Question

If $\mathit{f}\in {\mathit{S}}_n$, prove that $f^k=I$ for some positive integer k, where $f^k$means $f◦f◦f◦\cdots ◦f$(ktimes) and I is the identity permutation.

Short answer

It is proved that $f^k=I$.

Step-by-step solution

Identity permutation

The identity permutation which maps every element of the set maps to itself.

Show that fk=I

Let$f\in S_n$and let $a,b,c,\cdots \in S$.

Apply the permutation in element of S as follows.

$$\begin{gathered} f\left(a\right)=a \\ f\left(b\right)=b \\ f\left(c\right)=c \end{gathered}$$

Apply the permutation on element of S(Ktimes).

$\begin{array}{rcl}f\left(a\right)& =& a\\ f◦f\left(a\right)& =& a\\ f\left(f\left(a\right)\right)& =& f\left(a\right)\\ & =& a\end{array}$

And,

$\begin{array}{rcl}f◦f◦f\left(a\right)& =& f◦\left(f\left(f\left(a\right)\right)\right)\\ & =& f\left(f\left(a\right)\right)\\ & =& f\left(a\right)\\ & =& a\end{array}$

Repeat this process k times to get $\left(f◦f◦f◦f◦f◦f◦f◦f\dots \left(a\right)\right)$(ktimes) = a.

Therefore, this is an identity permutation because every element of S maps to itself. This implies $\left(f◦f◦f◦f◦f◦f◦f◦f\dots \left(a\right)\right)$(ktimes)=I, where is the identity permutation.

Thus, it is proved that $f^k=I$.