# ast1 - Math 3124 Tuesday September 20 First Sample Test...

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Unformatted text preview: Math 3124 Tuesday, September 20 First Sample Test Solutions 1. We can define f and g more simply by f ( x ) = 0 for all x ∈ S and g ( x ) = x for all x ∈ S . (a) From the above, it is now obvious that f is not onto and g is onto. (b) Again using the above, we see that f ◦ g ( x ) = f ( x ) = 0 and g ◦ f ( x ) = g ( ) = for all x ∈ S . Therefore f ◦ g = g ◦ f . 2. α = ( 1 6 )( 2 4 3 5 7 )( 8 9 ) . For example 4 → 4 → 4 → 7 → 3, 6 → 9 → 1 → 1 → 1 etc. To get α- 1 , write out in the reverse order; thus α- 1 = ( 1 6 )( 7 5 3 4 2 )( 8 9 ) . 3. G T is all those elements of S 5 which when written as a product of disjoint cycles do not involve 2 and 4. Thus G T = { ( 1 ) , ( 1 3 ) , ( 3 5 ) , ( 5 1 ) , ( 1 3 5 ) , ( 1 5 3 ) } . The elements of G T which are in A 5 are (1), (1 3 5) and (1 5 3). G ( T ) consists of those elements of S 5 which when written as a product of disjoint cycles, each cycle uses only the numbers 1,3,5 or only the numbers 2,4. Thus G T = {...
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ast1 - Math 3124 Tuesday September 20 First Sample Test...

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