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

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- Sep 6th 2008, 07:32 AMmandy123Verify
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

- Sep 6th 2008, 04:19 PMThePerfectHacker
We can assume that . Let . Let be the map and and leaves everything else fixed. Let be the map . Show that . Now if then it has a finite subseteq with . The functions on can be regarded as being embedded in . Thus, since it is not commutative on it is not commutative on .

- Sep 7th 2008, 06:08 PMHallsofIvy
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.