## Cyclic groups, Dihedral groups Dn, Proofs and Alternating groups

Last update: 5 September 2013

## Cyclic groups

 a) Let $H$ be a subset of the integers $ℤ\text{.}$ Then $H$ is a subgroup of $ℤ$ if and only if $H=mℤ$ for some $m\in ℙ\text{.}$ b) Let $m$ and $n$ be positive integers. Then $mℤ\subseteq nℤ$ if and only if $n$ divides $m\text{.}$ c) Let $n$ be a positive integer. Then the quotient group $ℤ/nℤ\cong {𝒵}_{n}\text{.}$

 Proof. To show: a) If $H$ is a subgroup of $Z$ then $H=mZ$ for some positive integer $m\text{.}$ b) If $m$ is a positive integer then $mℤ$ is a subgroup of $ℤ\text{.}$ $\square$

 a) The subgroups of the cyclic group ${𝒵}_{n}$ are $⟨{g}^{m}⟩,$ $0\le m\le n-1\text{.}$ b) Let $0\le m\le n-1$ and let $d=\text{gcd}\left(m,n\right)\text{.}$ Then $⟨{g}^{m}⟩=⟨{g}^{d}⟩$ where $d=\text{gcd}\left(m,n\right)$ and $|⟨{g}^{d}⟩|=n/d\text{.}$ c) Let $0\le m,k\le n-1\text{.}$ Then $⟨{g}^{m}⟩\subseteq ⟨{g}^{k}⟩$ if and only if $\text{gcd}\left(k,n\right)$ divides $\text{gcd}\left(m,n\right)\text{.}$ d) Let $0\le d\le n$ and suppose that $d$ divides $n\text{.}$ Then the quotient group $𝒵n/ ⟨gd⟩≃ 𝒵n/d.$

Let ${ℂ}^{×}=ℂ-\left\{0\right\}$ with the operation of multiplication. Every homomorphism from $ℤ$ to ${ℂ}^{×}$ is of the form $φk: ℤn ⟶ ℂ× g ⟼ ξk ,where ξ=e2πin.$

Let $S$ be a circular necklace with $n$ equally spaced beads ${b}_{0},{b}_{1},\dots ,{b}_{n-1},$ numbered counterclockwise around $S\text{.}$

 a) There is an action of the cyclic group ${𝒵}_{n}$ on the necklace $S$ such that $g$ acts by rotating $S$ counterclockwise by an angle of $2\pi /n\text{.}$ b) This action has one orbit, ${𝒵}_{n}{b}_{0}=\left\{{b}_{0},{b}_{1},\dots ,{b}_{n-1}\right\}$ and the stabilizer of each bead is the subgroup (1).

Picture

Every homomorphism from $ℤ$ to $ℤ$ is of the form $φm: ℤ ⟶ ℤ n ⟼ mn,$ for some positive integer $m\text{.}$

## The dihedral groups ${D}_{n}$

 a) The conjugacy classes of ${D}_{2}$ are $𝒞1={1}, 𝒞x={x}, 𝒞y={y}, 𝒞xy={xy}.$ b) If $n$ is even and $n\ne 2,$ then the conjugacy classes of ${D}_{n}$ are the sets $𝒞1={1}, 𝒞xn/2= {xn/2}, 𝒞xk= {xk,x-k}, 0 c) If $n$ is odd then the conjugacy classes of ${D}_{n}$ are the sets $𝒞1={1} 𝒞xk= {xk,x-k}, 0

Proof.

Sketch of Proof.

a) The group ${D}_{2}$ abelian, so each element is in a conjugacy class by itself.
b) and c)

 By the multiplication rule, $x(xk)x-1 = xk, y(xk)y = x-ky2= x-k, and x(xky) x-1 = xk+2y, y(xky) = yxk= x-ky,$ Thus, if ${x}^{k}$ is in a conjugacy class then ${x}^{-k}$ is also in the conjugacy class, and $\phantom{\rule{2em}{0ex}}$ If ${x}^{k}y$ is in a conjugacy class then ${x}^{k+2}y$ and ${x}^{-k}y$ are also in the conjugacy class. One checks case by case that the sets given in the statement of the proposition satisfy these two properties. Since these sets partition the group ${D}_{n},$ they must be the conjugacy classes.

$\square$

 a) ${D}_{n}$ is generated by the elements $x$ and $y\text{.}$ b) The elements $x$ and $y$ in ${D}_{n}$ satisfy the relations $xn=1, y2=1, yx=x-1y.$

 Proof. Both parts follow directly from the definition of the dihedral group ${D}_{n}\text{.}$ $\square$

The dihedral group ${D}_{n}$ has a presentation by generators $x$ and $y$ and relations $xn=1, y2=1, yx=x-1y.$

Let $⟨a,b,\cdots ⟩$ denotes the subgroup generated by elements $a,b,\dots \text{.}$

 a) The normal subgroups of the dihedral group ${D}_{2}$ we the subgroups $⟨x⟩, ⟨y⟩, ⟨xy⟩,$ b) If $n$ is even and $n\ne 2$ then the normal subgroups of the dihedral group ${D}_{n}$ are the subgroups $⟨xk⟩, 0≤k≤n-1, ⟨x2,y⟩, ⟨x2,xy⟩.$ c) If $n$ is odd then the normal subgroups of the dihedral group ${D}_{n}$ are the subgroups $⟨xk⟩, 1≤k≤n-1.$

 Proof. The subgroups given in the statement of the proposition are unions of conjugacy classes of ${D}_{n}$ as follows. $⟨xk⟩ = ⋃𝒞xjk ⟨x2,y⟩ = 𝒞y∪⟨x2⟩ ⟨x2,xy⟩ = 𝒞xy∪ ⟨x2⟩$ Thus these subgroups are normal. It remains to show that these are all the normal subgroups. $\square$

The orders of the elements in the dihedral group ${D}_{n}$ are $o(1)=1, o(xk)=gcd (k,n), o(xky)=2, 0

 Proof. This follows from the definition of the multiplication in ${D}_{n}\text{.}$ $\square$

Let $F$ be an $n\text{-gon}$ with vertices ${v}_{i}$ numbered $0$ to $n-1$ counterclockwise around $F\text{.}$  There is an action of the group ${D}_{n}$ on the $n\text{-gon}$ $F$ such that

 $x$ acts by rotating the $n\text{-gon}$ by an angle of $2\pi /n\text{.}$ $y$ acts by reflecting about the line which contains the vertex ${v}_{0}$ and the center of $F\text{.}$

 Proof. $\square$

## Proofs

For each permutation $\sigma \in {S}_{m},$ let $\text{det}\left(\sigma \right)$ denote the determinant of the matrix which represents the permutation $\sigma \text{.}$ The map $ε: Sm ⟶ ±1 σ ⟼ det(σ)$ is a homomorphism from the symmetric group ${S}_{m}$ to the group ${ℤ}_{2}=±1\text{.}$

Proof.
To show: a) If $\sigma$ and $\tau$ are permutation matrices then $\text{det}\left(\sigma \tau \right)=\text{det}\left(\sigma \right)\text{det}\left(\tau \right)\text{.}$
b) If $\sigma$ is a permutation matrix then $\text{det}\left(\sigma \right)=±1\text{.}$

 a) This follows from Proposition (). b) Any permutation matrix is an orthogonal matrix, i.e. $\sigma {\sigma }^{t}=1\text{.}$ Thus, $1=\text{det}\left(\sigma {\sigma }^{t}\right)=\text{det}\left(\sigma \right)\text{det}\left({\sigma }^{t}\right)=\text{det}{\left(\sigma \right)}^{2}\text{.}$ Thus $\text{det}\left(\sigma \right)=±1\text{.}$

$\square$

Suppose $\sigma \in {S}_{m}$ has cycle type $\lambda =\left({\lambda }_{1},{\lambda }_{2},\dots \right)$ and let ${\gamma }_{\lambda }$ be the permutation in ${S}_{m}$ which is given, in cycle notation, by $γλ= (1,2,⋯,λ1) (λ1+1,λ1+2,…,λ1+λ2) (λ1+λ2+1,⋯) ⋯.$

 a) Then $\sigma$ is conjugate to ${\gamma }_{\lambda }\text{.}$ b) If $\tau \in {S}_{m}$ is conjugate to $\sigma$ then $\tau$ has cycle type $\lambda \text{.}$ c) Suppose that $\lambda =\left({1}^{{m}_{1}}{2}^{{m}_{2}}\cdots \right)\text{.}$ Then the order of the stabilizer of the permutation ${\gamma }_{\lambda },$ under the action of ${S}_{m}$ on itself by conjugation, is $1m1m1! 2m2m2!⋯.$

Proof.
a)
 To show: $\sigma$ is conjugate to ${\gamma }_{\lambda }=\left(1,2,\cdots ,{\lambda }_{1}\right)\left({\lambda }_{1}+1,{\lambda }_{1}+2,\dots ,{\lambda }_{1}+{\lambda }_{2}\right)\left({\lambda }_{1}+{\lambda }_{2}+1,\cdots \right)\cdots \text{.}$ Suppose that, in cycle notation, $\sigma =\left({i}_{1},{i}_{2},\dots ,{i}_{{\lambda }_{1}}\right)\left({i}_{{\lambda }_{1}+1},\dots ,{i}_{{\lambda }_{1}+{\lambda }_{2}}\right)\cdots \text{.}$ Let $\pi$ be the permutation given by $\pi \left({i}_{j}\right)=j\text{.}$
Then $\pi \sigma {\pi }^{-1}={\gamma }_{\lambda }\text{.}$
b) Suppose that $\tau \in {S}_{m}$ is conjugate to $\sigma \text{.}$
Then $\tau =\pi \sigma {\pi }^{-1}$ for some $\pi \in {S}_{m}\text{.}$
 To show: The lengths of the cycles in $\pi \sigma {\pi }^{-1}$ are the same as the lengths of the cycles in $\sigma \text{.}$ Suppose that, in cycle notation, $\sigma =\left({i}_{1},{i}_{2},\dots ,{i}_{{\lambda }_{1}}\right)\left({i}_{{\lambda }_{1}+1},\dots ,{i}_{{\lambda }_{1}+{\lambda }_{2}}\right)\cdots \text{.}$ Then $\pi \sigma {\pi }^{-1}\left(\pi \left({i}_{j}\right)\right)=\pi \left(\sigma \left({i}_{j}\right)\right)=\pi \left({i}_{j+1}\right)\text{.}$ Thus, in cycle notation, $\pi \sigma {\pi }^{-1}=\left(\pi \left({i}_{1}\right),\pi \left({i}_{2}\right),\cdots ,\pi \left({i}_{{\lambda }_{1}}\right)\right)\left(\pi \left({i}_{{\lambda }_{1}+1}\right),\dots ,\pi \left({i}_{{\lambda }_{1}+{\lambda }_{2}}\right)\right)\cdots \text{.}$ So, the lengths of the cycles in $\pi \sigma {\pi }^{-1}$ are the same as the lengths of the cycles in $\sigma \text{.}$
So, $\tau$ has cycle type $\lambda \text{.}$
c) Suppose that $\pi \in {S}_{m}$ is in the stabilizer of ${\gamma }_{\lambda }\text{.}$
Then $\pi {\gamma }_{\lambda }{\pi }^{-1}={\gamma }_{\lambda }\text{.}$
In cycle notation, $\pi {\gamma }_{\lambda }{\pi }^{-1}=\left(\pi \left(1\right),\pi \left(2\right),\cdots ,\pi \left({\lambda }_{1}\right)\right)\left(\pi \left({\lambda }_{1}+1\right),\dots ,\pi \left({\lambda }_{1}+{\lambda }_{2}\right)\right)\cdots \text{.}$
Since $\pi {\gamma }_{\lambda }{\pi }^{-1}={\gamma }_{\lambda },$ it follows that each of the sequences $\left(\pi \left({\lambda }_{j}+1\right),\dots ,\pi \left({\lambda }_{j}+{\lambda }_{j+1}\right)\right)$ must be a cyclic rearrangement of some cycle of ${\gamma }_{\lambda }\text{.}$
This means that $\pi$ must be a permutation that
 1) permutes cycles of ${\gamma }_{\lambda }$ of the same length and/or 2) cyclically rearranges the elements of the cycles of ${\gamma }_{\lambda }\text{.}$
Note that,
 1) Each cycle of length $k$ in ${\gamma }_{\lambda }$ can be cyclically rearranged in $k$ ways. Thus, there are a total of ${k}^{{m}_{k}}$ ways of cyclically rearranging the elements of the ${m}_{k}$ cycles of length $k$ in ${\gamma }_{\lambda }\text{.}$ 2) The ${m}_{k}$ cycles of length $k$ in ${\gamma }_{\lambda }$ can be permuted in ${m}_{k}!$ different ways.
Thus, there are a total of ${1}^{{m}_{1}}{m}_{1}!{2}^{{m}_{2}}{m}_{2}!\cdots$ permutations $\pi$ which stabilize ${\gamma }_{\lambda }$ under the action of conjugation.

$\square$

 a) The conjugacy classes of ${S}_{m}$ are the sets $𝒞λ= {permutations σ with cycle type λ} ,$ where $\lambda$ is a partition of $m\text{.}$ b) If $\lambda =\left({1}^{{m}_{1}}{2}^{{m}_{2}}\cdots \right)$ then the size of the conjugacy class ${𝒞}_{\lambda }$ is $|𝒞λ|= m! m1!1m1 m2!2m2 m3!3m3⋯ ,$

Proof.
a)
To show: aa) If $\lambda ⊢m$ then ${𝒞}_{\lambda }$ is a conjugacy class of ${S}_{m}\text{.}$
ab) Every conjugacy class is equal to ${𝒞}_{\lambda }$ for some $\lambda ⊢m\text{.}$

aa) Let $\lambda$ be a partition of $m\text{.}$
Let ${𝒪}_{\gamma \lambda }$ denote the conjugacy class of ${\gamma }_{\lambda }\text{.}$
 To show: ${𝒪}_{\lambda }={𝒞}_{\lambda }\text{.}$
To show: aaa) ${𝒞}_{\lambda }\subseteq {𝒪}_{{\gamma }_{\lambda }}\text{.}$
aab) ${𝒪}_{{\gamma }_{\lambda }}\subseteq {𝒞}_{\lambda }\text{.}$

 aaa) Suppose that $\sigma \in {𝒞}_{\lambda }\text{.}$ Then $\sigma$ has cycle type $\lambda \text{.}$ Thus, by Lemma (), $\sigma$ is conjugate to ${\gamma }_{\lambda }\text{.}$ So, $\sigma \in {𝒪}_{{\gamma }_{\lambda }}\text{.}$ Thus, ${𝒞}_{\lambda }\subseteq {𝒪}_{{\gamma }_{\lambda }}\text{.}$ aab) Suppose that $\sigma \in {𝒪}_{{\gamma }_{\lambda }}\text{.}$ Then, $\sigma$ is conjugate to ${\gamma }_{\lambda }\text{.}$ Thus, by Lemma (), $\sigma$ has cycle type $\lambda \text{.}$ So, $\sigma \in {𝒞}_{\lambda }\text{.}$ So, ${𝒪}_{{\gamma }_{\lambda }}\subseteq {𝒞}_{\lambda }\text{.}$
So ${𝒞}_{\lambda }={𝒪}_{{\gamma }_{\lambda }}\text{.}$
So ${𝒞}_{\lambda }$ is a conjugacy class of ${S}_{m}\text{.}$
ab) Let $\sigma \in {S}_{m}$ and let ${𝒪}_{\sigma }$ be the conjugacy class of $\sigma \text{.}$
Suppose that $\sigma$ has cycle type $\lambda \text{.}$
Then, by Lemma (), $\sigma$ is conjugate to ${\gamma }_{\lambda }\text{.}$
Thus, by Proposition (), ${𝒪}_{\sigma }={𝒪}_{{\gamma }_{\lambda }}\text{.}$
So, by part a), ${𝒪}_{\sigma }={𝒪}_{{\gamma }_{\lambda }}={𝒞}_{\lambda }\text{.}$
So every conjugacy class is equal to ${𝒞}_{\lambda }$ for some $\lambda ⊢m\text{.}$
So the sets ${𝒞}_{\lambda },$ $\lambda ⊢m,$ are the conjugacy classes of ${S}_{m}\text{.}$
b) Let $\lambda =\left({1}^{{m}_{1}}{2}^{{m}_{2}}\cdots \right)$ be a partition of $m\text{.}$
By, Lemma (), the stabilizer of the permutation ${\gamma }_{\lambda },$ has order ${1}^{{m}_{1}}{m}_{1}!{2}^{{m}_{2}}{m}_{2}!\cdots \text{.}$
Thus, by Proposition (), the order of the conjugacy class ${𝒞}_{\lambda }$ is $|𝒞λ|= |Sm| 1m1m1! 2m2m2! ⋯ = m! 1m1m1! 2m2m2! ⋯ .$

$\square$

 a) ${S}_{m}$ is generated by the simple transpositions ${s}_{i},$ $1\le i\le m-1\text{.}$ b) The simple transpositions ${s}_{i},$ $1\le i\le m-1,$ in ${S}_{m}$ satisfy the relations $sisj = sjsi,if |i-j|>1, sisi+1si = si+1sisi+1 ,1≤i≤m-2, si2 = 1,1≤i≤m-1.$

Proof.
a)
 To show: Every permutation $\sigma$ can be written as a product of simple tranpositions. This is most easily seen by "stretching out" the function diagram of $\sigma \text{.}$ $PICTUREstretchout.ps$ We must give some argument to show that this can always be done, for an arbitrary permutation $\sigma \text{.}$ $PICTUREsigma.ps$ The set of inversions of $\sigma$ is the set $\text{inv}\left(\sigma \right)=\left\{\left(i,j\right) | 1\le i\sigma \left(j\right)\right\}\text{.}$ Let ${k}_{i}$ be the number of inversions of $\sigma$ that have first coordinate $i\text{.}$ Then define $γ(i)= { sisi+1⋯ si+ki-1, if ki≥1; 1, if ki=0.$ Then $\sigma =\gamma \left(m-1\right)\gamma \left(m-2\right)\cdots \gamma \left(1\right)\text{.}$ $PICTUREgammadec.ps$ Thus $\sigma$ can be written as a product of simple transpositions.
b)
To show: ba) ${s}_{i}{s}_{j}={s}_{j}{s}_{i},\phantom{\rule{2em}{0ex}}\text{if} |i-j|>1\text{.}$
bb) ${s}_{i}{s}_{i+1}{s}_{i}={s}_{i+1}{s}_{i}{s}_{i+1},\phantom{\rule{2em}{0ex}}1\le i\le m-2\text{.}$
bc) ${s}_{i}^{2}=1,\phantom{\rule{2em}{0ex}}1\le i\le m-1\text{.}$

 ba) $PICTUREsisjsjsi$ bb) $PICTUREsisip1$ bc) $PICTUREsi2$

$\square$

## Alternating group

Suppose that $\sigma \in {A}_{m}\text{.}$ Let ${𝒞}_{\sigma }$ denote the conjugacy class of $\sigma$ in ${S}_{m}$ and let ${§}_{\sigma }$ denote the conjugacy class of $\sigma$ in ${A}_{m}\text{.}$

 a) Then $\sigma$ has an even number of cycles of even length. b) $|𝒜σ|= { |𝒞σ|2, if all cycles σ are of different odd lengths; |𝒞σ|, otherwise;$

Proof.
a) Suppose that $\sigma$ has cycle type $\lambda =\left({\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{k}\right)\text{.}$
 To show: An even number of the ${\lambda }_{j},$ $1\le j\le k,$ are even. Let ${\gamma }_{\lambda }$ be the permutation given, in cycle notation, by $γλ= (1,2,⋯,λ1) (λ1+1,λ1+2,…,λ1+λ2) (λ1+λ2+1,⋯)⋯.$ Since ${A}_{m}$ is a normal subgroup of ${S}_{m}$ and $\sigma \in {A}_{m}$ it follows that ${C}_{\sigma }={𝒞}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$ So ${\gamma }_{\lambda }\in {A}_{m}\text{.}$ So the length $\ell \left({\gamma }_{\lambda }\right)$ of ${\gamma }_{\lambda }$ is even. So $\ell \left({\gamma }_{\lambda }\right)={\sum }_{i=1}^{k}\left({\lambda }_{i}-1\right)$ is even. So there are an even number of $1\le j\le k$ such that ${\lambda }_{j}-1$ is odd. So there are an even number of $1\le j\le k$ such that ${\lambda }_{j}$ is even. So $\sigma$ has an even number of cycles of even length.
b) Let ${S}_{\sigma }$ be the stabilizer of $\sigma$ under the action of ${S}_{m}$ on itself by conjugation.
Let ${A}_{\sigma }$ be the stabilizer of $\sigma$ under the action of ${A}_{m}$ on itself by conjugation.
Then, by Proposition (), $12 |Sσ| |𝒞σ|= |Sm|2= |Am|= |Aσ| |𝒜σ|.$ So, $|𝒜σ|= { |𝒞σ|, if |Aσ| ≠|Sσ|; |𝒞σ|2, if |Aσ| =|Sσ|.$ Since ${A}_{\sigma }\subseteq {S}_{\sigma },$ $|𝒜σ|= { |𝒞σ|, if Sσ⊆Aσ; |𝒞σ|2, if Sσ⊈Aσ.$ So, $|𝒜σ|= { |𝒞σ|, if Sσ⊆Am; |𝒞σ|2, if Sσ⊈Am.$ Then, by Lemma (), $|𝒜σ|= { |𝒞σ|, if Sγλ⊆Am; |𝒞σ|2, if Sγλ⊈Am.$ By Lemma (), $|𝒜σ|= { |𝒞σ|2, if all cycles σ are of different odd lengths; |𝒞σ|, otherwise;$

$\square$

Let $\sigma \in {A}_{m}$ and let $\lambda =\left({\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{k}\right)$ be the cycle type of $\sigma \text{.}$ Let ${\gamma }_{\lambda }$ be the permutation given, in cycle notation, by $γλ= (1,2,⋯,λ1) (λ1+1,λ1+2,…,λ1+λ2) (λ1+λ2+1,⋯)⋯.$ Let ${S}_{\sigma }$ denote the stabilizer of $\sigma$ under the action of ${S}_{m}$ on itself by conjugation. Then,

 a) ${S}_{\sigma }\subseteq {A}_{m}$ if and only if ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$ b) ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}$ if and only if ${\gamma }_{\lambda }$ has all odd cycles of different lengths.

Proof.
a)
 To show: ${S}_{\sigma }\subseteq {A}_{m}$ if and only if ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$
To show: aa) If ${S}_{\sigma }\subseteq {A}_{m}$ then ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$
ab) If ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}$ then ${S}_{\sigma }\subseteq {A}_{m}\text{.}$
Then, by Proposition (), there exists $\pi \in {S}_{m}$ such that $\pi \sigma {\pi }^{-1}={\gamma }_{\lambda }\text{.}$
Thus, ${S}_{{\gamma }_{\lambda }}=\pi {S}_{\sigma }{\pi }^{-1}\text{.}$
 aa) Assume ${S}_{\sigma }\subseteq {A}_{m}\text{.}$ Let $\tau \in {S}_{{\gamma }_{\lambda }}\text{.}$ Then ${\pi }^{-1}\tau \pi \in {S}_{\sigma }\text{.}$ So ${\pi }^{-1}\tau \pi \in {A}_{m}\text{.}$ So $1=\epsilon \left({\pi }^{-1}\tau \pi \right)\text{.}$ Since $\epsilon$ is a homomorphism, $\epsilon \left(\tau \right)=\epsilon {\left(\pi \right)}^{-1}\epsilon \left(\tau \right)\epsilon \left(\pi \right)=\epsilon \left({\pi }^{-1}\tau \pi \right)=1\text{.}$ So $\tau \in {A}_{m}\text{.}$ So ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$ ab) Assume ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$ Let $\tau \in {S}_{\sigma }\text{.}$ Then $\pi \tau {\pi }^{-1}\in {S}_{{\gamma }_{\lambda }}\text{.}$ So $\pi \tau {\pi }^{-1}\in {A}_{m}\text{.}$ So $1=\epsilon \left(\pi \tau {\pi }^{-1}\right)\text{.}$ Since $\epsilon$ is a homomorphism, $\epsilon \left(\tau \right)=\epsilon \left(\pi \right)\epsilon \left(\tau \right)\epsilon {\left(\pi \right)}^{-1}=\epsilon \left(\pi \tau {\pi }^{-1}\right)=1\text{.}$ So $\tau \in {A}_{m}\text{.}$ So ${S}_{\sigma }\subseteq {A}_{m}\text{.}$
So ${S}_{\sigma }\subseteq {A}_{m}$ if and only if ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$
b)
 To show: ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}$ if and only if ${\gamma }_{\lambda }$ has all odd cycles of different lengths.
To show: ba) If ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}$ then ${\gamma }_{\lambda }$ has all odd cycles of different lengths.
bb) If ${\gamma }_{\lambda }$ has all odd cycles of different lengths then ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$

ba) Proof by contradiction.
Assume $\lambda$ does not have all odd parts of different lengths.
To show: ${S}_{{\gamma }_{\lambda }}⊈{A}_{m}\text{.}$
 Case 1: Assume ${\gamma }_{\lambda }$ has an even cycle, say $\left(k+1,\dots ,k+2n\right)\text{.}$ Let $\pi$ be the permutation which cyclically permutes this cycle, $\pi =\left(k+1,\dots ,k+2n\right)\text{.}$ Then $\pi \in {S}_{{\gamma }_{\lambda }}\text{.}$ But $\epsilon \left(\pi \right)={\left(-1\right)}^{2n-1}=-1\text{.}$ So $\pi \notin {A}_{m}\text{.}$ So ${S}_{{\gamma }_{\lambda }}⊈{A}_{m}\text{.}$ Case 2: Assume ${\gamma }_{\lambda }$ has two cycles of the same odd length, say $\left(k+1,\dots ,k+n\right)$ and $\left(k+n+1,\dots ,k+n+1\right)\text{.}$ Let $\pi$ be the permutation which switches these two cycles, $\pi =\left(k+1,k+1+n\right)\left(k+2,k+2+n\right)\cdots \left(k+n,k+n+n\right)\text{.}$ Then $\pi \in {S}_{{\gamma }_{\lambda }}\text{.}$ But $\epsilon \left(\pi \right)={\left(-1\right)}^{{n}^{2}}=-1,$ since $n$ is odd. So $\pi \notin {A}_{m}\text{.}$ So ${S}_{{\gamma }_{\lambda }}⊈{A}_{m}\text{.}$
bb) Assume ${\gamma }_{\lambda }$ has all different odd cycles.
Suppose that $\tau \in {S}_{{\gamma }_{\lambda }}\text{.}$
This means that $\tau$ must be a permutation that
 1) permutes cycles of ${\gamma }_{\lambda }$ of the same length and/or 2) cyclically rearranges the elements of the cycles of ${\gamma }_{\lambda }\text{.}$
Since all cycles of ${\gamma }_{\lambda }$ are different lengths, $\tau$ cyclically permutes the elements of the cycles of ${\gamma }_{\lambda }\text{.}$
Define permutations $c1=(1,2,…,λ1), c2=(λ1+1,λ1++2,…,λ1+λ2), etc.$ Then $\tau ={c}_{1}^{{n}_{1}}{c}_{2}^{{n}_{2}}\cdots {c}_{k}^{{n}_{k}}$ for some positive integers ${n}_{1},{n}_{2},\dots ,{n}_{k}\text{.}$
Then $\epsilon \left({c}_{j}\right)={\left(-1\right)}^{{\lambda }_{j}-1}=1,$ since ${\lambda }_{j}$ is odd.
So $\epsilon \left(\tau \right)=\epsilon {\left({c}_{1}\right)}^{{n}_{1}}\epsilon \left({c}_{2}\right){n}_{2}\cdots \epsilon {\left({c}_{k}\right)}^{{n}_{k}}=1\text{.}$
So $\tau \in {A}_{m}\text{.}$
So ${S}_{{\gamma }_{\lambda }}\subseteq {A}_{m}\text{.}$

$\square$

 a) If $m\ne 4$ then ${A}_{m}$ is simple. b) The alternating group ${A}_{4}$ has a single nontrivial proper normal subgroup given by ${ (1234), (2143), (3412), (4321) }$

Proof.
Case a: $n=1,2,3\text{.}$
The groups ${A}_{1}=\left\{1\right\},$ ${A}_{2}=\left\{1\right\}$ and ${A}_{3}=\left\{\left(123\right),\left(213\right),\left(312\right)\right\}$ have no nontrivial proper subgroups.
So ${A}_{1},$ ${A}_{2}$ and ${A}_{3}$ have no nontrivial proper normal subgroups.
Case b: $n=4\text{.}$
The conjugacy classes of ${A}_{4}$ are ${1}, {(123),(134),(243),(142)}, {(132)(124),(234),(143)}, {(12)(34),(13)(24),(14)(23)}.$ Let $N$ be a normal subgroup of ${A}_{4}\text{.}$
 Case ba: $\pi =\left(123\right)\in N\text{.}$ Then ${\pi }^{-1}=\left(132\right)$ and $\left(123\right)\left(124\right)=\left(12\right)\left(34\right)$ are in $N\text{.}$ So $N$ contains all the conjugacy classes. So $N={A}_{n}\text{.}$ Case bb: $\pi =\left(132\right)\in N\text{.}$ Then ${\pi }^{-1}=\left(123\right)$ and $\left(123\right)\left(124\right)=\left(12\right)\left(34\right)$ are in $N\text{.}$ So $N$ contains all the conjugacy classes. So $N={A}_{n}\text{.}$
Thus, the only possible union of conjugacy classes which could be a normal subgroup is $N= { 1,(12)(34), (13)(24), (14)(23) } .$
It is easy to check that this is a subgroup of ${A}_{4}\text{.}$
Case c: $n\ge 5\text{.}$
Let $N$ be a normal subgroup of ${A}_{n}$ such that $N\ne \left(1\right)\text{.}$
To show: $N={A}_{n}\text{.}$
Let $\sigma \in N$ and suppose that $\sigma$ has cycle type $\lambda \text{.}$
Let ${\gamma }_{\lambda }$

 Case ca: $\sigma$ has a cycle $\left({i}_{1}{i}_{2}\cdots {i}_{r}\right)$ of length $r>3\text{.}$ Then ${\sigma }^{-1}\in N$ and $\left({i}_{2}{i}_{3}{i}_{4}\right)\sigma \left({i}_{4}{i}_{3}{i}_{2}\right)\in N\text{.}$ So ${\sigma }^{-1}\left(\left({i}_{2}{i}_{3}{i}_{4}\right)\sigma \left({i}_{4}{i}_{3}{i}_{2}\right)\right)=\left({\sigma }^{-1}\left({i}_{1}{i}_{2}{i}_{3}\right)\sigma \right)\left({i}_{4}{i}_{3}{i}_{2}\right)=\left({i}_{1}{i}_{2}{i}_{3}\right)\left({i}_{4}{i}_{3}{i}_{2}\right)=\left({i}_{1}{i}_{2}{i}_{4}\right)\in N\text{.}$ Thus, by Lemma (), $N={A}_{n}\text{.}$ Case cb: $\sigma$ does not have all odd cycles of different lengths and $\sigma$ has a cycle of length $>2\text{.}$ Then, by Propositions () and (), ${𝒜}_{\sigma }={𝒞}_{\sigma }={𝒞}_{{\gamma }_{\lambda }}\text{.}$ Since $N$ is normal, ${C}_{{\gamma }_{\lambda }}={𝒜}_{\sigma }\subseteq N\text{.}$ So ${\gamma }_{\lambda }\in N$ and ${s}_{1}{\gamma }_{\lambda }{s}_{1}\in N\text{.}$ Since $N$ is a subgroup ${\gamma }_{\lambda }^{-1}\in N\text{.}$ So ${\gamma }_{\lambda }^{-1}\left({s}_{1}{\gamma }_{\lambda }{s}_{1}\right)=\left({\gamma }_{\lambda }^{-1}{s}_{1}{\gamma }_{\lambda }\right){s}_{1}={s}_{2}{s}_{1}=\left(123\right)\in N\text{.}$ Thus, by Lemma (), $N={A}_{n}\text{.}$ Case cc: $\sigma$ has all cycles of length 2 or 1. Since $\sigma \in {A}_{n},$ $\sigma$ has at least two cycles of length 2. Thus, by Proposition (), ${𝒜}_{\sigma }={𝒞}_{\sigma }={𝒞}_{{\gamma }_{\lambda }}\text{.}$ Since $N$ is normal, ${𝒞}_{{\gamma }_{\lambda }}={𝒜}_{\sigma }\subseteq N\text{.}$ So ${\gamma }_{\lambda }\in N$ and ${s}_{2}{\gamma }_{\lambda }{s}_{2}\in N\text{.}$ Since $N$ is a subgroup ${\gamma }_{\lambda }^{-1}\in N\text{.}$ So ${\gamma }_{\lambda }^{-1}{s}_{2}{\gamma }_{\lambda }{s}_{2}=\left(14\right)\left(23\right)\in N\text{.}$ So ${\pi }_{1}=\left(12\right)\left(34\right)\left(5\right)$ and ${\pi }_{2}=\left(12\right)\left(3\right)\left(45\right)\in N\text{.}$ So ${\pi }_{1}{\pi }_{2}=\left(345\right)\in N\text{.}$ Thus, by Lemma (), $N={A}_{n}\text{.}$

$\square$

Suppose $N$ is a normal subgroup of ${A}_{n},$ $n>4,$ and $N$ contains a $3\text{-cycle.}$ Then $N={A}_{n}\text{.}$

Proof.
 To show: ${A}_{n}\subseteq N\text{.}$ Let $\pi =\left({i}_{1},{i}_{2},{i}_{3}\right)$ be a 3-cycle in $N\text{.}$ Since $n>4,$ $\pi$ has more than one 1-cycle and it follows from Proposition (), that ${𝒜}_{\pi }={𝒞}_{\pi }\text{.}$ Thus, since $N$ is normal, ${𝒞}_{\pi }\subseteq N\text{.}$ So $\left(123\right)$ and $\left(143\right)$ are elements of $N\text{.}$ Then $\sigma =\left(143\right)\left(123\right)=\left(12\right)\left(34\right)={s}_{1}{s}_{3}\in N\text{.}$ Since $\sigma$ has an even cycle, it follows from Proposition (), that ${𝒜}_{\sigma }={𝒞}_{\sigma }\subseteq N\text{.}$ Then $sisj∈ { 𝒞π, if j=i+1; 𝒞σ, otherwise.$ So ${s}_{i}{s}_{j}\in N$ for all $i,j\text{.}$ By, Proposition () and Proposition (), the elements ${s}_{i}j$ generate ${A}_{n}\text{.}$ So ${A}_{n}\subseteq N\text{.}$

$\square$