Question

Let $\mathit{H}$consist of all permutations in${\mathit{S}}_{\mathbf{n}}$that fix 1 and $\mathit{n}$, that is, $\mathit{H}\mathbf{=}\left\{\alpha \in S_n|\alpha \left(1\right)=1\text{and}\alpha \left(n\right)=n\right\}$. Prove that $\mathit{H}$is a subgroup of ${\mathit{S}}_{\mathbf{n}}$

Short answer

It is proved that H is a subgroup of S_n.

Step-by-step solution

Consider

Assume that the identity $\left(1\right)\in S_n$fixes all elements so, in particular $\left(1\right)\in H$. This implies that $H$ is a non-empty subset of $S_n$.

Prove the result

Assume that$\alpha ,\beta \in H$, this implies that

$$\begin{gathered} \left(\alpha \beta \right)=\alpha \left(1\right) \\ =1 \end{gathered}$$

$$\begin{gathered} \left(\alpha \beta \right)\left(n\right)=\alpha \left(n\right) \\ =n \end{gathered}$$

Therefore, $\left(\alpha \beta \right)\in H$. This implies that $H$is closed under composition.

Assume that $\alpha \in H$, this implies that $1=\alpha \left(1\right)$yields ${\alpha }^{-1}\left(1\right)=1$, similarly we have ${\alpha }^{-1}\left(n\right)=n$, this implies that ${\alpha }^{-1}\in H$.

This implies that $H$is closed under inverses.

Hence, we can say that $H$is a subgroup of $S_n$.