I have already shown that function composition is closed on Bij(A) but I am having trouble showing associativity.
The identity element is pretty obvious, and I have some ideas about inverses.
[Math] Prove that set of all permutations on a set is a group under function composition
abstract-algebra
Best Answer
To show that the $\circ$ is associative, we need to show that $$(f \circ g) \circ h=f \circ (g \circ h)$$ for all $f,g,h \in \mathrm{Bij}(A)$. Equivalently, we need to show that $$((f \circ g) \circ h)(x)=(f \circ (g \circ h))(x)$$ for all $x \in A$ and all $f,g,h \in \mathrm{Bij}(A)$.
By definition of $\circ$: \begin{align*} ((f \circ g) \circ h)(x) &= (f \circ g)(h(x))=f(g(h(x)) \\ \\ (f \circ (g \circ h))(x) &= f((g \circ h)(x))=f(g(h(x)) \\ \end{align*} for all $x \in A$ and all $f,g,h \in \mathrm{Bij}(A)$. So $\circ$ is associative.
(Note, different texts use a different definition of $\circ$. Here we use $(f \circ g)(x)=f(g(x))$, but sometimes it's defined as $(f \circ g)(x)=g(f(x))$. The proof is much the same either way.)