$3$-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




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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















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




share|cite|improve this question



















  • 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













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




share|cite|improve this question












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





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









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

















  • 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











1 Answer
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3
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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:



$$H:=left; a,b, cinBbb F_3;right$$






share|cite|improve this answer





















  • 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











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1 Answer
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active

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1 Answer
1






active

oldest

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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:



$$H:=left; a,b, cinBbb F_3;right$$






share|cite|improve this answer





















  • 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















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:



$$H:=left; a,b, cinBbb F_3;right$$






share|cite|improve this answer





















  • 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













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:



$$H:=left; a,b, cinBbb F_3;right$$






share|cite|improve this answer













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:



$$H:=left; a,b, cinBbb F_3;right$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











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

















  • 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













 

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