How many subgroups of order 17 does $S_{17}$ have?
$begingroup$
How many subgroups of order 17 does $S_{17}$ have ?
My attempt :
An order 17 group is of prime order, hence cyclic and each element in it is a generator and of order 17.
In $S_{17}$ group we can get an order 17 element only through a 17-cycle.
Number of elements of order 17 in $S_{17}$ is $frac{17!}{17} = 16!$.
Now given that two sylow 17 subgroups have only a trivial intersection. We can conclude that 16 of these elements fall into each sylow 17 subgroup.
Hence the number of sylow 17 subgroups would be $frac{16!}{16} = 15!$
abstract-algebra group-theory proof-verification finite-groups sylow-theory
asked Mar 19 '17 at 14:34
spaceman_spiffspaceman_spiff
61339
$endgroup$
add a comment |
$begingroup$
How many subgroups of order 17 does $S_{17}$ have ?
My attempt :
An order 17 group is of prime order, hence cyclic and each element in it is a generator and of order 17.
In $S_{17}$ group we can get an order 17 element only through a 17-cycle.
Number of elements of order 17 in $S_{17}$ is $frac{17!}{17} = 16!$.
Now given that two sylow 17 subgroups have only a trivial intersection. We can conclude that 16 of these elements fall into each sylow 17 subgroup.
Hence the number of sylow 17 subgroups would be $frac{16!}{16} = 15!$
abstract-algebra group-theory proof-verification finite-groups sylow-theory
asked Mar 19 '17 at 14:34
spaceman_spiffspaceman_spiff
61339
$endgroup$
5
$begingroup$
Looks ok to me. But surely you meant to write $17!/17=16!$ Bonus question: Can you think of another (=non-Sylow) way of proving that $(p-2)!equiv1pmod p$ for an odd prime $p$?
$endgroup$
– Jyrki Lahtonen
Mar 19 '17 at 14:43
2
$begingroup$
@JyrkiLahtonen Isn't it wilsons theorem in case of odd prime?.. thanks for the insight ...
$endgroup$
– spaceman_spiff
Mar 20 '17 at 4:09
2
$begingroup$
Correct (at least that's the idea I had). Goes nicely together with Sylow, doesn't it!
$endgroup$
– Jyrki Lahtonen
Mar 20 '17 at 5:33
add a comment |
$begingroup$
How many subgroups of order 17 does $S_{17}$ have ?
My attempt :
An order 17 group is of prime order, hence cyclic and each element in it is a generator and of order 17.
In $S_{17}$ group we can get an order 17 element only through a 17-cycle.
Number of elements of order 17 in $S_{17}$ is $frac{17!}{17} = 16!$.
Now given that two sylow 17 subgroups have only a trivial intersection. We can conclude that 16 of these elements fall into each sylow 17 subgroup.
Hence the number of sylow 17 subgroups would be $frac{16!}{16} = 15!$
abstract-algebra group-theory proof-verification finite-groups sylow-theory
asked Mar 19 '17 at 14:34
spaceman_spiffspaceman_spiff
61339
$endgroup$
How many subgroups of order 17 does $S_{17}$ have ?
My attempt :
An order 17 group is of prime order, hence cyclic and each element in it is a generator and of order 17.
In $S_{17}$ group we can get an order 17 element only through a 17-cycle.
Number of elements of order 17 in $S_{17}$ is $frac{17!}{17} = 16!$.
Now given that two sylow 17 subgroups have only a trivial intersection. We can conclude that 16 of these elements fall into each sylow 17 subgroup.
Hence the number of sylow 17 subgroups would be $frac{16!}{16} = 15!$
abstract-algebra group-theory proof-verification finite-groups sylow-theory
abstract-algebra group-theory proof-verification finite-groups sylow-theory
asked Mar 19 '17 at 14:34
spaceman_spiffspaceman_spiff
61339
asked Mar 19 '17 at 14:34
spaceman_spiffspaceman_spiff
61339
edited Mar 19 '17 at 15:57
spaceman_spiff
asked Mar 19 '17 at 14:34
spaceman_spiffspaceman_spiff
61339
asked Mar 19 '17 at 14:34
spaceman_spiffspaceman_spiff
61339
asked Mar 19 '17 at 14:34
spaceman_spiffspaceman_spiff
61339
61339
5
$begingroup$
Looks ok to me. But surely you meant to write $17!/17=16!$ Bonus question: Can you think of another (=non-Sylow) way of proving that $(p-2)!equiv1pmod p$ for an odd prime $p$?
$endgroup$
– Jyrki Lahtonen
Mar 19 '17 at 14:43
2
$begingroup$
@JyrkiLahtonen Isn't it wilsons theorem in case of odd prime?.. thanks for the insight ...
$endgroup$
– spaceman_spiff
Mar 20 '17 at 4:09
2
$begingroup$
Correct (at least that's the idea I had). Goes nicely together with Sylow, doesn't it!
$endgroup$
– Jyrki Lahtonen
Mar 20 '17 at 5:33
add a comment |
5
$begingroup$
Looks ok to me. But surely you meant to write $17!/17=16!$ Bonus question: Can you think of another (=non-Sylow) way of proving that $(p-2)!equiv1pmod p$ for an odd prime $p$?
$endgroup$
– Jyrki Lahtonen
Mar 19 '17 at 14:43
2
$begingroup$
@JyrkiLahtonen Isn't it wilsons theorem in case of odd prime?.. thanks for the insight ...
$endgroup$
– spaceman_spiff
Mar 20 '17 at 4:09
2
$begingroup$
Correct (at least that's the idea I had). Goes nicely together with Sylow, doesn't it!
$endgroup$
– Jyrki Lahtonen
Mar 20 '17 at 5:33
5
5
$begingroup$
Looks ok to me. But surely you meant to write $17!/17=16!$ Bonus question: Can you think of another (=non-Sylow) way of proving that $(p-2)!equiv1pmod p$ for an odd prime $p$?
$endgroup$
– Jyrki Lahtonen
Mar 19 '17 at 14:43
$begingroup$
Looks ok to me. But surely you meant to write $17!/17=16!$ Bonus question: Can you think of another (=non-Sylow) way of proving that $(p-2)!equiv1pmod p$ for an odd prime $p$?
$endgroup$
– Jyrki Lahtonen
Mar 19 '17 at 14:43
2
2
$begingroup$
@JyrkiLahtonen Isn't it wilsons theorem in case of odd prime?.. thanks for the insight ...
$endgroup$
– spaceman_spiff
Mar 20 '17 at 4:09
$begingroup$
@JyrkiLahtonen Isn't it wilsons theorem in case of odd prime?.. thanks for the insight ...
$endgroup$
– spaceman_spiff
Mar 20 '17 at 4:09
2
2
$begingroup$
Correct (at least that's the idea I had). Goes nicely together with Sylow, doesn't it!
$endgroup$
– Jyrki Lahtonen
Mar 20 '17 at 5:33
$begingroup$
Correct (at least that's the idea I had). Goes nicely together with Sylow, doesn't it!
$endgroup$
– Jyrki Lahtonen
Mar 20 '17 at 5:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your answer looks correct.
It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.
An alternative (but not necessarily better) proof is as follows:
Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.
For each subgroup $G$ fix $gin G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.
Write $g=(1,2,x_3,ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.
answered Mar 19 '17 at 15:10
Robert ChamberlainRobert Chamberlain
4,0971521
$endgroup$
add a comment |
$begingroup$
Prove that the number of $p-$Sylow subgroups in the symmetric group $S_p$ is
$(p − 2)!$.
Proof : Any $p-$Sylow subgroup is cyclic of order $p$ and has precisely $p − 1 $generators.
Moreover, if two $p-$Sylow subgroups share a generator, they are identical. So, the elements
of order p are partitioned according to which p-Sylow subgroup they belong to. We need to
count the number of elements of order exactly p. This is precisely the number of distinct
$p-$cycles, which is $p!/p = (p−1)!$. Grouping them into distinct $p-$Sylow subgroups (with $p−1$ in each clump), we see that the number of $p-$Sylow subgroups is $(p − 1)!/(p − 1) = (p − 2)!.$
Now take take $p= 17$ ,then
Number of $p-$Sylow subgroups in the symmetric group $S_{17}$ is
$(17 − 2)!=15!$.
answered Jan 4 at 21:38
jasminejasmine
1,701417
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Your answer looks correct.
It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.
An alternative (but not necessarily better) proof is as follows:
Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.
For each subgroup $G$ fix $gin G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.
Write $g=(1,2,x_3,ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.
answered Mar 19 '17 at 15:10
Robert ChamberlainRobert Chamberlain
4,0971521
$endgroup$
add a comment |
$begingroup$
Your answer looks correct.
It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.
An alternative (but not necessarily better) proof is as follows:
Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.
For each subgroup $G$ fix $gin G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.
Write $g=(1,2,x_3,ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.
answered Mar 19 '17 at 15:10
Robert ChamberlainRobert Chamberlain
4,0971521
$endgroup$
add a comment |
$begingroup$
Your answer looks correct.
It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.
An alternative (but not necessarily better) proof is as follows:
Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.
For each subgroup $G$ fix $gin G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.
Write $g=(1,2,x_3,ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.
answered Mar 19 '17 at 15:10
Robert ChamberlainRobert Chamberlain
4,0971521
$endgroup$
Your answer looks correct.
It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.
An alternative (but not necessarily better) proof is as follows:
Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.
For each subgroup $G$ fix $gin G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.
Write $g=(1,2,x_3,ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.
answered Mar 19 '17 at 15:10
Robert ChamberlainRobert Chamberlain
4,0971521
answered Mar 19 '17 at 15:10
Robert ChamberlainRobert Chamberlain
4,0971521
answered Mar 19 '17 at 15:10
Robert ChamberlainRobert Chamberlain
4,0971521
answered Mar 19 '17 at 15:10
Robert ChamberlainRobert Chamberlain
4,0971521
4,0971521
add a comment |
add a comment |
$begingroup$
Prove that the number of $p-$Sylow subgroups in the symmetric group $S_p$ is
$(p − 2)!$.
Proof : Any $p-$Sylow subgroup is cyclic of order $p$ and has precisely $p − 1 $generators.
Moreover, if two $p-$Sylow subgroups share a generator, they are identical. So, the elements
of order p are partitioned according to which p-Sylow subgroup they belong to. We need to
count the number of elements of order exactly p. This is precisely the number of distinct
$p-$cycles, which is $p!/p = (p−1)!$. Grouping them into distinct $p-$Sylow subgroups (with $p−1$ in each clump), we see that the number of $p-$Sylow subgroups is $(p − 1)!/(p − 1) = (p − 2)!.$
Now take take $p= 17$ ,then
Number of $p-$Sylow subgroups in the symmetric group $S_{17}$ is
$(17 − 2)!=15!$.
answered Jan 4 at 21:38
jasminejasmine
1,701417
$endgroup$
add a comment |
$begingroup$
Prove that the number of $p-$Sylow subgroups in the symmetric group $S_p$ is
$(p − 2)!$.
Proof : Any $p-$Sylow subgroup is cyclic of order $p$ and has precisely $p − 1 $generators.
Moreover, if two $p-$Sylow subgroups share a generator, they are identical. So, the elements
of order p are partitioned according to which p-Sylow subgroup they belong to. We need to
count the number of elements of order exactly p. This is precisely the number of distinct
$p-$cycles, which is $p!/p = (p−1)!$. Grouping them into distinct $p-$Sylow subgroups (with $p−1$ in each clump), we see that the number of $p-$Sylow subgroups is $(p − 1)!/(p − 1) = (p − 2)!.$
Now take take $p= 17$ ,then
Number of $p-$Sylow subgroups in the symmetric group $S_{17}$ is
$(17 − 2)!=15!$.
answered Jan 4 at 21:38
jasminejasmine
1,701417
$endgroup$
add a comment |
$begingroup$
Prove that the number of $p-$Sylow subgroups in the symmetric group $S_p$ is
$(p − 2)!$.
Proof : Any $p-$Sylow subgroup is cyclic of order $p$ and has precisely $p − 1 $generators.
Moreover, if two $p-$Sylow subgroups share a generator, they are identical. So, the elements
of order p are partitioned according to which p-Sylow subgroup they belong to. We need to
count the number of elements of order exactly p. This is precisely the number of distinct
$p-$cycles, which is $p!/p = (p−1)!$. Grouping them into distinct $p-$Sylow subgroups (with $p−1$ in each clump), we see that the number of $p-$Sylow subgroups is $(p − 1)!/(p − 1) = (p − 2)!.$
Now take take $p= 17$ ,then
Number of $p-$Sylow subgroups in the symmetric group $S_{17}$ is
$(17 − 2)!=15!$.
answered Jan 4 at 21:38
jasminejasmine
1,701417
$endgroup$
Prove that the number of $p-$Sylow subgroups in the symmetric group $S_p$ is
$(p − 2)!$.
Proof : Any $p-$Sylow subgroup is cyclic of order $p$ and has precisely $p − 1 $generators.
Moreover, if two $p-$Sylow subgroups share a generator, they are identical. So, the elements
of order p are partitioned according to which p-Sylow subgroup they belong to. We need to
count the number of elements of order exactly p. This is precisely the number of distinct
$p-$cycles, which is $p!/p = (p−1)!$. Grouping them into distinct $p-$Sylow subgroups (with $p−1$ in each clump), we see that the number of $p-$Sylow subgroups is $(p − 1)!/(p − 1) = (p − 2)!.$
Now take take $p= 17$ ,then
Number of $p-$Sylow subgroups in the symmetric group $S_{17}$ is
$(17 − 2)!=15!$.
answered Jan 4 at 21:38
jasminejasmine
1,701417
answered Jan 4 at 21:38
jasminejasmine
1,701417
answered Jan 4 at 21:38
jasminejasmine
1,701417
answered Jan 4 at 21:38
jasminejasmine
1,701417
1,701417
add a comment |
add a comment |
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5
$begingroup$
Looks ok to me. But surely you meant to write $17!/17=16!$ Bonus question: Can you think of another (=non-Sylow) way of proving that $(p-2)!equiv1pmod p$ for an odd prime $p$?
$endgroup$
– Jyrki Lahtonen
Mar 19 '17 at 14:43
2
$begingroup$
@JyrkiLahtonen Isn't it wilsons theorem in case of odd prime?.. thanks for the insight ...
$endgroup$
– spaceman_spiff
Mar 20 '17 at 4:09
2
$begingroup$
Correct (at least that's the idea I had). Goes nicely together with Sylow, doesn't it!
$endgroup$
– Jyrki Lahtonen
Mar 20 '17 at 5:33