Question

Prove that ${\mathit{S}}_{\mathbf{n}}$is isomorphic to a subgroup of ${\mathit{A}}_{\mathbf{n}\mathbf{+}\mathbf{2}}$.

Short answer

It is proved that$S_n$ is isomorphic to the subgroup of $A_{n+2}$.

Step-by-step solution

Define the map

As all the elements of $S_n$are disjoint, this implies that $\left(n+1n+2\right)$ commutes with all the elements of $\left(n+1n+2\right)$.

Define the map as:

$\left(n+1n+2\right)$defined by$$\begin{gathered} \varphi \left(\sigma \right)=\left\{\sigma ,\text{if}\sigma \text{is}\text{even} \\ \sigma \left(n+1n+2\right)=\left(n+1n+2\right)\sigma ,\text{if}\sigma \text{is}\text{odd} \\ \right. \end{gathered}$$

The defined map is homomorphism as:

$$\begin{gathered} \varphi \left(\sigma \tau \right)=\left\{\sigma \tau ,\text{if}\sigma \text{and}\tau \text{have}\text{the}\text{same}\text{parity} \\ \sigma \tau \left(n+1n+2\right),\text{if}\sigma \text{and}\tau \text{have}\text{distinct}\text{parities} \\ \right. \end{gathered}$$

Prove the result 

Show that homomorphism $\varphi$is injective as $\varphi \left(\sigma \right)=\varphi \left(\tau \right)$then in particular $\sigma$and$\tau$ have the same parity. When it is multiplied by $\tau$when their parity is odd, this implies that$\sigma =\tau$ .

Therefore,homomorphism$\varphi$ is injective.

$S_n$is isomorphic to the subgroup of$A_{n+2}$ as the images of homomorphism are subgroups by the image of $\varphi$.

Hence, it is proved that $S_n$is isomorphic to the subgroup of $A_{n+2}$.