If $X$ is a set, what does $S(X) = \{f\colon X \to X | \text{where f is a bijection}\}$ mean?

Question

If $X$ is a set, what does $S(X) = \{f\colon X \to X | \text{where $f$ is a bijection}\}$ mean. From my understanding it means the Set containing the function $f$ that maps elements of $X$ onto elements of $X$, where $f$ is a bijection. However, in the proof to see whether $S(X)$ is a group under composition, they said that composition is a closed binary operation for $S(X)$ because if $f$ is a bijection, then $f(g)$ is also a bijection so $f(g)$ is in $S(X)$

But where did they get $g$ from? Why can you randomly talk about other functions if $S(X)$ only contains the function $f$ that is bijective.

Thanks in advance

Answer 1

This notation means $S(X)$ is the set of all bijective mappings from $X$ to $X$.

You can easily check that this is a group:

• closed under $\circ$, since compositions of bijective mappings are bijective

• neutral element: $id\colon X\to X, x\mapsto x$.

• inverse element: $f\circ f^{-1} = id$

Answer 2

The set $S(X)$ is defined by set-builder notation: it is the set of all bijections from $X$ to $X$. The letter $f$ is a dummy variable used in the definition. Typically (when $X$ has more than one element) there will be more than one bijection $X \to X$, and so $S(X)$ will have more than one element.

Answer 3

$S(X)$ is a group with the operation of function composition

$(f\circ g)(x) = f(g(x)),\quad x\in X$,

and identity element $id_X :X\rightarrow X:x\mapsto x$.

Answer 4

Given a set $X$ and some property elements of $X$ might satisfy, say is green. We can filter out the subset of all elements of $X$, which are green by writing $$\{x \in X \mid x \text{ is green}\}.$$ The appearance of $x$ does not mean that there can only be one element of $X$, which is green. Rather, if we want to speak of an element which is green, we have to give it a name: In this case the variable $x$.

So in your specific instance we have the set $\operatorname{Fun}(X,X)$ of functions from $X$ to $X$ and want to filter out the subset of all elements (which are functions!), which are bijective. This is what $S(X)$ does: $$S(X) = \{f \in \operatorname{Fun}(X,X) \mid f \text{ bijective}\}$$

Answer 5

In a sense, $S(X)$ "comes prior" to (abstract) groups: are precisely the four properties fulfilled by $S(X)$ endowed with map composition as operation (closure, associativity, identity, inverses), that the definition of (abstract) group just mimics and axiomatizes for any set endowed with a binary operation.