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Question about p-Sylow subgroups being maximal
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I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G
Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?
abstract-algebra group-theory sylow-theory
edited Jan 1 at 20:19
Shaun
8,818113681
asked Dec 8 '18 at 4:22
Richard VillalobosRichard Villalobos
1567
$endgroup$
add a comment |
$begingroup$
I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G
Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?
abstract-algebra group-theory sylow-theory
edited Jan 1 at 20:19
Shaun
8,818113681
asked Dec 8 '18 at 4:22
Richard VillalobosRichard Villalobos
1567
$endgroup$
$begingroup$
Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
$endgroup$
– Chris Custer
Dec 8 '18 at 4:55
add a comment |
$begingroup$
I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G
Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?
abstract-algebra group-theory sylow-theory
edited Jan 1 at 20:19
Shaun
8,818113681
asked Dec 8 '18 at 4:22
Richard VillalobosRichard Villalobos
1567
$endgroup$
I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G
Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?
abstract-algebra group-theory sylow-theory
abstract-algebra group-theory sylow-theory
edited Jan 1 at 20:19
Shaun
8,818113681
asked Dec 8 '18 at 4:22
Richard VillalobosRichard Villalobos
1567
edited Jan 1 at 20:19
Shaun
8,818113681
asked Dec 8 '18 at 4:22
Richard VillalobosRichard Villalobos
1567
edited Jan 1 at 20:19
Shaun
8,818113681
edited Jan 1 at 20:19
Shaun
8,818113681
edited Jan 1 at 20:19
Shaun
8,818113681
8,818113681
asked Dec 8 '18 at 4:22
Richard VillalobosRichard Villalobos
1567
asked Dec 8 '18 at 4:22
Richard VillalobosRichard Villalobos
1567
asked Dec 8 '18 at 4:22
Richard VillalobosRichard Villalobos
1567
1567
$begingroup$
Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
$endgroup$
– Chris Custer
Dec 8 '18 at 4:55
add a comment |
$begingroup$
Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
$endgroup$
– Chris Custer
Dec 8 '18 at 4:55
$begingroup$
Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
$endgroup$
– Chris Custer
Dec 8 '18 at 4:55
$begingroup$
Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
$endgroup$
– Chris Custer
Dec 8 '18 at 4:55
add a comment |
2 Answers
2
active
oldest
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$begingroup$
An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.
In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.
answered Dec 8 '18 at 4:25
RandallRandall
9,31111129
$endgroup$
add a comment |
$begingroup$
I would take almost the same exemple than Randall,in S4 (symmetric group)
You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.
answered Jan 1 at 21:54
Cauchy is my masterCauchy is my master
92
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2 Answers
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oldest
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2 Answers
2
active
oldest
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$begingroup$
An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.
In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.
answered Dec 8 '18 at 4:25
RandallRandall
9,31111129
$endgroup$
add a comment |
$begingroup$
An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.
In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.
answered Dec 8 '18 at 4:25
RandallRandall
9,31111129
$endgroup$
add a comment |
$begingroup$
An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.
In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.
answered Dec 8 '18 at 4:25
RandallRandall
9,31111129
$endgroup$
An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely ${e, (12)}$, ${e, (13)}$ and ${e, (23)}$. All of them are isomorphic, of course.
In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.
answered Dec 8 '18 at 4:25
RandallRandall
9,31111129
edited Dec 8 '18 at 4:28
answered Dec 8 '18 at 4:25
RandallRandall
9,31111129
answered Dec 8 '18 at 4:25
RandallRandall
9,31111129
answered Dec 8 '18 at 4:25
RandallRandall
9,31111129
9,31111129
add a comment |
add a comment |
$begingroup$
I would take almost the same exemple than Randall,in S4 (symmetric group)
You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.
answered Jan 1 at 21:54
Cauchy is my masterCauchy is my master
92
$endgroup$
add a comment |
$begingroup$
I would take almost the same exemple than Randall,in S4 (symmetric group)
You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.
answered Jan 1 at 21:54
Cauchy is my masterCauchy is my master
92
$endgroup$
add a comment |
$begingroup$
I would take almost the same exemple than Randall,in S4 (symmetric group)
You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.
answered Jan 1 at 21:54
Cauchy is my masterCauchy is my master
92
$endgroup$
I would take almost the same exemple than Randall,in S4 (symmetric group)
You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.
answered Jan 1 at 21:54
Cauchy is my masterCauchy is my master
92
answered Jan 1 at 21:54
Cauchy is my masterCauchy is my master
92
answered Jan 1 at 21:54
Cauchy is my masterCauchy is my master
92
answered Jan 1 at 21:54
Cauchy is my masterCauchy is my master
92
92
add a comment |
add a comment |
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$begingroup$
Note that since the Sylow subgroup is of order $p^n$ for the largest power of $p$ that divides the order of the group, it couldn't be a proper subgroup of any other $p$-subgroup (by order considerations)- hence the maximality feature.
$endgroup$
– Chris Custer
Dec 8 '18 at 4:55