Exercise 13.3.11

Prove that GL ( 3 , 𝔽 2 ) = SL ( 3 , 𝔽 2 ) PGL ( 3 , 𝔽 2 ) = PSL ( 3 , 𝔽 2 ) .

Proof. A matrix M GL ( 3 , 𝔽 2 ) is such that det ( M ) is invertible in 𝔽 2 , but only 1 is invertible in 𝔽 2 , so that det ( M ) = 1 and M SL ( 3 , 𝔽 2 ) . Therefore GL ( 3 , 𝔽 2 ) SL ( 3 , 𝔽 2 ) , and SL ( 3 , F ) GL ( 3 , F ) is true in every field F , thus

GL ( 3 , 𝔽 2 ) = SL ( 3 , 𝔽 2 ) .

By definition, in any field F , PGL ( n , F ) = GL ( n , F ) F I and PSL ( n , F ) = SL ( n , F ) ( F I SL ( n , F ) ) . When F = 𝔽 2 , then 𝔽 2 = { 1 } , i.e., 𝔽 2 I = ( 𝔽 2 I SL ( n , 𝔽 2 ) ) = { I } . Therefore, for n=3,

GL ( 3 , 𝔽 2 ) = SL ( 3 , 𝔽 2 ) PGL ( 3 , 𝔽 2 ) = PSL ( 3 , 𝔽 2 ) .

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2022-07-19 00:00
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