Question

Let $T$ be a set with three or more elements and let $A\left(T\right)$ be the group of all permutations of $T$. If $a\in T$, let $H_a=\{f\in A\left(T\right)|f\left(a\right)=a\}$. Prove that $H_a$ is a subgroup of $A\left(T\right)$ that is not normal.

Short answer

It has been proved that $H_a$ is a subgroup of$A\left(T\right)$ that is not normal.

Step-by-step solution

Given that

Let$T$be a set with three or more elements.

Let $A\left(T\right)$be the group of all permutations of $T$.

If $a\in T$, $H_a=\{f\in A\left(T\right)|f\left(a\right)=a\}$.

Prove that  Ha is a subgroup of  AT

Let $f,g\in H_a$, then

$\begin{array}{c}\left(f\circ g\right)\left(a\right)=f\left(g\left(a\right)\right)\\ =g\left(a\right)\\ =a\end{array}$

So $f\circ g\in H_a$ …….(1)

Also, $f\left(a\right)=a$. This implies $a=f^{-1}\left(a\right)$

Thus $f^{-1}\in H_a$ …….(2)

From (1) and (2), $H_a$is a subgroup of $A\left(T\right)$.

Prove that Ha  is not normal

Claim that: $\phi \in A\left(T\right)$then $\phi H_a{\phi }^{-1}=H_{\phi \left(a\right)}$. (This can be proved easily)

Let $a,b,c$be three distinct elements of $T$.

Define $f\in A\left(T\right)$ such that for every$x\ne b,c$

$\begin{array}{l}f\left(b\right)=c\\ f\left(c\right)=b\\ f\left(x\right)=x\end{array}$

Thus $f\in H_a$ but $f\notin H_b$ .

Let any $\phi \in A\left(T\right)$, such that$\phi \left(a\right)=b$ .

From the claim, it can be seen that

$\begin{array}{c}\phi H_a{\phi }^{-1}=H_b\\ \ne H_a\end{array}$

Thus $H_a$ is not normal.

Conclusion

Thus it can be concluded that $H_a$ is a subgroup of $A\left(T\right)$that is not normal.