Let $X$ (finite or infinite) be a set and $G=$ Sym$(X)$ be the group of bijections $f$ such that $f: X\to X$. Here the operation is composition and the identity element is the identity function $id: X\to X$. If $X$ has 3 or more elements, show that $G$ is non-commutative.
My approach: My idea was that if I can just construct two functions $f$ and $g$ such that $f$∘$g$ ≠ $g$∘$f$ for any $n$ (finite or infinite) then I would be done.
I tried constructing my $f$ as a function that fixes $n-1$ but moves all the rest of the positions and $g$ as a function that fixes $1$ but moves all the other positions. I am not sure if these will work or if I am on the right track.
I am not sure if my proof/functions will work or if I am on the right track.
Any help would be greatly appreciated!
HINT: The real clue is the condition $|X|\ge 3$: this suggests that you can do it for a $3$-element subset of $X$, and once you’ve done that, you can extend the functions to any larger set by setting $f(x)=g(x)=x$ for any other element of $X$. If $x_0,x_1$ and $x_2$ are distinct elements of $X$, consider letting $f$ permute them cyclically, while $g$ interchanges two of them and does nothing to the third.