Question

Let A(T) be the group of permutations of the set T and let T₁ be a nonempty subset of T. Prove $\mathit{H}\mathbf{=}\left\{f\in A\left(T\right)|f\left(t\right)=tforeveryt\in T_1\right\}$that is a subgroup of A(T).

Short answer

Itis proved that $H=\left\{\mathrm{f}\in \mathrm{A}\left(\mathrm{T}\right)|\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}forevery\mathrm{t}\in {\mathrm{T}}_1\right\}$is a subgroup of A(T).

Step-by-step solution

Subgroup Theorem

A non-empty subset H of a group G is a subgroup of the latter if the elements a and b belong to H, then the element ab belongs to H , and if the element a belongs to H, then the element a^-1 belongs toH.

Show that a non-empty subset H of a group G is a subgroup of G 

Consider the given set:

$H=\left\{\mathrm{f}\in \mathrm{A}\left(\mathrm{T}\right)|\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}forevery\mathrm{t}\in {\mathrm{T}}_1\right\}$

Consider two elements f and g for this set.

Find the composition of these two elements:

$$\begin{gathered} fog=f\left(g\left(t\right)\right) \\ =f\left(t\right) \\ =t \end{gathered}$$

This implies that the element f o g fixes every element of the set T₁ .

Therefore, it can be concluded that the composition of these elements is also present in this set.

Now, consider the inverse f^-1 of the element f:

$f^{-1}\left(t\right)=f^{-1}f\left(t\right)=t$

Therefore, the inverse element is also present in this set.

From the above results, it is proved that this set is a subgroup of $A\left(T\right)$.