> [!NOTE] **Theorem** ($\text{Sym}(A)$ is group) > > Given a set $A$, $(\text{Sym}(A), \circ)$ is a [[Groups|group]] with the [[Identity Function|identity function]] $\text{id}_{A}$ as the identity element. *Proof*. ($G 0$) Closure in this group means that if $f: A \to A$ and $g: A \to A$ are bijections then $(g \circ f): A \to A$ is also a bijection. Let $c \in A$. Since $g$ is surjective, there exists $b \in A$ such that $g(b) =c$. Since $f$ is surjective, there exists $a \in A$ such that $f(a) = b$. Then $(g \circ f)(a) = g(f(a)) = g(b)= c$so $(g \circ f)$ is surjective. Let $a_{1}, a_{2} \in A$ be such that $(g \circ f)(a_{1}) = (g \circ f)(a_{2})$. Then $g(f(a_{1}))= g(f(a_{2}))$. By injectivity of $g$, $f(a_{1}) = f(a_{2})$. Then by Injectivity of $f$, $a_{1}= a_{2}$. So $(g \circ f)$ is injective. Thus $(g \circ f)$ is a bijection on $A$. $(G1)$ Follows from [[Function Composition#^f4ab9a|associativity of function composition]]. $(G2)$ Clearly $\text{id}_{A}$ is a bijection so it is in $\text{Sym}(A).$ Now $(f \circ \text{id}_{A})(x)= f(\text{id}_{A}(x)) = f(x)$ and $(\text{id}_{A} \circ f) = \text{id}_{A}(f(x)) = f(x)$ which shows that the existence of an identity element $(G3)$ Since [[Bijection#^db677e|bijections are necessarily invertible]], every element of $\text{Sym}(A)$ has an inverse in $\text{Sym}(A).$