Problem:
Let T be a nonempty set and A(T) the set of all permuations of T. Show that A(T) is a group under the operation of composition of functions.
Permutation of the set T is the bijective function $f: T \rightarrow T $
For A(T) to be a group it must be closed, associate have an identity for every element and an inverse.
Since f is bijection the composition of two bijective functions is itself bijective so A(T) is closed. Every bijective function has an inverse, and composition of functions is associative. What about the identity? Are we to take $ A(T) = S_n $ here, which would be assuming that $T$={$1, …, n$}? In which case $ I = \begin{bmatrix}1&2&…&n\\1&2&…&n\end{bmatrix} $
Not sure what I'm supposed to be showing since $T$ and A(T) are arbitrarly specified here.
Best Answer
The identity element would just be the identity function $i_{D}:T\to T$ in this case, since given any bijection $f:T\to T$, and $t\in T$, we have $(f\circ i_{D})(t)=f(i_{D}(t))=f(t)=i_{D}(f(t))=(i_{D}\circ f)(t)$.