Proof of Monoids of Functions and Matrices, Permutation Groups, and the Matrix Exponential as a One-Parameter Group

Routine facts about nonempty classes and restrictions of binary operations are used without comment (Toolkit for AL1).

For (i), recall that $X^{X}$ is a set whose elements are the functions from $X$ to $X$ (The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set (ii), The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set (i)). The product of $g$ and $f$ in $X^{X}$ is the composite $g\circ f$, first $f$ and then $g$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii), the composite of two functions from $X$ to $X$ is again such a function, with

$$(g\circ f)(u)=g(f(u))\qquad(u\in X),$$

so $(g,f)\mapsto g\circ f$ is a binary operation on $X^{X}$. For $f,g,h\in X^{X}$ the functions $(h\circ g)\circ f$ and $h\circ(g\circ f)$ have the same value $h(g(f(u)))$ at every $u\in X$, so they are equal by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (i). Finally $\mathrm{id}_{X}\in X^{X}$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (iii), and it is neutral by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (iv). So $X^{X}$ is a monoid with identity $\mathrm{id}_{X}$.

For (ii) and (iii), a permutation $f$ of $X$ has the two-sided inverse $f^{-1}\in X^{X}$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vi). Conversely, if $g\circ f=f\circ g=\mathrm{id}_{X}$ for some $g\in X^{X}$, then $f$ is bijective by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vii). So the units of $X^{X}$ are exactly the permutations of $X$ (Permutations of a Set and the Set of Permutations of [n] (i)), and they form a group by Units of a Monoid: Unique Inverses, Products, and the Group of Units (v).

For (iv), $S_{n}$ is the set of permutations of $[n]$ (Permutations of a Set and the Set of Permutations of [n] (ii)), so this is (iii) with $X=[n]$.

For (v) and (vi), the set $\{1,-1\}$ is nonempty, and for $z,w\in\{1,-1\}$ also $zw^{-1}=zw\in\{1,-1\}$. Hence it is a subgroup of $\mathbb{R}\setminus\{0\}$, which is a group by Additive and Multiplicative Groups of Numbers (ii), by Subgroup Criterion and Basic Examples (i), and it is a group with multiplication by Subgroup Criterion and Basic Examples (ii). The sign $\operatorname{sgn}$ takes values in $\{1,-1\}$ by Rules for the Sign of a Permutation (i) and is multiplicative by Rules for the Sign of a Permutation (iii), so it is a homomorphism by Group Homomorphisms and Isomorphisms (i).

For (vii), (viii) and (ix), the product of two $n\times n$ matrices is an $n\times n$ matrix, so the matrix product $(A,B)\mapsto AB$ is a binary operation on $\mathcal{M}_{n}(\mathbb{R})$ (Binary Operations on a Set (i)). It is associative by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i), and $I_{n}$ is neutral by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv), so $\mathcal{M}_{n}(\mathbb{R})$ is a monoid with identity $I_{n}$. A unit of this monoid is a matrix with a two-sided inverse, which is exactly an invertible matrix by Inverse Matrix and Invertible Real Square Matrix (i) and Inverse Matrix and Invertible Real Square Matrix (ii). Since $\mathrm{GL}_{n}(\mathbb{R})$ is the set of invertible matrices in $\mathcal{M}_{n}(\mathbb{R})$ (Euclidean Space and Real Matrices: Standing Notation (Algebra) (x)), it is the set of units of this monoid, and hence a group with identity $I_{n}$ by Units of a Monoid: Unique Inverses, Products, and the Group of Units (v).

For (x), let $A\in\mathrm{GL}_{n}(\mathbb{R})$. Then $\det A\neq0$ by A Square Real Matrix Is Invertible if and only if Its Determinant Is Nonzero (i), so $\det$ takes values in $\mathbb{R}\setminus\{0\}$, a group by Additive and Multiplicative Groups of Numbers (ii). It preserves products by The Determinant is Multiplicative (i), and so it is a homomorphism by Group Homomorphisms and Isomorphisms (i).

For (xi), let $A\in\mathcal{M}_{n}(\mathbb{R})$. For each $t\in\mathbb{R}$ the matrix $e^{tA}$ is invertible by Properties of the Matrix Exponential (v), so the function $t\mapsto e^{tA}$ takes values in $\mathrm{GL}_{n}(\mathbb{R})$. Since $0A=0_{n}$, Properties of the Matrix Exponential (i) and Properties of the Matrix Exponential (iv) give

$$e^{0A}=I_{n}\qquad\text{and}\qquad e^{(s+t)A}=e^{sA}e^{tA}\qquad(s,t\in\mathbb{R}).$$

As $I_{n}$ is the identity of $\mathrm{GL}_{n}(\mathbb{R})$ by (vii) and Units of a Monoid: Unique Inverses, Products, and the Group of Units (v), the function is a one-parameter group by One-Parameter Groups in a Group (i).