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$3$-Sylow subgroup of $SL(n, mathbbF_3)$
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Find a $3$-Sylow subgroup of $SL(n, mathbbF_3)$.
For $n=3$, how many elements are there in the center of this subgroup?
I tried to understand the order of the $3$-Sylow subgroups, but since we have a general $n$, I couldn't get to anything specific( I know the formula for $|SL(n, mathbbF_3)|$).
Without understanding the order of that sylow subgroup, I can't move any further, as that is the most basic assumption to continue, unless I don't see something.
Any ideas?
group-theory sylow-theory
asked Aug 6 at 20:36
ChikChak
650216
add a comment |Â
up vote
1
down vote
favorite
Find a $3$-Sylow subgroup of $SL(n, mathbbF_3)$.
For $n=3$, how many elements are there in the center of this subgroup?
I tried to understand the order of the $3$-Sylow subgroups, but since we have a general $n$, I couldn't get to anything specific( I know the formula for $|SL(n, mathbbF_3)|$).
Without understanding the order of that sylow subgroup, I can't move any further, as that is the most basic assumption to continue, unless I don't see something.
Any ideas?
group-theory sylow-theory
asked Aug 6 at 20:36
ChikChak
650216
The upper unitriangular matrices form a Sylow $p$-subgroup of $rm SL(n,p)$ for any $n ge 1$ and any prime $p$. That should enable you to work out its order. Its centre has the same order for all $n$.
– Derek Holt
Aug 6 at 21:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find a $3$-Sylow subgroup of $SL(n, mathbbF_3)$.
For $n=3$, how many elements are there in the center of this subgroup?
I tried to understand the order of the $3$-Sylow subgroups, but since we have a general $n$, I couldn't get to anything specific( I know the formula for $|SL(n, mathbbF_3)|$).
Without understanding the order of that sylow subgroup, I can't move any further, as that is the most basic assumption to continue, unless I don't see something.
Any ideas?
group-theory sylow-theory
asked Aug 6 at 20:36
ChikChak
650216
Find a $3$-Sylow subgroup of $SL(n, mathbbF_3)$.
For $n=3$, how many elements are there in the center of this subgroup?
I tried to understand the order of the $3$-Sylow subgroups, but since we have a general $n$, I couldn't get to anything specific( I know the formula for $|SL(n, mathbbF_3)|$).
Without understanding the order of that sylow subgroup, I can't move any further, as that is the most basic assumption to continue, unless I don't see something.
Any ideas?
group-theory sylow-theory
asked Aug 6 at 20:36
ChikChak
650216
asked Aug 6 at 20:36
ChikChak
650216
asked Aug 6 at 20:36
ChikChak
650216
asked Aug 6 at 20:36
ChikChak
650216
650216
The upper unitriangular matrices form a Sylow $p$-subgroup of $rm SL(n,p)$ for any $n ge 1$ and any prime $p$. That should enable you to work out its order. Its centre has the same order for all $n$.
– Derek Holt
Aug 6 at 21:01
add a comment |Â
The upper unitriangular matrices form a Sylow $p$-subgroup of $rm SL(n,p)$ for any $n ge 1$ and any prime $p$. That should enable you to work out its order. Its centre has the same order for all $n$.
– Derek Holt
Aug 6 at 21:01
The upper unitriangular matrices form a Sylow $p$-subgroup of $rm SL(n,p)$ for any $n ge 1$ and any prime $p$. That should enable you to work out its order. Its centre has the same order for all $n$.
– Derek Holt
Aug 6 at 21:01
The upper unitriangular matrices form a Sylow $p$-subgroup of $rm SL(n,p)$ for any $n ge 1$ and any prime $p$. That should enable you to work out its order. Its centre has the same order for all $n$.
– Derek Holt
Aug 6 at 21:01
add a comment |Â
1 Answer
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We have
$$left|SL(3,Bbb F_3)right|=frac13-1(3^3-1)(3^3-3)(3^3-3^2)=13cdot24cdot18=3^3cdot2^4cdot13cdot$$
and thus you need a subgroup of order $;3^3=27;$ (any such subgroup. Why?).
Check the following:
$$H:=left;beginpmatrix
1&a&b\
0&1&c\
0&0&1
endpmatrix;$$
answered Aug 6 at 20:59
DonAntonio
173k1484218
The assumption of $n=3$ is for the second question only.
– ChikChak
Aug 6 at 21:01
@ChikChak It doesn't matter: generalize the above and work on it.
– DonAntonio
Aug 6 at 21:02
Ho w? As the first thing you did, is calculating assuming some $n$.
– ChikChak
Aug 6 at 21:08
You need to calculate the order of $SL(n, mathbbF_3)$ for a general $n$. Start from calculating the order of $GL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:12
1
Correct. So how many times can you divide it by $3$? $3$ doesn't divide $(3^n-1)$ at all, divides $(3^n-3)$ one time, divides $(3^n-3^2)$ twice, and so on. Hence you get that the order of the 3-Sylow subgroups are $3^1+2+3+...+(n-1)$. As for really finding a Sylow subgroup-yes, you have to guess here, but there are reasons to think about diagonal matrices because we don't know too much subgroups of $SL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:19
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
We have
$$left|SL(3,Bbb F_3)right|=frac13-1(3^3-1)(3^3-3)(3^3-3^2)=13cdot24cdot18=3^3cdot2^4cdot13cdot$$
and thus you need a subgroup of order $;3^3=27;$ (any such subgroup. Why?).
Check the following:
$$H:=left;beginpmatrix
1&a&b\
0&1&c\
0&0&1
endpmatrix;$$
answered Aug 6 at 20:59
DonAntonio
173k1484218
The assumption of $n=3$ is for the second question only.
– ChikChak
Aug 6 at 21:01
@ChikChak It doesn't matter: generalize the above and work on it.
– DonAntonio
Aug 6 at 21:02
Ho w? As the first thing you did, is calculating assuming some $n$.
– ChikChak
Aug 6 at 21:08
You need to calculate the order of $SL(n, mathbbF_3)$ for a general $n$. Start from calculating the order of $GL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:12
1
Correct. So how many times can you divide it by $3$? $3$ doesn't divide $(3^n-1)$ at all, divides $(3^n-3)$ one time, divides $(3^n-3^2)$ twice, and so on. Hence you get that the order of the 3-Sylow subgroups are $3^1+2+3+...+(n-1)$. As for really finding a Sylow subgroup-yes, you have to guess here, but there are reasons to think about diagonal matrices because we don't know too much subgroups of $SL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:19
 |Â
show 1 more comment
up vote
3
down vote
We have
$$left|SL(3,Bbb F_3)right|=frac13-1(3^3-1)(3^3-3)(3^3-3^2)=13cdot24cdot18=3^3cdot2^4cdot13cdot$$
and thus you need a subgroup of order $;3^3=27;$ (any such subgroup. Why?).
Check the following:
$$H:=left;beginpmatrix
1&a&b\
0&1&c\
0&0&1
endpmatrix;$$
answered Aug 6 at 20:59
DonAntonio
173k1484218
The assumption of $n=3$ is for the second question only.
– ChikChak
Aug 6 at 21:01
@ChikChak It doesn't matter: generalize the above and work on it.
– DonAntonio
Aug 6 at 21:02
Ho w? As the first thing you did, is calculating assuming some $n$.
– ChikChak
Aug 6 at 21:08
You need to calculate the order of $SL(n, mathbbF_3)$ for a general $n$. Start from calculating the order of $GL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:12
1
Correct. So how many times can you divide it by $3$? $3$ doesn't divide $(3^n-1)$ at all, divides $(3^n-3)$ one time, divides $(3^n-3^2)$ twice, and so on. Hence you get that the order of the 3-Sylow subgroups are $3^1+2+3+...+(n-1)$. As for really finding a Sylow subgroup-yes, you have to guess here, but there are reasons to think about diagonal matrices because we don't know too much subgroups of $SL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:19
 |Â
show 1 more comment
up vote
3
down vote
up vote
3
down vote
We have
$$left|SL(3,Bbb F_3)right|=frac13-1(3^3-1)(3^3-3)(3^3-3^2)=13cdot24cdot18=3^3cdot2^4cdot13cdot$$
and thus you need a subgroup of order $;3^3=27;$ (any such subgroup. Why?).
Check the following:
$$H:=left;beginpmatrix
1&a&b\
0&1&c\
0&0&1
endpmatrix;$$
answered Aug 6 at 20:59
DonAntonio
173k1484218
We have
$$left|SL(3,Bbb F_3)right|=frac13-1(3^3-1)(3^3-3)(3^3-3^2)=13cdot24cdot18=3^3cdot2^4cdot13cdot$$
and thus you need a subgroup of order $;3^3=27;$ (any such subgroup. Why?).
Check the following:
$$H:=left;beginpmatrix
1&a&b\
0&1&c\
0&0&1
endpmatrix;$$
answered Aug 6 at 20:59
DonAntonio
173k1484218
answered Aug 6 at 20:59
DonAntonio
173k1484218
answered Aug 6 at 20:59
DonAntonio
173k1484218
answered Aug 6 at 20:59
DonAntonio
173k1484218
173k1484218
The assumption of $n=3$ is for the second question only.
– ChikChak
Aug 6 at 21:01
@ChikChak It doesn't matter: generalize the above and work on it.
– DonAntonio
Aug 6 at 21:02
Ho w? As the first thing you did, is calculating assuming some $n$.
– ChikChak
Aug 6 at 21:08
You need to calculate the order of $SL(n, mathbbF_3)$ for a general $n$. Start from calculating the order of $GL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:12
1
Correct. So how many times can you divide it by $3$? $3$ doesn't divide $(3^n-1)$ at all, divides $(3^n-3)$ one time, divides $(3^n-3^2)$ twice, and so on. Hence you get that the order of the 3-Sylow subgroups are $3^1+2+3+...+(n-1)$. As for really finding a Sylow subgroup-yes, you have to guess here, but there are reasons to think about diagonal matrices because we don't know too much subgroups of $SL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:19
 |Â
show 1 more comment
The assumption of $n=3$ is for the second question only.
– ChikChak
Aug 6 at 21:01
@ChikChak It doesn't matter: generalize the above and work on it.
– DonAntonio
Aug 6 at 21:02
Ho w? As the first thing you did, is calculating assuming some $n$.
– ChikChak
Aug 6 at 21:08
You need to calculate the order of $SL(n, mathbbF_3)$ for a general $n$. Start from calculating the order of $GL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:12
1
Correct. So how many times can you divide it by $3$? $3$ doesn't divide $(3^n-1)$ at all, divides $(3^n-3)$ one time, divides $(3^n-3^2)$ twice, and so on. Hence you get that the order of the 3-Sylow subgroups are $3^1+2+3+...+(n-1)$. As for really finding a Sylow subgroup-yes, you have to guess here, but there are reasons to think about diagonal matrices because we don't know too much subgroups of $SL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:19
The assumption of $n=3$ is for the second question only.
– ChikChak
Aug 6 at 21:01
The assumption of $n=3$ is for the second question only.
– ChikChak
Aug 6 at 21:01
@ChikChak It doesn't matter: generalize the above and work on it.
– DonAntonio
Aug 6 at 21:02
@ChikChak It doesn't matter: generalize the above and work on it.
– DonAntonio
Aug 6 at 21:02
Ho w? As the first thing you did, is calculating assuming some $n$.
– ChikChak
Aug 6 at 21:08
Ho w? As the first thing you did, is calculating assuming some $n$.
– ChikChak
Aug 6 at 21:08
You need to calculate the order of $SL(n, mathbbF_3)$ for a general $n$. Start from calculating the order of $GL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:12
You need to calculate the order of $SL(n, mathbbF_3)$ for a general $n$. Start from calculating the order of $GL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:12
1
1
Correct. So how many times can you divide it by $3$? $3$ doesn't divide $(3^n-1)$ at all, divides $(3^n-3)$ one time, divides $(3^n-3^2)$ twice, and so on. Hence you get that the order of the 3-Sylow subgroups are $3^1+2+3+...+(n-1)$. As for really finding a Sylow subgroup-yes, you have to guess here, but there are reasons to think about diagonal matrices because we don't know too much subgroups of $SL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:19
Correct. So how many times can you divide it by $3$? $3$ doesn't divide $(3^n-1)$ at all, divides $(3^n-3)$ one time, divides $(3^n-3^2)$ twice, and so on. Hence you get that the order of the 3-Sylow subgroups are $3^1+2+3+...+(n-1)$. As for really finding a Sylow subgroup-yes, you have to guess here, but there are reasons to think about diagonal matrices because we don't know too much subgroups of $SL(n, mathbbF_3)$.
– Mark
Aug 6 at 21:19
 |Â
show 1 more comment
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The upper unitriangular matrices form a Sylow $p$-subgroup of $rm SL(n,p)$ for any $n ge 1$ and any prime $p$. That should enable you to work out its order. Its centre has the same order for all $n$.
– Derek Holt
Aug 6 at 21:01