Show that A(T) is a group under the operation of composition of functions. Let T be the non-empty set and A(T) the set of all permutations of T. Show that A(T) is a group under the operation of composition of functions. 

Let's try and verify each axiom after writing out the sets:
Let T = { 1,2,...,n} where $n\in$ $\Bbb Z$
Let A(T) = { f: T-->T, all functions that are bijections from T to T }
Now, let's try and go axiom by axiom.
Closure. Let f , g $\in$ A(T). Then, f $\circ$ g$\in$ A(T), since the composition of two bijective functions from T-->T is also a bijective function from T --> T. 
Identity. Lets find an identity element e in our set. Well, this isn't to hard because it is basically the "do nothing" operation. Let f $\in$ A(T), some arbitrary bijective map. Then, just let e be the identity map, such that e: x --> x, where x $\in$ T. Then, f $\circ$ e = e $\circ$ f = f  for all f $\in A(T)$.
Associative. This one is kind of easy. We know that (f $\circ$ g) $\circ$ h = f $\circ$ (g $\circ$ h) for all f,g,h $\in$ A(T) Not much to do here but if you want to be pedantic you could prove this. 
Inverse map. We need to find some function d that, when composed with f $\in$ A(T) gives us the identity map e. This is a little more tricky I think, but it basically boils down to finding a map that "undoes" everything that f does.  Now, since f is a bijection we are guaranteed d exists and that d is a bijection from T to T. Now let f: x --> k, for some x , k  $\in$ T. So f maps arbitrary elements to each other. Define the inverse of f as d: k-->x. Then, we have that f  $\circ$ d = d $\circ$ f = e. I hope this helps.
