Exercise 8.4.3

Proof. Let $H\ne \{e\}$ a normal subgroup of $A_n$. It is sufficient to prove that $H=A_n$. We show first that $H$ contains a 3-cycle. As $H\ne \{e\}$, $H$ contains an even permutation $\sigma \ne e$. For any 3-cycle $(j_1{j}_2{j}_3)$, $(j_1{j}_2{j}_3)\in A_n$, and $H◃A_n$, so

(a)

Suppose that $j\notin \{j_1,j_2,j_3\}$ and $\sigma (j)\notin \{j_1,j_2,j_3\}$.

We show then that ${\sigma }^{-1}{(j_1{j}_2{j}_3)}^{-1}\sigma (j_1{j}_2{j}_3)$ fixes $j$.

As $j\notin \{j_1,j_2,j_3\}$, then $(j_1{j}_2{j}_3)\cdot j=j$, and $[\sigma (j_1{j}_2{j}_3)]\cdot j=\sigma (j)$.

As $\sigma (j)\notin \{j_1,j_2,j_3\}$, and ${(j_1{j}_2{j}_3)}^{-1}=(j_3{j}_2{j}_1)$,

$$[{(j_1{j}_2{j}_3)}^{-1}\sigma (j_1{j}_2{j}_3)]\cdot j={(j_1{j}_2{j}_3)}^{-1}\cdot \sigma (j)=\sigma (j).$$

Therefore

$$[{\sigma }^{-1}{(j_1{j}_2{j}_3)}^{-1}\sigma (j_1{j}_2{j}_3)]\cdot j={\sigma }^{-1}(\sigma (j))=j.$$

This proves that this commutator is a permutation in $H$ that moves at most 6 elements of $\{1,\dots ,n\}$, the elements $j_1,j_2,j_3,{\sigma }^{-1}(j_1),{\sigma }^{-1}(j_2),{\sigma }^{-1}(j_3)$.

(b)

$\text{∙}$

Case 1. First suppose that one of the cycles in the cycle decomposition of $\sigma$ has a length $\ge 4$, say $$\sigma =(i_1{i}_2{i}_3{i}_4\cdots )(\cdots )\cdots .$$

In this case we claim that

$$\lambda ={\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)=(i_1{i}_3{i}_4).$$

We already know that $\lambda$ fixes every element $k$ which is not in

$$A=\{i_2,i_3,i_4,{\sigma }^{-1}(i_2),{\sigma }^{-1}(i_3),{\sigma }^{-1}(i_4)\}=\{i_1,i_2,i_3,i_4\}.$$

$\text{∙}$

If $k\notin A$, then $\lambda (k)=k=(i_1{i}_3{i}_4)\cdot k$.

$\text{∙}$

If $k=i_1$, $\lambda (i_1)=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)]\cdot i_1=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}]\cdot i_2={\sigma }^{-1}(i_4)=i_3=(i_1{i}_3{i}_4)\cdot i_1$.

$\text{∙}$

If $k=i_2$, $\lambda (i_2)=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)]\cdot i_2=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}]\cdot i_4=i_2=(i_1{i}_3{i}_4)\cdot i_2$

$\text{∙}$

If $k=i_3$, $\lambda (i_3)=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)]\cdot i_3=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}]\cdot \sigma (i_4)={\sigma }^{-1}(\sigma (i_4))=i_4=(i_1{i}_3{i}_4)\cdot i_3$ (since $\sigma (i_4)\notin \{i_2,i_3,i_4\}$).

$\text{∙}$

If $k=i_4$, $\lambda (i_4)=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)]\cdot i_4=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}]\cdot i_3={\sigma }^{-1}(i_2)=i_1=(i_1{i}_3{i}_4)\cdot i_4$.

$\forall <!--\; FUNCTION\; APPLICATION\; -->k\in \{1,\dots ,n\},\lambda \cdot k=(i_1{i}_3{i}_4)\cdot k$, so $\lambda =(i_1{i}_3{i}_4)$.

In this case, $H$ contains the 3-cycle $(i_1{i}_3{i}_4)$.

$\text{∙}$

Case 2. Next suppose that $\sigma$ has a 3-cycle. If $\sigma$ is a 3-cycle, then we are done. Hence we may assume that $$\sigma =(i_1{i}_2{i}_3)(i_4{i}_5\cdots )\cdots .$$

We show that

$$\mu ={\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}\sigma (i_2{i}_3{i}_5)=(i_1{i}_4{i}_2{i}_3{i}_5).$$

We know that $\mu$ fixes every element not in the set

$$B=\{i_2,i_3,i_5,{\sigma }^{-1}(i_2),{\sigma }^{-1}(i_3),{\sigma }^{-1}(i_5)\}=\{i_1,i_2,i_3,i_4,i_5\}.$$

$\text{∙}$

If $k\notin B$, $\mu (k)=k=(i_1{i}_4{i}_2{i}_3{i}_5)\cdot k$.

$\text{∙}$

If $k=i_1$, $[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}\sigma (i_2{i}_3{i}_5)]\cdot i_1=[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}]\cdot i_2={\sigma }^{-1}(i_5)=i_4=(i_1{i}_4{i}_2{i}_3{i}_5)\cdot i_1$

$\text{∙}$

If $k=i_2$, $[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}\sigma (i_2{i}_3{i}_5)]\cdot i_2=[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}]\cdot i_1={\sigma }^{-1}(i_1)=i_3=(i_1{i}_4{i}_2{i}_3{i}_5)\cdot i_2$

$\text{∙}$

If $k=i_3$, $[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}\sigma (i_2{i}_3{i}_5)]\cdot i_3=[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}]\cdot \sigma (i_5)={\sigma }^{-1}(\sigma (i_5))=i_5=(i_1{i}_4{i}_2{i}_3{i}_5)\cdot i_3$ (since $\sigma (i_5)\notin \{i_2,i_3,i_5\}$).

$\text{∙}$

If $k=i_4$, $[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}\sigma (i_2{i}_3{i}_5)]\cdot i_4=[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}]\cdot i_5={\sigma }^{-1}(i_3)=i_2=(i_1{i}_4{i}_2{i}_3{i}_5)\cdot i_4$

$\text{∙}$

If $k=i_5$, $[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}\sigma (i_2{i}_3{i}_5)]\cdot i_5=[{\sigma }^{-1}{(i_2{i}_3{i}_5)}^{-1}]\cdot i_3={\sigma }^{-1}(i_2)=i_1=(i_1{i}_4{i}_2{i}_3{i}_5)\cdot i_5$

Hence $\mu =(i_1{i}_4{i}_2{i}_3{i}_5)$.

As $H$ contains a 5-cycle, by case 1, it contains also a 3-cycle.

$\text{∙}$

Case 3. Finally suppose that $\sigma$ is a product of disjoint 2-cycles. There must be at least two since $\sigma \in H\subset A_n$ : $$\sigma =(i_1{i}_2)(i_3{i}_4)(\cdots )(\cdots )\cdots .$$

This time, we have

$$\nu ={\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)=(i_1{i}_3)(i_2{i}_4).$$

Indeed, every element not in

$$C=\{i_2,i_3,i_4,{\sigma }^{-1}(i_2),{\sigma }^{-1}(i_3),{\sigma }^{-1}(i_4)\}=\{i_1,i_2,i_3,i_4\}$$

is fixed by $\nu$.

$\text{∙}$

If $k\notin C$, $\nu (k)=k=(i_1{i}_3)(i_2{i}_4)\cdot k$

$\text{∙}$

If $k=i_1$, $\nu (i_1)=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)]\cdot i_1=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}]\cdot i_2={\sigma }^{-1}(i_4)=i_3=(i_1{i}_3)(i_2{i}_4)\cdot i_1$.

$\text{∙}$

If $k=i_2$, $\nu (i_2)=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)]\cdot i_2=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}]\cdot i_4={\sigma }^{-1}(i_3)=i_4=(i_1{i}_3)(i_2{i}_4)\cdot i_2$.

$\text{∙}$

If $k=i_3$, $\nu (i_3)=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)]\cdot i_3=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}]\cdot i_3={\sigma }^{-1}(i_2)=i_1=(i_1{i}_3)(i_2{i}_4)\cdot i_3$.

$\text{∙}$

If $k=i_4$, $\nu (i_4)=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}\sigma (i_2{i}_3{i}_4)]\cdot i_4=[{\sigma }^{-1}{(i_2{i}_3{i}_4)}^{-1}]\cdot i_1={\sigma }^{-1}(i_1)=i_2=(i_1{i}_3)(i_2{i}_4)\cdot i_4$.

Hence $\nu =(i_1{i}_3)(i_2{i}_4)$, so $(i_1{i}_3)(i_2{i}_4)\in H$. To turn this into a 3-cycle, let $i_5\notin \{i_1,i_2,i_3,i_4\}$ (this is where we use $n\ge 5$).

Then

$${((i_1{i}_3)(i_2{i}_4))}^{-1}{(i_1{i}_3{i}_5)}^{-1}(i_1{i}_3)(i_2{i}_4)(i_1{i}_3{i}_5)=(i_1{i}_5{i}_3).$$

We verify this with more simple notations,

$${((13)(24))}^{-1}{(135)}^{-1}(13)(24)(135)=(153)$$

by computing the successive images of 1 2 3 4 5:

$$\begin{array}{llll}\hfill & 12345& \hfill & \\ \hfill & 32541\mathrm{by}(135)& \hfill & \\ \hfill & 14523\mathrm{by}(13)(24)& \hfill & \\ \hfill & 54321\mathrm{by}{(135)}^{-1}=(153)& \hfill & \\ \hfill & 52143\mathrm{by}{((13)(24))}^{-1}=(13)(24)& \hfill & \\ \hfill & & \hfill \end{array}$$

This is the permutation $(153)$.

Hence $H$ contains in this case the 3-cycle $(i_1{i}_5{i}_3)$.

As every $\sigma \ne e$ in $H$ satisfies one of these three cases, we can conclude that $H$ contains always a 3-cycle, say $(ijk)$.

(c)

We prove then that $H$ contains all 3-cycles.

If $(i^{\prime }{j}^{\prime }{k}^{\prime })$ (where $i^{\prime },j^{\prime },k^{\prime }$ are distincts) is any 3-cycle, there exists a permutation $𝜃\in S_n$ such that $𝜃(i)=i^{\prime },𝜃(j)=j^{\prime },𝜃(k)=k^{\prime }$.

Recall the following property, which is true for all cycle $(i_1{i}_2\cdots i_l)$:

$$𝜃(i_1{i}_2\cdots i_l){𝜃}^{-1}=(𝜃(i_1)𝜃(i_2)\cdots 𝜃(i_l)).$$

Indeed,

$\text{∙}$

if $1\le k<l,(𝜃(i_1{i}_2\cdots i_l){𝜃}^{-1})(𝜃(i_k))=𝜃(i_{k+1})$,

$\text{∙}$

if $k=l,(𝜃(i_1{i}_2\cdots i_l){𝜃}^{-1})(𝜃(i_l))=𝜃(i_1)$,

$\text{∙}$

if $x\notin \{𝜃(i_1),\cdots ,𝜃(i_l)\}$, then ${𝜃}^{-1}(x)\notin \{i_1,\cdots ,i_l\}$, hence $(𝜃(i_1{i}_2\cdots i_l){𝜃}^{-1})(x)=𝜃({𝜃}^{-1}(x))=x$.

This implies that

$$𝜃(ijk){𝜃}^{-1}=(i^{\prime }{j}^{\prime }{k}^{\prime })\in H.$$

$H$ contains all 3-cycles, and the 3-cycles generate $A_n$ (Exercise 2), so $H=A_n$. The group $A_n,n\ge 5$ is simple.