I just wanna know if my argument could be written down on my final exam paper as a correct formal argument.
This is the problem: If m<n, show that there is a 1-1 mapping such that for all .
well, my idea is that with every element f in I associate an element f' in defined in this way:
If then , otherwise: .
then I define as . It's easy to show that this operation is well defined. Now, I define the map by .
To see that this map is one-to-one, suppose . from the definition of , It's easy to see that .
Now, with the definition of the function F, It's clear that and we're done.
i assume by A(S) you mean the set of all bijections on a set S.
it is usually understood that the domain of the elements of is the set {1,2,...,m}.
if not, you can easily create a bijection between {1,2,...,m} and whatever m letters you are permuting with and induce an ordering on this set from that bijection.
EXAMPLE: suppose we represent S3 as:
A-->A
B-->B
C-->C for the identity
A-->B
B-->A
C-->C for the transposition (1 2), and so on.
here, the set S is {A,B,C}. so we have a bijection (i'll call it k, whatever).
1<-->A
2<-->B
3<-->C (that is k(1) = A, k(2) = B, k(3) = C).
for m,n in S, define m < n if and only if .
all of which is a rather long-winded way of saying, you may as well assume that the set that is acting on is (set-isomorphic to) {1,2,...,m}.
the "numbers" "permutes" are just "placeholders", in many practical applications these may be algebraic objects (such as subgroups), or hydrogen atoms (in certain symmetry group applications in chemistry), or seating arrangments at a table. so, in many cases, when one refers to " ", one is refering to a group of functions that acts "just like" the set of bijections on the set of natural numbers {1,2,...,m}.
the natural bijection between the set {1,2,..m} and some given set of m elements is so transparent as to be omitted from most discussions.