# Proof that (S(X),$\circ$) is a group.

$X$ is a non-empty set and $S$(X) := {f : X $\rightarrow$ X : f is bijective}. Investigate if (S(X),$\circ$) is a group. Notes: f is bijective which means that the function is also injective and surjective. My solution:

Let f, g and h $\in$ S(X) be arbitrary. (f $\circ$ g)$\circ$ h = f $\circ$(g $\circ$ h) (Associative). Let x $\in$ X be arbitrary so that: ((f $\circ$ g) $\circ$ h) (x) = (f $\circ$ (g $\circ$ h))(x) = f $\circ$ g(h(x)) = f (g(h(x)) and (f $\circ$ (g $\circ$ h))(x) = f (g(h(x)).

Now for the neutral element: Let f be arbitrary and let's assume that there exists I$_X$ (Identity of X). (f $\circ$ I$_X$) (x) = f (I$_X$(x)). And (I$_X$ $\circ$ f) (x) = I$_X$(f(x)). I don't know how to continue at this point. I somehow think that my solutions aren't correct and I still need to prove that (S(X),$\circ$) has an inverse element but I don't know how.

Any hints or solutions guiding to the right direction I much appreciate.

• Fix a bijection $f$ in $S(X)$, then the inverse function is the needed symmetrical element with respect to composition. Commented May 13, 2018 at 12:52

You are almost there. For the neutral element: indeed you need the identity map. Then $\mathrm{Id}\circ f(x)=\mathrm{Id}(f(x))=f(x)=f(\mathrm{Id}(x))$ for all $x\in X$, by definition of the identity map.
Then for the inverse, every bijective function has an inverse, so because $f$ is a bijection there exists a two-sided inverse $f^{-1}$.
Note that you also need to check that the composite of two bijections $X\to X$ is again a bijection $X\to X$.
Then, together with associativity, which you already checked, you conclude $S(X)$ is a group, using composition of maps $\circ$ as the operation.