Verify that if set S contains more than one element, then composition is not a commutative operation on set M(S) of mappings alph: S-S
We can assume that $\displaystyle |S| < \infty$. Let $\displaystyle S = \{a_1,...,a_n\}$. Let $\displaystyle \alpha $ be the map $\displaystyle a_1\mapsto a_2$ and $\displaystyle a_2\mapsto a_1$ and leaves everything else fixed. Let $\displaystyle \beta$ be the map $\displaystyle a_1\mapsto a_2 , a_2\mapsto a_3, ... , a_n\mapsto a_1$. Show that $\displaystyle \alpha \circ \beta \not = \beta \circ \alpha$. Now if $\displaystyle |S| = \infty$ then it has a finite subseteq $\displaystyle T$ with $\displaystyle |T| \geq 2$. The functions on $\displaystyle T$ can be regarded as being embedded in $\displaystyle S$. Thus, since it is not commutative on $\displaystyle T$ it is not commutative on $\displaystyle S$.
The simplest way to do that is to look at a set with 2 elements, say {a, b}
Then we can write out M(S). Since S contains 2 elements M(S) contains 2^2= 4 mappings:
f1(a)= a, f1(b)= a
f2(a)= a, f2(b)= b
f3(a)= b, f3(b)= a
f4(a)= b, f4(b)= b
Now simply note that f4(f1(a))= f4(a)= b and f4(f1(b))= f3(b)= b while f1(f4(a)= f1(b)= a and f1(f4(b))= f1(b)= a.
If S contains more than two elements, M(S) will still contain mappings that take whatever corresponds to a and b as above and maps all other members of S to themselves. Those will not commute.