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.