Question

If $\mathit{N}$is a normal subgroup of ${\mathbf{S}}_{\mathbf{n}}$and $\mathit{N}\mathbf{\cap }{\mathit{A}}_{\mathbf{n}}\mathbf{=}{\mathit{A}}_{\mathbf{n}}$, prove that$\mathit{N}\mathbf{=}{\mathit{A}}_{\mathbf{n}}$ or ${\mathit{S}}_{\mathbf{n}}$.

Short answer

It is proved that,$N=S_n$ or$N=A_n$ .

Step-by-step solution

Determine given functions

Consider that the given function, $N$is a normal subgroup of $S_n$so that, $N\cap A_n=A_n$.

As , $N\cap A_n=A_n$it follows$A_n\subset N$ .

Indeed, for$\sigma \in A_n,\sigma \in N\cap A_n$ .

In particular,$\sigma \in N$ .

Determine N=An or Sn

As $N$is a normal subgroup of $S_n$, then by using Lagrange’s Theorem, $\left|N\right|$divides $\left|S\right|$and we have:

$\frac{\left|S_n\right|}{\left|A_n\right|}=2$

And $A_n\subset N$follows that,

$\frac{\left|S_n\right|}{\left|N\right|}\le 2$

When it is equal to 1 then$N=S_n$ and when it is 2 then,$N=A_n$ .

Hence, it is proved that$N=S_n$ or$N=A_n$ .