Question

Let T be an infinite set and let A(T) be the group of permutations of T (Exercise 12). Let

$M=\{f\in A\left(T\right)|f\left(t\right)\ne t$for only a finite number of$t\in T\}$Prove that M is a group.

Short answer

It is proved that M is a group.

Step-by-step solution

Definition of group

A group is a nonempty set equipped with a binary operation${}_{°}$that must be satisfies the axioms of group.

Show that M is a group

First, Let f ,g be arbitrary element in M, where$f=\left\{t\in T|f\left(t\right)\ne t\right\}$ and$g=\left\{t\in T|g\left(t\right)\ne t\right\}$ . Note that$\left(f◦g\right)\left(t\right)\ne t$ only on some subset of$f\cap g$ and hence on finite set.

Second, composition of functions is associative in general.

Third, define $g\left(x\right)=x,x\in T$. It is known that $f◦g=g◦f=f$for arbitrary f in M.

Fourth, as per the set theory each bijection has an inverse. So, just verify that the inverse of arbitrary element f in M is also element of M.

Since,$f^{-1}\left(t\right)=t$ for every t on which$f\left(t\right)=t$ and it follows that$f^{-1}\left(t\right)\ne t$ only on finitely many elements of T.

Thus, it is proved that M is a group.