Chapter 2 Review: Solved problems

1. In S10, let = (1,3,5,7,9), = (1,2,6), and = (1,2,5,3). For = , write as a product of disjoint cycles, and use this to find its order and its inverse. Is even or odd?

Solution: We have = (1,6,3,2,7,9), so has order 6, and -1 = (1,9,7,2,3,6). Since has length 6, it can be written as a product of 5 transpositions, so it is an odd permutation.

2. Define the function f : Z17× -> Z17× by f(x) = x-1, for all x in Z17×. Is f one to one? Is f onto? If possible, find the inverse function f-1.

Solution: For all x in Z17× we have f(f(x)) = f(x-1) = (x-1)-1 = x, so f = f-1, which also shows that f is one-to-one and onto.

3. (a) Let be a fixed element of Sn. Show that f : Sn -> Sn defined by f() = -1, for all in Sn, is a one-to-one and onto function.

Solution: If f() = f(), for and in Sn, then -1 = -1.
We can multiply on the left by -1 and on the right by , to get = , so f is one-to-one.
Finally, given in Sn, we have f() = for = -1 , and so f is onto.

Another way to show that f is one-to-one and onto is to show that it has an inverse function. A short computation shows that f-1 is defined by f-1() = -1 , for all in Sn,

(b) In S3, let = (1,2). Compute the corresponding function f defined in part (a).

Solution: Since (1,2) is its own inverse, direct computations show that
f( (1) ) = (1),
f( (1,2) ) = (1,2),
f( (1,3) ) = (2,3),
f( (2,3) ) = (1,3),
f( (1,2,3) ) = (1,3,2), and
f( (1,3,2) ) = (1,2,3).