Special Linear Group is Subgroup of General Linear Group
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 subgroup of the general linear group $\GL {n, K}$.
Proof
Because the determinants of the elements of $\SL {n, K}$ are not $0_K$, they are invertible.
So $\SL {n, K}$ is a subset of $\GL {n, K}$.
Now we need to show that $\SL {n, K}$ is a subgroup of $\GL {n, K}$.
Let $\mathbf A$ and $\mathbf B$ be elements of $\SL {n, K}$.
As $\mathbf A$ is invertible we have that it has an inverse $\mathbf A^{-1} \in \GL {n, K}$.
From Determinant of Inverse Matrix:
- $\map \det {\mathbf A^{-1} } = \dfrac 1 {\map \det {\mathbf A} }$
and so:
- $\map \det {\mathbf A^{-1} } = 1$
So $\mathbf A^{-1} \in \SL {n, K}$.
Also, from Determinant of Matrix Product:
- $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B} = 1$
Hence the result from the Two-Step Subgroup Test.
$\blacksquare$
Sources
- 1974: Robert Gilmore: Lie Groups, Lie Algebras and Some of their Applications ... (previous) ... (next): Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $6$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 36$. Subgroups: Worked Example
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $4$: Subgroups: Example $4.4$