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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.

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  • $\begingroup$ As a function on $\{1,\dots,n\}$, what is the permutation you just described ? $\endgroup$
    – Captain Lama
    Mar 31, 2016 at 20:32
  • $\begingroup$ As long as you can show that $A(T)$ satisfies the axioms of the group, you're fine. However, you don't know necessarily $A(T) = S_n$ since $T$ is arbitrary. But, this does not change the fact that you can permute the elements of $T$ in precisely the same manner the elements of $S_n$ permute the elements of $\{1,2,\ldots,n\}$. $\endgroup$
    – Brian Yao
    Mar 31, 2016 at 20:35

2 Answers 2

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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 $(f\circ i_{D})(t)=f(i_{D}(t))=f(t)=i_{D}(f(t))=(i_{D}\circ f)(t)$.

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You are right about the composition of bijections to be a bijection and that the composition is associative. Note: Be careful, you state the every bijective function $f$ has an inverse function. It might seem ambiguous to talk about inverses before you don't have an idea what the identity element in the group is. It is actually the case that the inverse function is the inverse of the element in this group, but it need not be in general the case. There might be some obscure operation as composition of functions which makes $A(T)$ into a group, but the inverse of a function in the group differs from the inverse for normal composition.

Well, the identity map is the identity of the group and the inverse of an element in $A(T)$ will just be the inverse function. This is because $f\circ id_T= f$ for all $f\in A(T)$.

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