Mixing
Finding a normal subgroup in $G := GL_2( mathbb F_7)$ [closed]
$begingroup$
For now, i know Sylow Groups and the theorem of structure for abelian groups.
$G := GL_2( mathbb F_7)$, $|G|=2^5 cdot 3^2 cdot 7 $
I m trying to show that:
There exists a normal sub group in $G$ of order $2^5 cdot 3 cdot 7$.
I have no clue at all. Any idea?
group-theory normal-subgroups
edited Jan 15 at 18:03
Surb
38.1k94375
asked Jan 15 at 17:58
Marine GalantinMarine Galantin
860316
$endgroup$
closed as off-topic by Derek Holt, Cesareo, ancientmathematician, José Carlos Santos, user91500 Jan 16 at 10:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Cesareo, ancientmathematician, José Carlos Santos, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
For now, i know Sylow Groups and the theorem of structure for abelian groups.
$G := GL_2( mathbb F_7)$, $|G|=2^5 cdot 3^2 cdot 7 $
I m trying to show that:
There exists a normal sub group in $G$ of order $2^5 cdot 3 cdot 7$.
I have no clue at all. Any idea?
group-theory normal-subgroups
edited Jan 15 at 18:03
Surb
38.1k94375
asked Jan 15 at 17:58
Marine GalantinMarine Galantin
860316
$endgroup$
closed as off-topic by Derek Holt, Cesareo, ancientmathematician, José Carlos Santos, user91500 Jan 16 at 10:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Cesareo, ancientmathematician, José Carlos Santos, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Hint: What is the most "obvious" subgroup (it will not be that one, but it will help).
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:00
$begingroup$
A kind of quotient? I tries to look at the matrix but it gave me no clue...
$endgroup$
– Marine Galantin
Jan 15 at 18:02
$begingroup$
What's the order of $SL_2(Bbb F_7)$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:04
$begingroup$
It is easier to find a normal subgroup of order 6=7-1
$endgroup$
– Thomas
Jan 15 at 18:31
add a comment |
$begingroup$
For now, i know Sylow Groups and the theorem of structure for abelian groups.
$G := GL_2( mathbb F_7)$, $|G|=2^5 cdot 3^2 cdot 7 $
I m trying to show that:
There exists a normal sub group in $G$ of order $2^5 cdot 3 cdot 7$.
I have no clue at all. Any idea?
group-theory normal-subgroups
edited Jan 15 at 18:03
Surb
38.1k94375
asked Jan 15 at 17:58
Marine GalantinMarine Galantin
860316
$endgroup$
For now, i know Sylow Groups and the theorem of structure for abelian groups.
$G := GL_2( mathbb F_7)$, $|G|=2^5 cdot 3^2 cdot 7 $
I m trying to show that:
There exists a normal sub group in $G$ of order $2^5 cdot 3 cdot 7$.
I have no clue at all. Any idea?
group-theory normal-subgroups
group-theory normal-subgroups
edited Jan 15 at 18:03
Surb
38.1k94375
asked Jan 15 at 17:58
Marine GalantinMarine Galantin
860316
edited Jan 15 at 18:03
Surb
38.1k94375
asked Jan 15 at 17:58
Marine GalantinMarine Galantin
860316
edited Jan 15 at 18:03
Surb
38.1k94375
edited Jan 15 at 18:03
Surb
38.1k94375
edited Jan 15 at 18:03
Surb
38.1k94375
38.1k94375
asked Jan 15 at 17:58
Marine GalantinMarine Galantin
860316
asked Jan 15 at 17:58
Marine GalantinMarine Galantin
860316
asked Jan 15 at 17:58
Marine GalantinMarine Galantin
860316
860316
closed as off-topic by Derek Holt, Cesareo, ancientmathematician, José Carlos Santos, user91500 Jan 16 at 10:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Cesareo, ancientmathematician, José Carlos Santos, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Derek Holt, Cesareo, ancientmathematician, José Carlos Santos, user91500 Jan 16 at 10:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Cesareo, ancientmathematician, José Carlos Santos, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Hint: What is the most "obvious" subgroup (it will not be that one, but it will help).
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:00
$begingroup$
A kind of quotient? I tries to look at the matrix but it gave me no clue...
$endgroup$
– Marine Galantin
Jan 15 at 18:02
$begingroup$
What's the order of $SL_2(Bbb F_7)$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:04
$begingroup$
It is easier to find a normal subgroup of order 6=7-1
$endgroup$
– Thomas
Jan 15 at 18:31
add a comment |
$begingroup$
Hint: What is the most "obvious" subgroup (it will not be that one, but it will help).
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:00
$begingroup$
A kind of quotient? I tries to look at the matrix but it gave me no clue...
$endgroup$
– Marine Galantin
Jan 15 at 18:02
$begingroup$
What's the order of $SL_2(Bbb F_7)$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:04
$begingroup$
It is easier to find a normal subgroup of order 6=7-1
$endgroup$
– Thomas
Jan 15 at 18:31
$begingroup$
Hint: What is the most "obvious" subgroup (it will not be that one, but it will help).
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:00
$begingroup$
Hint: What is the most "obvious" subgroup (it will not be that one, but it will help).
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:00
$begingroup$
A kind of quotient? I tries to look at the matrix but it gave me no clue...
$endgroup$
– Marine Galantin
Jan 15 at 18:02
$begingroup$
A kind of quotient? I tries to look at the matrix but it gave me no clue...
$endgroup$
– Marine Galantin
Jan 15 at 18:02
$begingroup$
What's the order of $SL_2(Bbb F_7)$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:04
$begingroup$
What's the order of $SL_2(Bbb F_7)$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:04
$begingroup$
It is easier to find a normal subgroup of order 6=7-1
$endgroup$
– Thomas
Jan 15 at 18:31
$begingroup$
It is easier to find a normal subgroup of order 6=7-1
$endgroup$
– Thomas
Jan 15 at 18:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Define $varphi:Gto mathbb{F_7}^times$ by $varphi(A)=detA$. This is a homomorphism, its image is $mathbb{F_7}^timescongmathbb{Z_6}$. Hence by an isomorphism theorem $G/Kervarphicongmathbb{Z_6}$. That means $G/Kervarphi$ has a normal subgroup of index $3$. And now from the correspondence theorem we get that $G$ must have a normal subgroup of index $3$. (a subgroup which contains $Kervarphi$, but it isn't important for your exercise)
answered Jan 15 at 19:19
MarkMark
8,379521
$endgroup$
add a comment |
$begingroup$
First consider $det:GL_2(7)rightarrow mathbb{F}_7^times$, where $det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$.
The kernel of $det$ is $SL_2(7)$ which contains matrices of determinant $1$. The map $det$ is a surjection onto $mathbb{F}_7^times$ which has order $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$. Clearly we need to find an overgroup of $SL_2(7)$ to find a subgroup of index $3$ of $GL_2(7)$.
Indeed we can, if we consider the group $G$ consisting of matrices with determinant equal to $1$ or $-1mod 7$. This is a proper subgroup (exercise) and is of index $3$.
Hint for exercise:
For the exercise consider what happens to the determinant when you multiply elements of determinant $1$ or $-1$.
It remains to show that $G$ is normal in $GL_2(7)$. We will do do this using the determinant map, let $hin GL_2(7)$ and $gin G$, then $det(hgh^{-1})=det(g)=1$ or $-1$ and so we conclude $gin G$. In particular $G$ is normal.
answered Jan 15 at 19:00
Sam HughesSam Hughes
526112
$endgroup$
1
$begingroup$
excuse me, could you please remind me how to ocunt the cardinal of $SL_n$ ? I know how to do it for the inversible matrices, but how are you doing this computation for matrices of det = 1?
$endgroup$
– Marine Galantin
Jan 15 at 19:34
1
$begingroup$
$det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$. The map is a surjection onto $mathbb{F}_7^times$ which is size $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$.
$endgroup$
– Sam Hughes
Jan 15 at 19:37
add a comment |
$begingroup$
Consider the set $H$ of elements $pi in GL_2(mathbb{F}_7)$ s.t. det$(pi) in {1,6}$ [note that ${1,6}$ is an index-3 subgroup of $(mathbb{F}_7)^{times}$. This is a group, is normal, and has index 3 in $GL_2(mathbb{F}_7)$.
We first show that $H$ is a group:
Let $pi in H$. Then $pi^{-1} in GL_2(mathbb{F}_7)$ and det($pi^{-1})$ must be det$^{-1}(pi)$, so det($pi^{-1})$ must be in ${1,6}$ as det$(pi)$ is. So $pi^{-1}$ is in $H$.
Likewise, if $pi$ and $pi'$ are in $H$, then $pipi'$ is in $GL_2(mathbb{F}_7)$. Furthermore, det$(pipi')$ $=$ det$(pi)$det$(pi')$ which is in ${1,6}$ as both det$(pi)$ and det$(pi')$ are in ${1,6}$.
So $H$ is a group. We now show that $H$ is normal. Letting $sigma$ be any element in $GL_2(mathbb{F}_7)$. Then $sigma^{-1}pisigma$ is also in $GL_2(mathbb{F}_7)$. Furthermore, det$(sigma^{-1}pisigma)$ $=$ det$(pi) in {1,6}$ so $sigma^{-1}pisigma$ is also in $H$. Thus $H$ is normal.
Can you check that $H$ has index 3 in $GL_2(mathbb{F}_7)$?
answered Jan 15 at 19:41
MikeMike
4,151412
$endgroup$
1
$begingroup$
Surely each of those statements requires a proof?
$endgroup$
– Sam Hughes
Jan 15 at 19:45
1
$begingroup$
Well, I did add the proofs even though I believe they really are straightforward.
$endgroup$
– Mike
Jan 15 at 19:57
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Define $varphi:Gto mathbb{F_7}^times$ by $varphi(A)=detA$. This is a homomorphism, its image is $mathbb{F_7}^timescongmathbb{Z_6}$. Hence by an isomorphism theorem $G/Kervarphicongmathbb{Z_6}$. That means $G/Kervarphi$ has a normal subgroup of index $3$. And now from the correspondence theorem we get that $G$ must have a normal subgroup of index $3$. (a subgroup which contains $Kervarphi$, but it isn't important for your exercise)
answered Jan 15 at 19:19
MarkMark
8,379521
$endgroup$
add a comment |
$begingroup$
Define $varphi:Gto mathbb{F_7}^times$ by $varphi(A)=detA$. This is a homomorphism, its image is $mathbb{F_7}^timescongmathbb{Z_6}$. Hence by an isomorphism theorem $G/Kervarphicongmathbb{Z_6}$. That means $G/Kervarphi$ has a normal subgroup of index $3$. And now from the correspondence theorem we get that $G$ must have a normal subgroup of index $3$. (a subgroup which contains $Kervarphi$, but it isn't important for your exercise)
answered Jan 15 at 19:19
MarkMark
8,379521
$endgroup$
add a comment |
$begingroup$
Define $varphi:Gto mathbb{F_7}^times$ by $varphi(A)=detA$. This is a homomorphism, its image is $mathbb{F_7}^timescongmathbb{Z_6}$. Hence by an isomorphism theorem $G/Kervarphicongmathbb{Z_6}$. That means $G/Kervarphi$ has a normal subgroup of index $3$. And now from the correspondence theorem we get that $G$ must have a normal subgroup of index $3$. (a subgroup which contains $Kervarphi$, but it isn't important for your exercise)
answered Jan 15 at 19:19
MarkMark
8,379521
$endgroup$
Define $varphi:Gto mathbb{F_7}^times$ by $varphi(A)=detA$. This is a homomorphism, its image is $mathbb{F_7}^timescongmathbb{Z_6}$. Hence by an isomorphism theorem $G/Kervarphicongmathbb{Z_6}$. That means $G/Kervarphi$ has a normal subgroup of index $3$. And now from the correspondence theorem we get that $G$ must have a normal subgroup of index $3$. (a subgroup which contains $Kervarphi$, but it isn't important for your exercise)
answered Jan 15 at 19:19
MarkMark
8,379521
answered Jan 15 at 19:19
MarkMark
8,379521
answered Jan 15 at 19:19
MarkMark
8,379521
answered Jan 15 at 19:19
MarkMark
8,379521
8,379521
add a comment |
add a comment |
$begingroup$
First consider $det:GL_2(7)rightarrow mathbb{F}_7^times$, where $det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$.
The kernel of $det$ is $SL_2(7)$ which contains matrices of determinant $1$. The map $det$ is a surjection onto $mathbb{F}_7^times$ which has order $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$. Clearly we need to find an overgroup of $SL_2(7)$ to find a subgroup of index $3$ of $GL_2(7)$.
Indeed we can, if we consider the group $G$ consisting of matrices with determinant equal to $1$ or $-1mod 7$. This is a proper subgroup (exercise) and is of index $3$.
Hint for exercise:
For the exercise consider what happens to the determinant when you multiply elements of determinant $1$ or $-1$.
It remains to show that $G$ is normal in $GL_2(7)$. We will do do this using the determinant map, let $hin GL_2(7)$ and $gin G$, then $det(hgh^{-1})=det(g)=1$ or $-1$ and so we conclude $gin G$. In particular $G$ is normal.
answered Jan 15 at 19:00
Sam HughesSam Hughes
526112
$endgroup$
1
$begingroup$
excuse me, could you please remind me how to ocunt the cardinal of $SL_n$ ? I know how to do it for the inversible matrices, but how are you doing this computation for matrices of det = 1?
$endgroup$
– Marine Galantin
Jan 15 at 19:34
1
$begingroup$
$det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$. The map is a surjection onto $mathbb{F}_7^times$ which is size $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$.
$endgroup$
– Sam Hughes
Jan 15 at 19:37
add a comment |
$begingroup$
First consider $det:GL_2(7)rightarrow mathbb{F}_7^times$, where $det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$.
The kernel of $det$ is $SL_2(7)$ which contains matrices of determinant $1$. The map $det$ is a surjection onto $mathbb{F}_7^times$ which has order $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$. Clearly we need to find an overgroup of $SL_2(7)$ to find a subgroup of index $3$ of $GL_2(7)$.
Indeed we can, if we consider the group $G$ consisting of matrices with determinant equal to $1$ or $-1mod 7$. This is a proper subgroup (exercise) and is of index $3$.
Hint for exercise:
For the exercise consider what happens to the determinant when you multiply elements of determinant $1$ or $-1$.
It remains to show that $G$ is normal in $GL_2(7)$. We will do do this using the determinant map, let $hin GL_2(7)$ and $gin G$, then $det(hgh^{-1})=det(g)=1$ or $-1$ and so we conclude $gin G$. In particular $G$ is normal.
answered Jan 15 at 19:00
Sam HughesSam Hughes
526112
$endgroup$
1
$begingroup$
excuse me, could you please remind me how to ocunt the cardinal of $SL_n$ ? I know how to do it for the inversible matrices, but how are you doing this computation for matrices of det = 1?
$endgroup$
– Marine Galantin
Jan 15 at 19:34
1
$begingroup$
$det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$. The map is a surjection onto $mathbb{F}_7^times$ which is size $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$.
$endgroup$
– Sam Hughes
Jan 15 at 19:37
add a comment |
$begingroup$
First consider $det:GL_2(7)rightarrow mathbb{F}_7^times$, where $det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$.
The kernel of $det$ is $SL_2(7)$ which contains matrices of determinant $1$. The map $det$ is a surjection onto $mathbb{F}_7^times$ which has order $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$. Clearly we need to find an overgroup of $SL_2(7)$ to find a subgroup of index $3$ of $GL_2(7)$.
Indeed we can, if we consider the group $G$ consisting of matrices with determinant equal to $1$ or $-1mod 7$. This is a proper subgroup (exercise) and is of index $3$.
Hint for exercise:
For the exercise consider what happens to the determinant when you multiply elements of determinant $1$ or $-1$.
It remains to show that $G$ is normal in $GL_2(7)$. We will do do this using the determinant map, let $hin GL_2(7)$ and $gin G$, then $det(hgh^{-1})=det(g)=1$ or $-1$ and so we conclude $gin G$. In particular $G$ is normal.
answered Jan 15 at 19:00
Sam HughesSam Hughes
526112
$endgroup$
First consider $det:GL_2(7)rightarrow mathbb{F}_7^times$, where $det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$.
The kernel of $det$ is $SL_2(7)$ which contains matrices of determinant $1$. The map $det$ is a surjection onto $mathbb{F}_7^times$ which has order $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$. Clearly we need to find an overgroup of $SL_2(7)$ to find a subgroup of index $3$ of $GL_2(7)$.
Indeed we can, if we consider the group $G$ consisting of matrices with determinant equal to $1$ or $-1mod 7$. This is a proper subgroup (exercise) and is of index $3$.
Hint for exercise:
For the exercise consider what happens to the determinant when you multiply elements of determinant $1$ or $-1$.
It remains to show that $G$ is normal in $GL_2(7)$. We will do do this using the determinant map, let $hin GL_2(7)$ and $gin G$, then $det(hgh^{-1})=det(g)=1$ or $-1$ and so we conclude $gin G$. In particular $G$ is normal.
answered Jan 15 at 19:00
Sam HughesSam Hughes
526112
edited Jan 15 at 20:06
answered Jan 15 at 19:00
Sam HughesSam Hughes
526112
answered Jan 15 at 19:00
Sam HughesSam Hughes
526112
answered Jan 15 at 19:00
Sam HughesSam Hughes
526112
526112
1
$begingroup$
excuse me, could you please remind me how to ocunt the cardinal of $SL_n$ ? I know how to do it for the inversible matrices, but how are you doing this computation for matrices of det = 1?
$endgroup$
– Marine Galantin
Jan 15 at 19:34
1
$begingroup$
$det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$. The map is a surjection onto $mathbb{F}_7^times$ which is size $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$.
$endgroup$
– Sam Hughes
Jan 15 at 19:37
add a comment |
1
$begingroup$
excuse me, could you please remind me how to ocunt the cardinal of $SL_n$ ? I know how to do it for the inversible matrices, but how are you doing this computation for matrices of det = 1?
$endgroup$
– Marine Galantin
Jan 15 at 19:34
1
$begingroup$
$det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$. The map is a surjection onto $mathbb{F}_7^times$ which is size $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$.
$endgroup$
– Sam Hughes
Jan 15 at 19:37
1
1
$begingroup$
excuse me, could you please remind me how to ocunt the cardinal of $SL_n$ ? I know how to do it for the inversible matrices, but how are you doing this computation for matrices of det = 1?
$endgroup$
– Marine Galantin
Jan 15 at 19:34
$begingroup$
excuse me, could you please remind me how to ocunt the cardinal of $SL_n$ ? I know how to do it for the inversible matrices, but how are you doing this computation for matrices of det = 1?
$endgroup$
– Marine Galantin
Jan 15 at 19:34
1
1
$begingroup$
$det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$. The map is a surjection onto $mathbb{F}_7^times$ which is size $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$.
$endgroup$
– Sam Hughes
Jan 15 at 19:37
$begingroup$
$det$ is just the determinate of the matrix so for $begin{bmatrix}a&b\ c&dend{bmatrix}$ it would be $ad-bc$. The map is a surjection onto $mathbb{F}_7^times$ which is size $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$.
$endgroup$
– Sam Hughes
Jan 15 at 19:37
add a comment |
$begingroup$
Consider the set $H$ of elements $pi in GL_2(mathbb{F}_7)$ s.t. det$(pi) in {1,6}$ [note that ${1,6}$ is an index-3 subgroup of $(mathbb{F}_7)^{times}$. This is a group, is normal, and has index 3 in $GL_2(mathbb{F}_7)$.
We first show that $H$ is a group:
Let $pi in H$. Then $pi^{-1} in GL_2(mathbb{F}_7)$ and det($pi^{-1})$ must be det$^{-1}(pi)$, so det($pi^{-1})$ must be in ${1,6}$ as det$(pi)$ is. So $pi^{-1}$ is in $H$.
Likewise, if $pi$ and $pi'$ are in $H$, then $pipi'$ is in $GL_2(mathbb{F}_7)$. Furthermore, det$(pipi')$ $=$ det$(pi)$det$(pi')$ which is in ${1,6}$ as both det$(pi)$ and det$(pi')$ are in ${1,6}$.
So $H$ is a group. We now show that $H$ is normal. Letting $sigma$ be any element in $GL_2(mathbb{F}_7)$. Then $sigma^{-1}pisigma$ is also in $GL_2(mathbb{F}_7)$. Furthermore, det$(sigma^{-1}pisigma)$ $=$ det$(pi) in {1,6}$ so $sigma^{-1}pisigma$ is also in $H$. Thus $H$ is normal.
Can you check that $H$ has index 3 in $GL_2(mathbb{F}_7)$?
answered Jan 15 at 19:41
MikeMike
4,151412
$endgroup$
1
$begingroup$
Surely each of those statements requires a proof?
$endgroup$
– Sam Hughes
Jan 15 at 19:45
1
$begingroup$
Well, I did add the proofs even though I believe they really are straightforward.
$endgroup$
– Mike
Jan 15 at 19:57
add a comment |
$begingroup$
Consider the set $H$ of elements $pi in GL_2(mathbb{F}_7)$ s.t. det$(pi) in {1,6}$ [note that ${1,6}$ is an index-3 subgroup of $(mathbb{F}_7)^{times}$. This is a group, is normal, and has index 3 in $GL_2(mathbb{F}_7)$.
We first show that $H$ is a group:
Let $pi in H$. Then $pi^{-1} in GL_2(mathbb{F}_7)$ and det($pi^{-1})$ must be det$^{-1}(pi)$, so det($pi^{-1})$ must be in ${1,6}$ as det$(pi)$ is. So $pi^{-1}$ is in $H$.
Likewise, if $pi$ and $pi'$ are in $H$, then $pipi'$ is in $GL_2(mathbb{F}_7)$. Furthermore, det$(pipi')$ $=$ det$(pi)$det$(pi')$ which is in ${1,6}$ as both det$(pi)$ and det$(pi')$ are in ${1,6}$.
So $H$ is a group. We now show that $H$ is normal. Letting $sigma$ be any element in $GL_2(mathbb{F}_7)$. Then $sigma^{-1}pisigma$ is also in $GL_2(mathbb{F}_7)$. Furthermore, det$(sigma^{-1}pisigma)$ $=$ det$(pi) in {1,6}$ so $sigma^{-1}pisigma$ is also in $H$. Thus $H$ is normal.
Can you check that $H$ has index 3 in $GL_2(mathbb{F}_7)$?
answered Jan 15 at 19:41
MikeMike
4,151412
$endgroup$
1
$begingroup$
Surely each of those statements requires a proof?
$endgroup$
– Sam Hughes
Jan 15 at 19:45
1
$begingroup$
Well, I did add the proofs even though I believe they really are straightforward.
$endgroup$
– Mike
Jan 15 at 19:57
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Consider the set $H$ of elements $pi in GL_2(mathbb{F}_7)$ s.t. det$(pi) in {1,6}$ [note that ${1,6}$ is an index-3 subgroup of $(mathbb{F}_7)^{times}$. This is a group, is normal, and has index 3 in $GL_2(mathbb{F}_7)$.
We first show that $H$ is a group:
Let $pi in H$. Then $pi^{-1} in GL_2(mathbb{F}_7)$ and det($pi^{-1})$ must be det$^{-1}(pi)$, so det($pi^{-1})$ must be in ${1,6}$ as det$(pi)$ is. So $pi^{-1}$ is in $H$.
Likewise, if $pi$ and $pi'$ are in $H$, then $pipi'$ is in $GL_2(mathbb{F}_7)$. Furthermore, det$(pipi')$ $=$ det$(pi)$det$(pi')$ which is in ${1,6}$ as both det$(pi)$ and det$(pi')$ are in ${1,6}$.
So $H$ is a group. We now show that $H$ is normal. Letting $sigma$ be any element in $GL_2(mathbb{F}_7)$. Then $sigma^{-1}pisigma$ is also in $GL_2(mathbb{F}_7)$. Furthermore, det$(sigma^{-1}pisigma)$ $=$ det$(pi) in {1,6}$ so $sigma^{-1}pisigma$ is also in $H$. Thus $H$ is normal.
Can you check that $H$ has index 3 in $GL_2(mathbb{F}_7)$?
answered Jan 15 at 19:41
MikeMike
4,151412
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Consider the set $H$ of elements $pi in GL_2(mathbb{F}_7)$ s.t. det$(pi) in {1,6}$ [note that ${1,6}$ is an index-3 subgroup of $(mathbb{F}_7)^{times}$. This is a group, is normal, and has index 3 in $GL_2(mathbb{F}_7)$.
We first show that $H$ is a group:
Let $pi in H$. Then $pi^{-1} in GL_2(mathbb{F}_7)$ and det($pi^{-1})$ must be det$^{-1}(pi)$, so det($pi^{-1})$ must be in ${1,6}$ as det$(pi)$ is. So $pi^{-1}$ is in $H$.
Likewise, if $pi$ and $pi'$ are in $H$, then $pipi'$ is in $GL_2(mathbb{F}_7)$. Furthermore, det$(pipi')$ $=$ det$(pi)$det$(pi')$ which is in ${1,6}$ as both det$(pi)$ and det$(pi')$ are in ${1,6}$.
So $H$ is a group. We now show that $H$ is normal. Letting $sigma$ be any element in $GL_2(mathbb{F}_7)$. Then $sigma^{-1}pisigma$ is also in $GL_2(mathbb{F}_7)$. Furthermore, det$(sigma^{-1}pisigma)$ $=$ det$(pi) in {1,6}$ so $sigma^{-1}pisigma$ is also in $H$. Thus $H$ is normal.
Can you check that $H$ has index 3 in $GL_2(mathbb{F}_7)$?
answered Jan 15 at 19:41
MikeMike
4,151412
edited Jan 15 at 22:56
answered Jan 15 at 19:41
MikeMike
4,151412
answered Jan 15 at 19:41
MikeMike
4,151412
answered Jan 15 at 19:41
MikeMike
4,151412
4,151412
1
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Surely each of those statements requires a proof?
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– Sam Hughes
Jan 15 at 19:45
1
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Well, I did add the proofs even though I believe they really are straightforward.
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– Mike
Jan 15 at 19:57
add a comment |
1
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Surely each of those statements requires a proof?
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– Sam Hughes
Jan 15 at 19:45
1
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Well, I did add the proofs even though I believe they really are straightforward.
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– Mike
Jan 15 at 19:57
1
1
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Surely each of those statements requires a proof?
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– Sam Hughes
Jan 15 at 19:45
$begingroup$
Surely each of those statements requires a proof?
$endgroup$
– Sam Hughes
Jan 15 at 19:45
1
1
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Well, I did add the proofs even though I believe they really are straightforward.
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– Mike
Jan 15 at 19:57
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Well, I did add the proofs even though I believe they really are straightforward.
$endgroup$
– Mike
Jan 15 at 19:57
add a comment |
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Hint: What is the most "obvious" subgroup (it will not be that one, but it will help).
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– Tobias Kildetoft
Jan 15 at 18:00
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A kind of quotient? I tries to look at the matrix but it gave me no clue...
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– Marine Galantin
Jan 15 at 18:02
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What's the order of $SL_2(Bbb F_7)$?
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– Lord Shark the Unknown
Jan 15 at 18:04
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It is easier to find a normal subgroup of order 6=7-1
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– Thomas
Jan 15 at 18:31