Question

Question:Let T be a set n elements and A(T) be the group of permutations of T. Prove that $A\left(T\right)\cong S_n$if the element of T in some order are relabeled as 1,2,……..n, then every permutation of T becomes a permutation of 1, 2, ………..n.

Short answer

Answer

It has been proved that$A\left(T\right)\cong S_n$

Step-by-step solution

Definition of permutation

Suppose is a non-empty set, a one-to-one mapping from to is called

a permutation.

Proving that  A(T)≅Sn 

Since T is a set of nelements, it can be written as T= {1,2,3................n}

Since we defined $f:A\left(T\right)\to S_n$, for some $\sigma ,\phi \in A\left(T\right)$

Suppose,

$$\begin{gathered} f\left(\sigma \right)=f\circ \sigma \circ f^{-1} \\ f\left(\phi \right)=f\circ \phi \circ f^{-1} \end{gathered}$$

The permutation set can be given as,

$$\begin{gathered} A\left(T\right)=\left(\begin{array}{ccccc}1& 2& 3& ...& n\\ f\circ \sigma \circ f^{-1}\left(1\right)& f\circ \sigma \circ f^{-1}\left(2\right)& f\circ \sigma \circ f^{-1}\left(3\right)& & f\circ \sigma \circ f^{-1}\left(n\right)\end{array}\right) \\ So, \\ f(\sigma \circ \phi )=f\circ \sigma \circ \phi \circ f^{-1} \\ =f\circ \sigma \circ f^{-1}off\circ \phi \circ f^{-1} \\ =f\left(\sigma \right)f\left(\phi \right) \end{gathered}$$

$$\begin{gathered} Since\varphi isinjective,so\varphi \left(\sigma \right)=\varphi \left(\phi \right). \\ Or,wecanalsosaythat\sigma =\phi \end{gathered}$$

This implies that in the permutation set, the elements from the right are mapped to the left. So, there is a one-to-one mapping between the elements of the set. Therefore, it is proved that $A\left(T\right)\cong S_n$