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The special linear group is normal in the general linear group

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.35

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 $M \cdot SL_n(F) \mapsto \mathsf{det}(M)$.

(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.

(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.

(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.

(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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