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.










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  • 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















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.










share|cite|improve this question


















  • 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













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.










share|cite|improve this question













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






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asked Nov 16 at 0:30









mathnoob

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  • 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




    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















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