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?
$\endgroup$ 82 Answers
$\begingroup$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.
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$\endgroup$ $\begingroup$One of these tests fails, so $T(S)$ is not actually a group under composition.
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 $fog=gof=identity~ map$
