Let $\Omega$ be a set and define Sym($\Omega$) be the set of all bijections from $\Omega$ to $\Omega$. Prove that Sym($\Omega$) with operation composition of functions is a group. Moreover, let $\Omega$ be $\Omega ^ {\Omega}$ be the set of all functions from $\Omega$ to $\Omega$. If $\Omega$ contains at least two distinct elements, is $\Omega$ be $\Omega ^ {\Omega}$ still a group?
I am fuzzy on the notion of operation composition here, and though I know that the set of all bijections of set $\Omega$ is a group under composition, I'm not sure what that really means and how to prove this.
My idea is that for every $f \in Sym(\Omega)$ there is a bijection, and there is an inverse map, g that is also a bijection, and thus $g \in Sym(\Omega)$, and therefore every element in $Sym(\Omega)$ has an inverse and is thus a group under function composition.
If $f, g$ are bijections then $f\circ g$ is a bijection.
There is an identity function $I(x) = x.$
For any bijection $f, I\circ f = f\circ I = f$
If $f$ is a bijection then $f$ as an inverse.
and function composition is associative.
Regarding $\Omega^\Omega$. If we consider all functions, (i.e. not just bijections) then there will be functions with no inverse.