Sorry about the title, I have no idea how to describe these types of problems.
Problem statement:
$A(S)$ is the set of 1-1 mappings of $S$ onto itself. Let $S \supset T$ and consider the subset $U(T) = $ { $f \in A(S)$ | $f(t) \in T$ for every $t \in T$ }. $S$ has $n$ elements and $T$ has $m$ elements. Show that there is a mapping $F:U(T) \rightarrow S_m$ such that $F(fg) = F(f)F(g)$ for $f, g \in U(T)$ and $F$ is onto $S_m$.
How do I write up this reasoning: When I look at the sets $S$ = { 1, 2, ..., $n$} and $T$ = { 1, 2, ..., $m$}, I can see that there are a bunch of permutations of the elements of $T$ within $S$. I can see there are $(n - m)!$ members of $S$ for each permutation of $T$'s elements. But there needs to be some way to get a handle on the positions of the elements in $S$ and $T$ in order to compare them to each other. But $S$ isn't any particular set, like a set of integers, so how can I relate the positions of the elements to one another? Or, is this the wrong way to go about it?
Example:
$U(T_3 \subset S_6) = \left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 6 & 5 \\ 1 & 2 & 3 & 5 & 4 & 6 \\ 1 & 2 & 3 & 5 & 6 & 4 \\ 1 & 2 & 3 & 6 & 4 & 5 \\ 1 & 2 & 3 & 6 & 5 & 4 \\ 1 & 3 & 2 & 4 & 5 & 6 \\ 1 & 3 & 2 & 4 & 6 & 5 \\ 1 & 3 & 2 & 5 & 4 & 6 \\ 1 & 3 & 2 & 5 & 6 & 4 \\ 1 & 3 & 2 & 6 & 4 & 5 \\ 1 & 3 & 2 & 6 & 5 & 4 \\ 2 & 1 & 3 & 4 & 5 & 6 \\ 2 & 1 & 3 & 4 & 6 & 5 \\ 2 & 1 & 3 & 5 & 4 & 6 \\ 2 & 1 & 3 & 5 & 6 & 4 \\ 2 & 1 & 3 & 6 & 4 & 5 \\ 2 & 1 & 3 & 6 & 5 & 4 \\ 2 & 3 & 1 & 4 & 5 & 6 \\ 2 & 3 & 1 & 4 & 6 & 5 \\ 2 & 3 & 1 & 5 & 4 & 6 \\ 2 & 3 & 1 & 5 & 6 & 4 \\ 2 & 3 & 1 & 6 & 4 & 5 \\ 2 & 3 & 1 & 6 & 5 & 4 \\ 3 & 1 & 2 & 4 & 5 & 6 \\ 3 & 1 & 2 & 4 & 6 & 5 \\ 3 & 1 & 2 & 5 & 4 & 6 \\ 3 & 1 & 2 & 5 & 6 & 4 \\ 3 & 1 & 2 & 6 & 4 & 5 \\ 3 & 1 & 2 & 6 & 5 & 4 \\ 3 & 2 & 1 & 4 & 5 & 6 \\ 3 & 2 & 1 & 4 & 6 & 5 \\ 3 & 2 & 1 & 5 & 4 & 6 \\ 3 & 2 & 1 & 5 & 6 & 4 \\ 3 & 2 & 1 & 6 & 4 & 5 \\ 3 & 2 & 1 & 6 & 5 & 4 \end{array} \right),A(T_3) = \left( \begin{array}{ccc} 1 & 2 & 3 \\ 1 & 3 & 2 \\ 2 & 1 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \\ 3 & 2 & 1 \end{array} \right)$