Find the number of $2$-Sylow subgroups in $GL_2(mathbb F_5)$
$begingroup$
First, we can calculate the order of $GL_2(mathbb F_5)$ to find the order of the $2$-Sylow subgroups.
$$|GL_2(mathbb F_5)|=(5^2-1)(5^2-5)=24cdot 20=2^5cdot 3cdot 5$$
Thus we have $|P|=2^5$.
Now by using Sylow theorems, we can try to find $m$, the number of $2$-Sylow subgroups.
$$m | frac{|GL_2(mathbb F_5)|}{|P|}, m equiv 1 mod p$$
$$m | 3cdot 5,m equiv 1 mod 2 $$
The solution given in my textbook involves the third Sylow theorem, $m = left[ G : N _ { G } ( P ) right]$ and $m=6$. While this solution would be simpler since $N _ { G } ( P )$ is the group of upper triangular matrices, I was wondering where was the error in my reasoning. $6$ does not divide $3cdot 5$ so there must be an error.
EDIT: Here is the correction given by my teacher:
Taking as P the subgroup, we claim that $N _ { G } ( P ) = { T | T in mathrm { GL } _ { 2 } left( mathbb { F } _ { 5 } right)$ is upper triangular $}$.
$$
A U A ^ { - 1 } = ( a d - b c ) ^ { - 1 } left[ begin{array} { l l } { a } & { b } \ { c } & { d } end{array} right] left[ begin{array} { l l } { 1 } & { x } \ { 0 } & { 1 } end{array} right] left[ begin{array} { c c } { d } & { - b } \ { - c } & { a } end{array} right]
$$
$$
= ( a d - b c ) ^ { - 1 } left[ begin{array} { c c } { a d - b c - a c x } & { a ^ { 2 } x } \ { - c ^ { 2 } x } & { a d - b c + a c x } end{array} right]
$$
$$
= left[ begin{array} { c c } { 1 - acx( a d - b c ) ^ { - 1 } } & { a ^ { 2 } x ( a d - b c ) ^ { - 1 } } \ { - c ^ { 2 } x ( a d - b c ) ^ { - 1 } } & { 1 + a c x ( a d - b c ) ^ { - 1 } } end{array} right]
$$
$text { Therefore, } A U A ^ { - 1 } in P text { is equivalent to: }$
$$
left{ begin{array} { l l } { a c x = 0 } & { forall x in mathbb { F } _ { 5 } } \ { c ^ { 2 } x = 0 } & { forall x in mathbb { F } _ { 5 } } end{array} right.
$$
which, in turn, is equivalent to $c = 0$. $
text { Therefore, } N _ { G } ( P ) = { T | T in mathrm { G } mathrm { L } _ { 2 } left( mathbb { F } _ { 5 } right) text { is upper triangular } }
$
$$
m = left[ G : N _ { G } ( P ) right] = frac { | G | } { left| N _ { G } ( P ) right| } = frac { 480 } { 80 } = 6
$$
abstract-algebra group-theory finite-groups sylow-theory
edited Jan 8 at 1:43
the_fox
2,90031538
asked Jan 7 at 18:05
NotAbelianGroupNotAbelianGroup
18711
$endgroup$
|
show 5 more comments
$begingroup$
First, we can calculate the order of $GL_2(mathbb F_5)$ to find the order of the $2$-Sylow subgroups.
$$|GL_2(mathbb F_5)|=(5^2-1)(5^2-5)=24cdot 20=2^5cdot 3cdot 5$$
Thus we have $|P|=2^5$.
Now by using Sylow theorems, we can try to find $m$, the number of $2$-Sylow subgroups.
$$m | frac{|GL_2(mathbb F_5)|}{|P|}, m equiv 1 mod p$$
$$m | 3cdot 5,m equiv 1 mod 2 $$
The solution given in my textbook involves the third Sylow theorem, $m = left[ G : N _ { G } ( P ) right]$ and $m=6$. While this solution would be simpler since $N _ { G } ( P )$ is the group of upper triangular matrices, I was wondering where was the error in my reasoning. $6$ does not divide $3cdot 5$ so there must be an error.
EDIT: Here is the correction given by my teacher:
Taking as P the subgroup, we claim that $N _ { G } ( P ) = { T | T in mathrm { GL } _ { 2 } left( mathbb { F } _ { 5 } right)$ is upper triangular $}$.
$$
A U A ^ { - 1 } = ( a d - b c ) ^ { - 1 } left[ begin{array} { l l } { a } & { b } \ { c } & { d } end{array} right] left[ begin{array} { l l } { 1 } & { x } \ { 0 } & { 1 } end{array} right] left[ begin{array} { c c } { d } & { - b } \ { - c } & { a } end{array} right]
$$
$$
= ( a d - b c ) ^ { - 1 } left[ begin{array} { c c } { a d - b c - a c x } & { a ^ { 2 } x } \ { - c ^ { 2 } x } & { a d - b c + a c x } end{array} right]
$$
$$
= left[ begin{array} { c c } { 1 - acx( a d - b c ) ^ { - 1 } } & { a ^ { 2 } x ( a d - b c ) ^ { - 1 } } \ { - c ^ { 2 } x ( a d - b c ) ^ { - 1 } } & { 1 + a c x ( a d - b c ) ^ { - 1 } } end{array} right]
$$
$text { Therefore, } A U A ^ { - 1 } in P text { is equivalent to: }$
$$
left{ begin{array} { l l } { a c x = 0 } & { forall x in mathbb { F } _ { 5 } } \ { c ^ { 2 } x = 0 } & { forall x in mathbb { F } _ { 5 } } end{array} right.
$$
which, in turn, is equivalent to $c = 0$. $
text { Therefore, } N _ { G } ( P ) = { T | T in mathrm { G } mathrm { L } _ { 2 } left( mathbb { F } _ { 5 } right) text { is upper triangular } }
$
$$
m = left[ G : N _ { G } ( P ) right] = frac { | G | } { left| N _ { G } ( P ) right| } = frac { 480 } { 80 } = 6
$$
abstract-algebra group-theory finite-groups sylow-theory
edited Jan 8 at 1:43
the_fox
2,90031538
asked Jan 7 at 18:05
NotAbelianGroupNotAbelianGroup
18711
$endgroup$
$begingroup$
If they describe the subgroup of upper triangular matrices as the normalizer of a Sylow subgroup my educated guess is that there is a typo and they wanted you to figure out the number of Sylow $5$-subgroups. Anyway, you are right, the number of Sylow $2$-subgroups is always odd.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:15
$begingroup$
One Sylow $2$-subgroup of this group is the group of monomial matrices $M$. That is, invertible matrices with a single non-zero entry on both rows and columns. There will be $4cdot4=16$ diagonal matrices in $M$, and $16$ matrices with zeros on the diagonal.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:18
$begingroup$
@JyrkiLahtonen I added the correction given by the teacher. Your comment makes more sense, with $32$ we would have $480/32=15=3 cdot 5$ which is precisely the result required by both Sylow theorems
$endgroup$
– NotAbelianGroup
Jan 7 at 18:27
$begingroup$
Ok, the edit clinches it. The matrices of the form $$pmatrix{1&xcr0&1cr}$$ form a Sylow $5$-subgroup. After all, there are five choices for $x$. Your teacher is looking for the normalizer of that group.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:27
$begingroup$
But do check with your teacher. Is the question about Sylow-2 or Sylow-5 subgroups?
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:31
|
show 5 more comments
$begingroup$
First, we can calculate the order of $GL_2(mathbb F_5)$ to find the order of the $2$-Sylow subgroups.
$$|GL_2(mathbb F_5)|=(5^2-1)(5^2-5)=24cdot 20=2^5cdot 3cdot 5$$
Thus we have $|P|=2^5$.
Now by using Sylow theorems, we can try to find $m$, the number of $2$-Sylow subgroups.
$$m | frac{|GL_2(mathbb F_5)|}{|P|}, m equiv 1 mod p$$
$$m | 3cdot 5,m equiv 1 mod 2 $$
The solution given in my textbook involves the third Sylow theorem, $m = left[ G : N _ { G } ( P ) right]$ and $m=6$. While this solution would be simpler since $N _ { G } ( P )$ is the group of upper triangular matrices, I was wondering where was the error in my reasoning. $6$ does not divide $3cdot 5$ so there must be an error.
EDIT: Here is the correction given by my teacher:
Taking as P the subgroup, we claim that $N _ { G } ( P ) = { T | T in mathrm { GL } _ { 2 } left( mathbb { F } _ { 5 } right)$ is upper triangular $}$.
$$
A U A ^ { - 1 } = ( a d - b c ) ^ { - 1 } left[ begin{array} { l l } { a } & { b } \ { c } & { d } end{array} right] left[ begin{array} { l l } { 1 } & { x } \ { 0 } & { 1 } end{array} right] left[ begin{array} { c c } { d } & { - b } \ { - c } & { a } end{array} right]
$$
$$
= ( a d - b c ) ^ { - 1 } left[ begin{array} { c c } { a d - b c - a c x } & { a ^ { 2 } x } \ { - c ^ { 2 } x } & { a d - b c + a c x } end{array} right]
$$
$$
= left[ begin{array} { c c } { 1 - acx( a d - b c ) ^ { - 1 } } & { a ^ { 2 } x ( a d - b c ) ^ { - 1 } } \ { - c ^ { 2 } x ( a d - b c ) ^ { - 1 } } & { 1 + a c x ( a d - b c ) ^ { - 1 } } end{array} right]
$$
$text { Therefore, } A U A ^ { - 1 } in P text { is equivalent to: }$
$$
left{ begin{array} { l l } { a c x = 0 } & { forall x in mathbb { F } _ { 5 } } \ { c ^ { 2 } x = 0 } & { forall x in mathbb { F } _ { 5 } } end{array} right.
$$
which, in turn, is equivalent to $c = 0$. $
text { Therefore, } N _ { G } ( P ) = { T | T in mathrm { G } mathrm { L } _ { 2 } left( mathbb { F } _ { 5 } right) text { is upper triangular } }
$
$$
m = left[ G : N _ { G } ( P ) right] = frac { | G | } { left| N _ { G } ( P ) right| } = frac { 480 } { 80 } = 6
$$
abstract-algebra group-theory finite-groups sylow-theory
edited Jan 8 at 1:43
the_fox
2,90031538
asked Jan 7 at 18:05
NotAbelianGroupNotAbelianGroup
18711
$endgroup$
First, we can calculate the order of $GL_2(mathbb F_5)$ to find the order of the $2$-Sylow subgroups.
$$|GL_2(mathbb F_5)|=(5^2-1)(5^2-5)=24cdot 20=2^5cdot 3cdot 5$$
Thus we have $|P|=2^5$.
Now by using Sylow theorems, we can try to find $m$, the number of $2$-Sylow subgroups.
$$m | frac{|GL_2(mathbb F_5)|}{|P|}, m equiv 1 mod p$$
$$m | 3cdot 5,m equiv 1 mod 2 $$
The solution given in my textbook involves the third Sylow theorem, $m = left[ G : N _ { G } ( P ) right]$ and $m=6$. While this solution would be simpler since $N _ { G } ( P )$ is the group of upper triangular matrices, I was wondering where was the error in my reasoning. $6$ does not divide $3cdot 5$ so there must be an error.
EDIT: Here is the correction given by my teacher:
Taking as P the subgroup, we claim that $N _ { G } ( P ) = { T | T in mathrm { GL } _ { 2 } left( mathbb { F } _ { 5 } right)$ is upper triangular $}$.
$$
A U A ^ { - 1 } = ( a d - b c ) ^ { - 1 } left[ begin{array} { l l } { a } & { b } \ { c } & { d } end{array} right] left[ begin{array} { l l } { 1 } & { x } \ { 0 } & { 1 } end{array} right] left[ begin{array} { c c } { d } & { - b } \ { - c } & { a } end{array} right]
$$
$$
= ( a d - b c ) ^ { - 1 } left[ begin{array} { c c } { a d - b c - a c x } & { a ^ { 2 } x } \ { - c ^ { 2 } x } & { a d - b c + a c x } end{array} right]
$$
$$
= left[ begin{array} { c c } { 1 - acx( a d - b c ) ^ { - 1 } } & { a ^ { 2 } x ( a d - b c ) ^ { - 1 } } \ { - c ^ { 2 } x ( a d - b c ) ^ { - 1 } } & { 1 + a c x ( a d - b c ) ^ { - 1 } } end{array} right]
$$
$text { Therefore, } A U A ^ { - 1 } in P text { is equivalent to: }$
$$
left{ begin{array} { l l } { a c x = 0 } & { forall x in mathbb { F } _ { 5 } } \ { c ^ { 2 } x = 0 } & { forall x in mathbb { F } _ { 5 } } end{array} right.
$$
which, in turn, is equivalent to $c = 0$. $
text { Therefore, } N _ { G } ( P ) = { T | T in mathrm { G } mathrm { L } _ { 2 } left( mathbb { F } _ { 5 } right) text { is upper triangular } }
$
$$
m = left[ G : N _ { G } ( P ) right] = frac { | G | } { left| N _ { G } ( P ) right| } = frac { 480 } { 80 } = 6
$$
abstract-algebra group-theory finite-groups sylow-theory
abstract-algebra group-theory finite-groups sylow-theory
edited Jan 8 at 1:43
the_fox
2,90031538
asked Jan 7 at 18:05
NotAbelianGroupNotAbelianGroup
18711
edited Jan 8 at 1:43
the_fox
2,90031538
asked Jan 7 at 18:05
NotAbelianGroupNotAbelianGroup
18711
edited Jan 8 at 1:43
the_fox
2,90031538
edited Jan 8 at 1:43
the_fox
2,90031538
edited Jan 8 at 1:43
the_fox
2,90031538
2,90031538
asked Jan 7 at 18:05
NotAbelianGroupNotAbelianGroup
18711
asked Jan 7 at 18:05
NotAbelianGroupNotAbelianGroup
18711
asked Jan 7 at 18:05
NotAbelianGroupNotAbelianGroup
18711
18711
$begingroup$
If they describe the subgroup of upper triangular matrices as the normalizer of a Sylow subgroup my educated guess is that there is a typo and they wanted you to figure out the number of Sylow $5$-subgroups. Anyway, you are right, the number of Sylow $2$-subgroups is always odd.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:15
$begingroup$
One Sylow $2$-subgroup of this group is the group of monomial matrices $M$. That is, invertible matrices with a single non-zero entry on both rows and columns. There will be $4cdot4=16$ diagonal matrices in $M$, and $16$ matrices with zeros on the diagonal.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:18
$begingroup$
@JyrkiLahtonen I added the correction given by the teacher. Your comment makes more sense, with $32$ we would have $480/32=15=3 cdot 5$ which is precisely the result required by both Sylow theorems
$endgroup$
– NotAbelianGroup
Jan 7 at 18:27
$begingroup$
Ok, the edit clinches it. The matrices of the form $$pmatrix{1&xcr0&1cr}$$ form a Sylow $5$-subgroup. After all, there are five choices for $x$. Your teacher is looking for the normalizer of that group.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:27
$begingroup$
But do check with your teacher. Is the question about Sylow-2 or Sylow-5 subgroups?
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:31
|
show 5 more comments
$begingroup$
If they describe the subgroup of upper triangular matrices as the normalizer of a Sylow subgroup my educated guess is that there is a typo and they wanted you to figure out the number of Sylow $5$-subgroups. Anyway, you are right, the number of Sylow $2$-subgroups is always odd.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:15
$begingroup$
One Sylow $2$-subgroup of this group is the group of monomial matrices $M$. That is, invertible matrices with a single non-zero entry on both rows and columns. There will be $4cdot4=16$ diagonal matrices in $M$, and $16$ matrices with zeros on the diagonal.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:18
$begingroup$
@JyrkiLahtonen I added the correction given by the teacher. Your comment makes more sense, with $32$ we would have $480/32=15=3 cdot 5$ which is precisely the result required by both Sylow theorems
$endgroup$
– NotAbelianGroup
Jan 7 at 18:27
$begingroup$
Ok, the edit clinches it. The matrices of the form $$pmatrix{1&xcr0&1cr}$$ form a Sylow $5$-subgroup. After all, there are five choices for $x$. Your teacher is looking for the normalizer of that group.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:27
$begingroup$
But do check with your teacher. Is the question about Sylow-2 or Sylow-5 subgroups?
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:31
$begingroup$
If they describe the subgroup of upper triangular matrices as the normalizer of a Sylow subgroup my educated guess is that there is a typo and they wanted you to figure out the number of Sylow $5$-subgroups. Anyway, you are right, the number of Sylow $2$-subgroups is always odd.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:15
$begingroup$
If they describe the subgroup of upper triangular matrices as the normalizer of a Sylow subgroup my educated guess is that there is a typo and they wanted you to figure out the number of Sylow $5$-subgroups. Anyway, you are right, the number of Sylow $2$-subgroups is always odd.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:15
$begingroup$
One Sylow $2$-subgroup of this group is the group of monomial matrices $M$. That is, invertible matrices with a single non-zero entry on both rows and columns. There will be $4cdot4=16$ diagonal matrices in $M$, and $16$ matrices with zeros on the diagonal.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:18
$begingroup$
One Sylow $2$-subgroup of this group is the group of monomial matrices $M$. That is, invertible matrices with a single non-zero entry on both rows and columns. There will be $4cdot4=16$ diagonal matrices in $M$, and $16$ matrices with zeros on the diagonal.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:18
$begingroup$
@JyrkiLahtonen I added the correction given by the teacher. Your comment makes more sense, with $32$ we would have $480/32=15=3 cdot 5$ which is precisely the result required by both Sylow theorems
$endgroup$
– NotAbelianGroup
Jan 7 at 18:27
$begingroup$
@JyrkiLahtonen I added the correction given by the teacher. Your comment makes more sense, with $32$ we would have $480/32=15=3 cdot 5$ which is precisely the result required by both Sylow theorems
$endgroup$
– NotAbelianGroup
Jan 7 at 18:27
$begingroup$
Ok, the edit clinches it. The matrices of the form $$pmatrix{1&xcr0&1cr}$$ form a Sylow $5$-subgroup. After all, there are five choices for $x$. Your teacher is looking for the normalizer of that group.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:27
$begingroup$
Ok, the edit clinches it. The matrices of the form $$pmatrix{1&xcr0&1cr}$$ form a Sylow $5$-subgroup. After all, there are five choices for $x$. Your teacher is looking for the normalizer of that group.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:27
$begingroup$
But do check with your teacher. Is the question about Sylow-2 or Sylow-5 subgroups?
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:31
$begingroup$
But do check with your teacher. Is the question about Sylow-2 or Sylow-5 subgroups?
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:31
|
show 5 more comments
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$begingroup$
If they describe the subgroup of upper triangular matrices as the normalizer of a Sylow subgroup my educated guess is that there is a typo and they wanted you to figure out the number of Sylow $5$-subgroups. Anyway, you are right, the number of Sylow $2$-subgroups is always odd.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:15
$begingroup$
One Sylow $2$-subgroup of this group is the group of monomial matrices $M$. That is, invertible matrices with a single non-zero entry on both rows and columns. There will be $4cdot4=16$ diagonal matrices in $M$, and $16$ matrices with zeros on the diagonal.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:18
$begingroup$
@JyrkiLahtonen I added the correction given by the teacher. Your comment makes more sense, with $32$ we would have $480/32=15=3 cdot 5$ which is precisely the result required by both Sylow theorems
$endgroup$
– NotAbelianGroup
Jan 7 at 18:27
$begingroup$
Ok, the edit clinches it. The matrices of the form $$pmatrix{1&xcr0&1cr}$$ form a Sylow $5$-subgroup. After all, there are five choices for $x$. Your teacher is looking for the normalizer of that group.
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:27
$begingroup$
But do check with your teacher. Is the question about Sylow-2 or Sylow-5 subgroups?
$endgroup$
– Jyrki Lahtonen
Jan 7 at 18:31