Matrix Groups
Consider $M_{n \times n}(F)$, the set of $n \times n$ matrices over $F = \mathbb{R}, \mathbb{C}$. Matrix multiplication is associative as they are some sort of linear transformation, but generally not commutative. If $I$ is the identity, the matrices have inverses iff they have non-zero determinants. Hence, we have the following definition of a group containing these matrices:
General Linear Group
Definition. The general linear group is defined by
\[\text{GL}_n(F) = \set{A \in M_{n \times n}(F) : \det(A) \not = 0}\]
Proposition. $\det: \text{GL}_n(F) \to F^\ast$ is a surjective homomorphism.
Proof.
As $\det AB = \det A \det B$, $\det$ is a group homomorphism. For any $x \in F^\ast$,
\[\det \begin{pmatrix} x \\ & 1 & 0 \\ & 0 & \ddots \\ & & & 1 \\ \end{pmatrix} = x\]so $\det$ is surjective.
Definition. The special linear group is the kernel of $\det$, i.e.
\[\text{SL}_n(F) = \set{A \in \text{GL}_n(F) : \det(A) = 1}\]
Proposition. The special linear group is a normal subgroup of general linear group, i.e. $\text{SL}_n(F) \trianglelefteq \text{GL}_n(F)$.
Proof.
Algebraically, kernel is always a normal subgroup.
Proposition. $\text{GL}_n(\mathbb{R}) / \text{SL}_n(\mathbb{R}) \cong \mathbb{R}^\ast$.
Proof.
By Isomorphism Theorem, $\text{GL}_n(\mathbb{R}) / \text{SL}_n(\mathbb{R})$ is isomorphic to $\text{im}(\det) = \mathbb{R}^\ast$.
Orthogonal Group
Recall the properties of transpose $A^\intercal$ of $A$,
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$(AB)^\intercal = B^\intercal A^\intercal$
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$(A^{-1})^\intercal = (A^\intercal)^{-1}$
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$A^\intercal A = I \iff A A^\intercal = I \iff A^\intercal = A^{-1}$
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$\det A^\intercal = \det A$
Definition. The orthogonal group is defined by
\[\text{O}_n = \set{A \in \text{GL}_n(\mathbb{R}) : A^\intercal A = I}\]
Proposition. Orthogonal group is a group with respect to matrix multiplication.
Proof.
For any $A, B \in \text{O}_n$, $(AB)^\intercal(AB) = B^\intercal A^\intercal A B = I$, so $AB \in \text{O}_n$ and it is closed under matrix multiplication. $I$ is the identity element. For $A \in \text{O}_n$, $(A^\intercal)^\intercal A^\intercal = A A^\intercal = I$, so $A^\intercal \in \text{O}_n$ and $A^{-1} = A^\intercal$ is the inverse. Matrix multiplication is associative.
Proposition. $\det: \text{O}_n \to \set{\pm 1}$ is a surjective homomorphism.
Proof.
For any $A \in \text{O}_n$, $\det A \det A^\intercal = (\det A)^2 = \det I = 1$, so $\det A = \pm 1$ and it is a homomorphism.
Similarily,
\[\det \begin{pmatrix} -1 \\ & 1 & 0 \\ & 0 & \ddots \\ & & & 1 \\ \end{pmatrix} = -1\]so it is surjective.
Definition. The special orthogonal group is the kernel of $\det: \text{O}_n \to \set{\pm 1}$, i.e.
\[\text{SO}_n = \set{A \in \text{O}_n : \det(A) = 1}\]
The geometric property of orthogonal matrices is the following:
Proposition. Orthogonal matrices are isometries, i.e.
$A$ preserves dot product, $(Ax)^\intercal (Ay) = x^\intercal y$.
$A$ preserves length, $\vert Ax \vert = \vert x \vert$.
Proof.
\[(Ax)^\intercal (Ay) = x^\intercal A^\intercal A y = x^\intercal I y = x^\intercal y\] \[\vert Ax \vert^2 = (Ax)^\intercal (Ax) = x^\intercal x = \vert x \vert\]
Unitary Group
Unitary group is the variation of orthogonal group over complex numbers, base on the Hermitian conjugate $(A^\dagger)_{ij} = (A^\ast)_{ji}$. We still have
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$(AB)^\dagger = B^\dagger A^\dagger$
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$(A^{-1})^\dagger = (A^\dagger)^{-1}$
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$A^\dagger A = I \iff A A^\dagger = I \iff A^\dagger = A^{-1}$
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$\det A^\dagger = (\det A)^\ast$
Definition. The unitary group is defined by
\[\text{U}_n = \set{A \in \text{GL}_n(\mathbb{C}) : A^\dagger A = I}\]
With similar proof as orthogonal group, $\text{U}_n$ is a group.
Let $S^1 = \set{z \in \mathbb{C} : \vert z \vert = 1}$, which is the unit circle in complex plane and forms a group under complex multiplication. We have
Lemma. $\det: \text{U}_n \to S^1$ is a surjective homomorphism.
Proof.
For any $A \in \text{O}_n$, $\det A \det A^\dagger = \vert \det A \vert^2 = 1$, so $\det A \in S^1 $ and it is a homomorphism.
Similarily, given $z \in S^1$,
\[\det \begin{pmatrix} z \\ & 1 & 0 \\ & 0 & \ddots \\ & & & 1 \\ \end{pmatrix} = z\]so it is surjective.
Definition. The special unitary group is the kernel of $\det: \text{U}_n \to S^1$, i.e.
\[\text{SU}_n = \set{A \in \text{U}_n : \det(A) = 1}\]
Equivalently, unitary matrices preserve the complex dot product $x^\dagger y = x_1^\ast y_1 + x_2^\ast y_2 + … + x_n^\ast y_n$.