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 : Z_{17}^{×} -> Z_{17}^{×} by f(x) = x^{-1}, for all x in Z_{17}^{×}. Is f one to one? Is f onto? If possible, find the inverse function f^{-1}.
Solution: For all x in Z_{17}^{×} 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 S_{n}. Show that f : S_{n} -> S_{n} defined by f() = ^{-1}, for all in S_{n}, is a one-to-one and onto function.
Solution:
If
f()
= f(),
for and
in S_{n}, 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 S_{n}, 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 S_{n},
(b) In S_{3}, 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).