Show $Gl_3(mathbb{F}_5)$ has a normal subgroup of index 4
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Show $Gl_3(mathbb{F}_5)$ has a normal subgroup of index 4.
Here is my attempt: $|Gl_3(mathbb{F}_5)|=(5^3-1)(5^3-5)(5^3-5^2)=124*120*100=1488000=2^7*3*5^3*31$. So I'm stuck.
abstract-algebra group-theory finite-groups
asked Nov 16 at 0:30
mathnoob
1,043115
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up vote
2
down vote
favorite
Show $Gl_3(mathbb{F}_5)$ has a normal subgroup of index 4.
Here is my attempt: $|Gl_3(mathbb{F}_5)|=(5^3-1)(5^3-5)(5^3-5^2)=124*120*100=1488000=2^7*3*5^3*31$. So I'm stuck.
abstract-algebra group-theory finite-groups
asked Nov 16 at 0:30
mathnoob
1,043115
3
Hint: Consider the determinant map $GL_3(mathbb{F}_5) to mathbb{F}_5^*$.
– Daniel Schepler
Nov 16 at 0:39
Ok! The determinant map is a surjective homomorphism hence will have a no trivial kernel. And the kernel is of index 4. That's it right?
– mathnoob
Nov 16 at 0:45
Yes, the kernel is $SL_3(mathbb{F}_5)$.
– Qiaochu Yuan
Nov 16 at 4:11
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Show $Gl_3(mathbb{F}_5)$ has a normal subgroup of index 4.
Here is my attempt: $|Gl_3(mathbb{F}_5)|=(5^3-1)(5^3-5)(5^3-5^2)=124*120*100=1488000=2^7*3*5^3*31$. So I'm stuck.
abstract-algebra group-theory finite-groups
asked Nov 16 at 0:30
mathnoob
1,043115
Show $Gl_3(mathbb{F}_5)$ has a normal subgroup of index 4.
Here is my attempt: $|Gl_3(mathbb{F}_5)|=(5^3-1)(5^3-5)(5^3-5^2)=124*120*100=1488000=2^7*3*5^3*31$. So I'm stuck.
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Nov 16 at 0:30
mathnoob
1,043115
asked Nov 16 at 0:30
mathnoob
1,043115
asked Nov 16 at 0:30
mathnoob
1,043115
asked Nov 16 at 0:30
mathnoob
1,043115
asked Nov 16 at 0:30
mathnoob
1,043115
1,043115
3
Hint: Consider the determinant map $GL_3(mathbb{F}_5) to mathbb{F}_5^*$.
– Daniel Schepler
Nov 16 at 0:39
Ok! The determinant map is a surjective homomorphism hence will have a no trivial kernel. And the kernel is of index 4. That's it right?
– mathnoob
Nov 16 at 0:45
Yes, the kernel is $SL_3(mathbb{F}_5)$.
– Qiaochu Yuan
Nov 16 at 4:11
add a comment |
3
Hint: Consider the determinant map $GL_3(mathbb{F}_5) to mathbb{F}_5^*$.
– Daniel Schepler
Nov 16 at 0:39
Ok! The determinant map is a surjective homomorphism hence will have a no trivial kernel. And the kernel is of index 4. That's it right?
– mathnoob
Nov 16 at 0:45
Yes, the kernel is $SL_3(mathbb{F}_5)$.
– Qiaochu Yuan
Nov 16 at 4:11
3
3
Hint: Consider the determinant map $GL_3(mathbb{F}_5) to mathbb{F}_5^*$.
– Daniel Schepler
Nov 16 at 0:39
Hint: Consider the determinant map $GL_3(mathbb{F}_5) to mathbb{F}_5^*$.
– Daniel Schepler
Nov 16 at 0:39
Ok! The determinant map is a surjective homomorphism hence will have a no trivial kernel. And the kernel is of index 4. That's it right?
– mathnoob
Nov 16 at 0:45
Ok! The determinant map is a surjective homomorphism hence will have a no trivial kernel. And the kernel is of index 4. That's it right?
– mathnoob
Nov 16 at 0:45
Yes, the kernel is $SL_3(mathbb{F}_5)$.
– Qiaochu Yuan
Nov 16 at 4:11
Yes, the kernel is $SL_3(mathbb{F}_5)$.
– Qiaochu Yuan
Nov 16 at 4:11
add a comment |
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Hint: Consider the determinant map $GL_3(mathbb{F}_5) to mathbb{F}_5^*$.
– Daniel Schepler
Nov 16 at 0:39
Ok! The determinant map is a surjective homomorphism hence will have a no trivial kernel. And the kernel is of index 4. That's it right?
– mathnoob
Nov 16 at 0:45
Yes, the kernel is $SL_3(mathbb{F}_5)$.
– Qiaochu Yuan
Nov 16 at 4:11