Question

Prove that $A_n$ is a normal subgroup of $S_n$.[Hint if $\sigma \in S_n$and $\tau \in A_n$is${\sigma }^{-1}\tau \sigma$ , even or odd? See Example 7 of section 7.5]

Short answer

It is Proved $A_n$that is a normal subgroup of $S_n$.

Step-by-step solution

Required Theorem

Theorem 8.11: The following conditions on a subgroup N of a group G are equivalent:

1. Nis a normal subgroup of G.
2. $a^{-1}Na\subseteq N$for every $a\in G$ , Where$a^{-1}Na\subseteq N$$\left\{a^{-1}Na| n\in N\right\}$
3. $aN{a}^{-1}\subseteq N$for every $a\in G$,Where $aN{a}^{-1}\subseteq N$$\left\{aN{a}^{-1}| n\in N\right\}$
4. $a^{-1}Na\subseteq N$for every $a\in G$.
5. $aN{a}^{-1}\subseteq N$for every $a\in G$.

Proving that  is a normal subgroup of  

According to the hint,

let’s take :$\sigma \in s_n$and $\tau \in A_n$

Now taking k -cycles $\left(a_1{a}_2.....a_k\right)$in$A_n$

We have,

$\sigma \tau {\sigma }^{-1}=\sigma \left(a_1{a}_2.....a_k\right){\sigma }^{-1}$

Since every cycle can be written as the transposition of its disjoints cycles,

$\sigma \tau {\sigma }^{-1}=\sigma \left(a_1{a}_2.....a_k\right){\sigma }^{-1}$

$=\sigma \left(a_1\right)\sigma \left(a_2\right)......\sigma \left(a_k\right)$

Let’s take any arbitrary b, such that $\sigma \left(a_i\right)= b_i$.

So, $\sigma \tau {\sigma }^{-1}=\sigma \left(a_1{a}_2.....a_k\right){\sigma }^{-1}$

$\begin{array}{l}=\sigma \left(a_1{a}_2.....a_k\right){\sigma }^{-1}{b}_i\\ =\sigma \left(a_1{a}_2.....a_k\right){\sigma }^{-1}\sigma \left(a_i\right)\\ =\sigma \left(a_{i+1}\right)=b_{i+1}\end{array}$

Since $\tau \in A_n$, it can be written as an even number of transpositions. We know that the nature of $\sigma \tau {\sigma }^{-1}$ depends on the$\tau$. Because$\tau$ has an even number of transpositions, must also have an even number of transpositions.$\sigma \tau {\sigma }^{-1}$Hence, it is proved that$A_n$is a normal subgroup of $S_n$.