Permutations are symmetries of a (not necessarily finite) set $X$, often denoted as Sym(T). That is, a permutation $p: X\to X$ is a bijective map from a set $X$ to itself.
I wish to prove the following and wish to check my approach:
The composition of two permutations is again a permutation.
Approach:
Proof 1
Given two permutations $p_1$ and $p_2$ we know that these are bijective, therefore by definition of a bijection:
$$\forall y \in X \quad \exists ! x\in X: p(x)=y$$
In English: Every element in the codomain, corresponds to a unique element in the domain. It thus uniquely pairs elements in the codomain to elements in the domain.
We now consider the composition of these two permutations and show it is bijection from $X$ to itself. To prove bijectivity one can prove that there exists one unique element in the domain, for every element in the codomain.
Consider an arbitrary $z\in X$, we then know since $p_1$ is permutation (bijection), so there is some unique $y$, such that
$$ p_1(y)=z$$
Now since $y\in X$, by there must exist a unique element $x$, such that we can write:
$$ p_2 (x)=y $$
We conclude that:
$$ p_1(p_2(x))=p_1(y)=z$$
Since for every element $z$ in the codomain $X$, there exists a unique element $x$ in the domain, we have a bijective map. This means that the composition $p_1 \circ p_2$ is again a permutation on $X$. $\square$
Proof 2:
Canonical approach:
Let $f$ and $g$ be bijective maps from a set $X$ to itself. We will prove that the composition $f \circ g$ is bijective.
Injectivity: (each element that is reached is reached once)
Consider arbitrary $a, b \in X$ such that:
$$ f(g(a))=f(g(b))$$
by injectivity of $f$ we know that $g(a)=g(b)$. Now by injectivity of $g$ we have that $a=b$, hence the composition $f \circ g$ is injective.
surjectivity: (each element is reached)
We have to prove that for every element $b \in X$, there exists some element $a$ sucht that $f(g(a))=b$. Indeed we have bijective maps so the inverse map is a well-defined bijective map for each of these maps $f, g$. Consider the element $a=g^{-1}( f^{-1}(b))$ and observe:
$$ f(g(a))=f(g(g^{-1}( f^{-1}(b))))=f(f^{-1}(b))=b$$
By the associativity property of maps (and the fact that composition of a map with its inverse yields the identity).
Injectivity and surjectivity both hold therefore the composition is bijective. (every element is reached exactly once)
Proof 3
Alternatively, by the pigeonhole principle we have that for finite sets of the same cardinality a map is surjective if and only if it is injective.
Consider arbitrary $a, b \in X$ such that:
$$ f(g(a))=f(g(b))$$
by injectivity of $f$ we know that $g(a)=g(b)$. Now by injectivity of $g$ we have that $a=b$, hence the composition $f \circ g$ is injective.
Now we also know the composition is surjective and therefore bijective.
Best Answer
This is a "I have a lovely proof but the margin is too large to fit it" type situation. If your proof is more than two lines and contains more than definitions, then it's too long and too complicated.
Claim: If $f:X\to Y$ and $g:Y \to Z$ are bijections then $g\circ f: X \to Z$ is a bijection (and if $X = Y =Z$ then $f,g,g\circ f$ are permutations).
Proof: As $g$ is a bijection, for any $z \in Z$ there is a unique $y_z \in Y$ so that $g(y_z) = z$; and as $f$ is a bijection, for that $y_z$ there is a unique $x_{y_z}$ so that $f(x_{y_z}) = y_z$, and therefore $g\circ f(x_{y_z}) = z$. And, if it's not immediately apparent, $x_{y_z}$ is uniquely such as no other $x \in X$ is such that $f(x) = y_z$ and no other $y \in Y$ is such that $g(y) = z$.
(The above proof is probably at least $85\%$ longer than it needs to be.)