Group under composition of functions Let $T(S)$ be the set of all functions on $S = \{\ 1,2,3 \}\ $. $T(S)$ is a group under composition of functions.
I am to prove that this is either true or not. I would like some help understanding exactly what the author means by "Group under composition of functions". And can you give me an example of a set of all functions on some other set?
 A: Test 0:  Is the operation associative?  If not, then you're done.  $T(S)$ is not a group.  If the operation is associative, then proceed to...
Test 1:  Does the set $T(S)$ contains an identity?  If you have a candidate function in $T(S)$, then what equations must it satisfy?  Does it?  If not, then $T(S)$ is not a group.  If you do have an identity, then proceed to...
Test 2:  Does every element of $T(S)$ have an inverse?  Given an arbitrary function in $f \in T(S)$, can you write down its inverse $f^{-1} \in T(S)$?  What equation must $f$ and $f^{-1}$ satisfy?  (Hint:  you need the identity function.)  If any function fails to have an inverse, then $T(S)$ is not a group.  If every function does have an inverse, then..
Congratulations!  Your set $T(S)$ is a group.
Mouse over the box to reveal a hint.

One of these tests fails, so $T(S)$ is not actually a group under composition.

A: As @Sammy Black pointed out: 
Identity element in our case is the identity map but
Inverse element property fails in this case. Take the function $f(1)=f(2)=f(3)=1$ then there's no function $g$ such that $fog=gof=identity~ map$
