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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?

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  • $\begingroup$ You have a binary operation $T(S)\times T(S) \to T(S)$, namely the composition of functions. Does that operation make $T(S)$ a group? $\endgroup$ – Daniel Fischer Feb 5 '14 at 22:29
  • $\begingroup$ When you write $ \times $ do you mean any operation or do you mean the multiplication (was it called multiplication?) of sets? $\endgroup$ – Paze Feb 5 '14 at 22:32
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    $\begingroup$ The $\times$ is the Cartesian product, $T(S)\times T(S)$ is the set of ordered pairs whose components belong to $T(S)$. $\endgroup$ – Daniel Fischer Feb 5 '14 at 22:33
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    $\begingroup$ To help clarify Daniel's point: the binary operation $T(S) \times T(S) \to T(S)$ that may (or may not) give $T(S)$ the structure of a group is composition. From a pair of functions $f, g \in T(S)$, form $f \circ g \in T(S)$. $\endgroup$ – Sammy Black Feb 5 '14 at 22:39
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    $\begingroup$ @Spock: if the operation is composition, then the range has to be a subset of S. $\endgroup$ – Hesky Cee Feb 5 '14 at 22:42
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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.

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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 $fog=gof=identity~ map$

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