= (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.
= (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).