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Math Help - Verify

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    Unhappy Verify

    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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    Quote Originally Posted by mandy123 View Post
    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 |S| < \infty. Let S = \{a_1,...,a_n\}. Let \alpha be the map a_1\mapsto a_2 and a_2\mapsto a_1 and leaves everything else fixed. Let \beta be the map a_1\mapsto a_2 , a_2\mapsto a_3, ... , a_n\mapsto a_1. Show that \alpha \circ \beta \not = \beta \circ \alpha. Now if |S| = \infty then it has a finite subseteq T with |T| \geq 2. The functions on T can be regarded as being embedded in S. Thus, since it is not commutative on T it is not commutative on S.
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  3. #3
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    Quote Originally Posted by mandy123 View Post
    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
    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.
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