The special linear group is normal in the general linear group

Let F be a field and n a positive integer. Prove that SL_n(F) is normal in GL_n(F) and describe the isomorphism type of GL_n(F)/SL_n(F).

  1. Let A \in SL_n(F) and B \in GL_n(F). Now \mathsf{det}(BAB^{-1}) = \mathsf{det}(B) \mathsf{det}(A) \mathsf{det}(B)^{-1} = \mathsf{det}(A) = 1, since multiplication in F is commutative. Thus BAB^{-1} \in SL_n(F). Hence SL_n(F) is normal in GL_n(F).
  2. Define a mapping \varphi : GL_n(F) / SL_n(F) \rightarrow F^\times by M \cdot SL_n(F) \mapsto \mathsf{det}(M).
    1. (Well-defined) Suppose A,B \in GL_n(F) such that AB^{-1} \in SL_n(F). Then \mathsf{det}(AB^{-1}) = 1, so that \varphi(A) = \mathsf{det}(A) = \mathsf{det}(B) = \varphi(B). Thus \varphi is well defined.
    2. (Homomorphism) We have \varphi(\overline{A} \overline{B}) = \varphi(\overline{AB}) = \mathsf{det}(AB) = \mathsf{det}(A) \mathsf{det}(B) = \varphi(\overline{A}) \varphi(\overline{B}). Thus \varphi is a homomorphism.
    3. (Injective) Suppose \varphi(\overline{A}) = \varphi(\overline{B}). Then \mathsf{det}(A) = \mathsf{det}(B), and we have \mathsf{det}(AB^{-1}) = 1, so that AB^{-1} \in SL_n(F). Hence \overline{A} = \overline{B}, and \varphi is injective.
    4. (Surjective) For all q \in F^\times, note that the matrix with q in the (1,1) entry, 1 in all other diagonal entries, and 0 in all off diagonal entries has determinant q. Thus \varphi is surjective.

    Thus \varphi is a group isomorphism, so that GL_n(F) / SL_n(F) \cong F^\times.

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