Question

Prove that ${\mathbf{A}}_{\mathbf{n}}$ is the only subgroup of index 2 in ${\mathbf{S}}_{\mathbf{n}}$.

Short answer

It is proved that,${\mathrm{A}}_{\mathrm{n}}$ is the only subgroup of index 2 in ${\mathrm{S}}_{\mathrm{n}}$.

Step-by-step solution

Determine An is the only subgroup of index 2 in Sn

Consider that $G$is the group and $N$is the subgroup of index 2.

Claim that $N$is a normal subgroup.

As $N$has the index 2 in $G$, the left co-sets are $N$in $G$ is $\{N,gN\}$.

As $g\in G$, the right co-sets are $\{N,Nh\}$.

For $h\in G$as $N$has index 2 in $G$is $\{N,gN\}=\{N,Nh\}$.

Further Simplification

As $N\ne gNandN\ne Nh$,$gN=Nh$.

By observing the $Ng$is right co-sets of $N$in $G$.

As $N$has only distinct co-sets and $g\ne N$, and it follows that:

$Ng=Nh$

Hence, it is shown that, $Ng=gN$and $N$is normal.

Now, applying the claim in above case with $G=S_n$, any subgroup of order 2 in $S_n$is normal.

As $A_n$is the only normal subgroup of $S_n$.

Therefore, the subgroup of order 2 in $S_n$is $A_n$.