec 2 '18 at 21:46
$begingroup$
If you have a unique Sylow $p$-subgroup then the $p$-subgroup is normal. So if this $p$-subgroup is proper, then $G$ won't be simple.
$endgroup$
– Morgan Rodgers
Dec 2 '18 at 21:46
$begingroup$
If you have a unique Sylow $p$-subgroup then the $p$-subgroup is normal. So if this $p$-subgroup is proper, then $G$ won't be simple.
$endgroup$
– Morgan Rodgers
Dec 2 '18 at 21:46
add a comment |
2 Answers
2
active
oldest
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By the Sylow theorems, all Sylow-$p$-subgroups of $G$ are conjugate. If there is only one Sylow-$p$-subgroup, then it is normal because it is equal to all its conjugates. So if $G$ has a unique Sylow-$p$-subgroup then it has a normal subgroup, hence it is not simple unless the Sylow-$p$-subgroup is all of $G$.
So your statement is not true; the simplest counterexample is the group of $2$ elements, and in general any group $G$ with $|G|$ a prime power.
answered Dec 2 '18 at 21:46
ServaesServaes
23.3k33893
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add a comment |
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Kind of a hint: Sylow's $3$rd theorem says that $n_p = [G : N_G(P)]$ where $P$ is a Sylow $p$-subgroup of $G$ and $n_p$ is the number of such subgroups. What does it mean that $n_p = 1$?
answered Dec 2 '18 at 21:47
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By the Sylow theorems, all Sylow-$p$-subgroups of $G$ are conjugate. If there is only one Sylow-$p$-subgroup, then it is normal because it is equal to all its conjugates. So if $G$ has a unique Sylow-$p$-subgroup then it has a normal subgroup, hence it is not simple unless the Sylow-$p$-subgroup is all of $G$.
So your statement is not true; the simplest counterexample is the group of $2$ elements, and in general any group $G$ with $|G|$ a prime power.
answered Dec 2 '18 at 21:46
ServaesServaes
23.3k33893
$endgroup$
add a comment |
$begingroup$
By the Sylow theorems, all Sylow-$p$-subgroups of $G$ are conjugate. If there is only one Sylow-$p$-subgroup, then it is normal because it is equal to all its conjugates. So if $G$ has a unique Sylow-$p$-subgroup then it has a normal subgroup, hence it is not simple unless the Sylow-$p$-subgroup is all of $G$.
So your statement is not true; the simplest counterexample is the group of $2$ elements, and in general any group $G$ with $|G|$ a prime power.
answered Dec 2 '18 at 21:46
ServaesServaes
23.3k33893
$endgroup$
add a comment |
$begingroup$
By the Sylow theorems, all Sylow-$p$-subgroups of $G$ are conjugate. If there is only one Sylow-$p$-subgroup, then it is normal because it is equal to all its conjugates. So if $G$ has a unique Sylow-$p$-subgroup then it has a normal subgroup, hence it is not simple unless the Sylow-$p$-subgroup is all of $G$.
So your statement is not true; the simplest counterexample is the group of $2$ elements, and in general any group $G$ with $|G|$ a prime power.
answered Dec 2 '18 at 21:46
ServaesServaes
23.3k33893
$endgroup$
By the Sylow theorems, all Sylow-$p$-subgroups of $G$ are conjugate. If there is only one Sylow-$p$-subgroup, then it is normal because it is equal to all its conjugates. So if $G$ has a unique Sylow-$p$-subgroup then it has a normal subgroup, hence it is not simple unless the Sylow-$p$-subgroup is all of $G$.
So your statement is not true; the simplest counterexample is the group of $2$ elements, and in general any group $G$ with $|G|$ a prime power.
answered Dec 2 '18 at 21:46
ServaesServaes
23.3k33893
answered Dec 2 '18 at 21:46
ServaesServaes
23.3k33893
answered Dec 2 '18 at 21:46
ServaesServaes
23.3k33893
answered Dec 2 '18 at 21:46
ServaesServaes
23.3k33893
23.3k33893
add a comment |
add a comment |
$begingroup$
Kind of a hint: Sylow's $3$rd theorem says that $n_p = [G : N_G(P)]$ where $P$ is a Sylow $p$-subgroup of $G$ and $n_p$ is the number of such subgroups. What does it mean that $n_p = 1$?
answered Dec 2 '18 at 21:47
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
$endgroup$
add a comment |
$begingroup$
Kind of a hint: Sylow's $3$rd theorem says that $n_p = [G : N_G(P)]$ where $P$ is a Sylow $p$-subgroup of $G$ and $n_p$ is the number of such subgroups. What does it mean that $n_p = 1$?
answered Dec 2 '18 at 21:47
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
$endgroup$
add a comment |
$begingroup$
Kind of a hint: Sylow's $3$rd theorem says that $n_p = [G : N_G(P)]$ where $P$ is a Sylow $p$-subgroup of $G$ and $n_p$ is the number of such subgroups. What does it mean that $n_p = 1$?
answered Dec 2 '18 at 21:47
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
$endgroup$
Kind of a hint: Sylow's $3$rd theorem says that $n_p = [G : N_G(P)]$ where $P$ is a Sylow $p$-subgroup of $G$ and $n_p$ is the number of such subgroups. What does it mean that $n_p = 1$?
answered Dec 2 '18 at 21:47
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
answered Dec 2 '18 at 21:47
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
answered Dec 2 '18 at 21:47
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
answered Dec 2 '18 at 21:47
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
2,83511029
add a comment |
add a comment |
$begingroup$
Do you know what simple means?
$endgroup$
– Fakemistake
Dec 2 '18 at 21:46
$begingroup$
If you have a unique Sylow $p$-subgroup then the $p$-subgroup is normal. So if this $p$-subgroup is proper, then $G$ won't be simple.
$endgroup$
– Morgan Rodgers
Dec 2 '18 at 21:46