Special Linear Group is Normal Subgroup of General Linear Group

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Theorem

Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.

Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.


Then $\SL {n, K}$ is a normal subgroup of the general linear group $\GL {n, K}$.


Proof

From Special Linear Group is Subgroup of General Linear Group, we have that $\SL {n, K}$ is a subgroup of $\GL {n, K}$.


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Let $\mathbf A \in \SL {n, K}$.

Let $\mathbf B \in \GL {n, K}$.

Then:

\(\ds \map \det {\mathbf B \mathbf A \mathbf B^{-1} }\) \(=\) \(\ds \map \det {\mathbf B} \map \det {\mathbf A} \map \det{\mathbf B^{-1} }\) Determinant of Matrix Product
\(\ds \) \(=\) \(\ds \map \det {\mathbf B} \map \det {\mathbf A} \map \det{\mathbf B}^{-1}\) Determinant of Inverse Matrix
\(\ds \) \(=\) \(\ds \map \det {\mathbf B} \map \det {\mathbf B}^{-1} \map \det {\mathbf A}\) Field Axiom $\text M2$: Commutativity of Product
\(\ds \) \(=\) \(\ds 1_K \map \det {\mathbf A}\) Field Axiom $\text M4$: Inverses for Product
\(\ds \) \(=\) \(\ds \map \det {\mathbf A}\) Field Axiom $\text M1$: Associativity of Product
\(\ds \) \(=\) \(\ds 1_K\) Definition of Special Linear Group

Therefore by definition of special linear group:

$\mathbf B \mathbf A \mathbf B^{-1} \in \SL {n, K}$


Thus $\SL {n, K}$ is a normal subgroup of $\GL {n, K}$

$\blacksquare$


Sources

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