Exercise 3.1.35 ($\mathrm{GL}_n(F)/ \mathrm{SL}_n(F) \simeq F^\times.$)

Proof. By Exercise 2.1.9, we know that ${\mathrm{SL}}_n(F)\le {\mathrm{GL}}_n(F)$.

Consider the map

• For all $A,B\in {\mathrm{GL}}_n(F)$,

  $$\phi (\mathit{𝐴𝐵})=\det <!--\; FUNCTION\; APPLICATION\; -->(\mathit{𝐴𝐵})=\det <!--\; FUNCTION\; APPLICATION\; -->(A)\det <!--\; FUNCTION\; APPLICATION\; -->(B)=\phi (A)\phi (B),$$

  so $\phi$ is an homomorphism.

• $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )={\mathrm{SL}}_n(F)$: For all $A\in {\mathrm{GL}}_n(F)$,

  $$A\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )⟺\det <!--\; FUNCTION\; APPLICATION\; -->(A)=1⟺A\in {\mathrm{SL}}_n(F).$$

  This shows that

  $${\mathrm{SL}}_n(F)⊴{\mathrm{GL}}_n(F)$$

  since ${\mathrm{SL}}_n(F)$ is the kernel of a homomorphism.

• $\phi$ is surjective: If $a\in F^{\times }$, then the matrix $A=\mathrm{diag}(a,1,1,\dots ,1)\in {\mathrm{GL}}_n(F)$, and $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=a$, so $\phi (A)=a$.

The First Isomorphism Theorem (Theorem 16, Section 3.3) shows that

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