Group theory

Table of contents

• Group Cosets & Lagrange’s Theorem
  • Cosets
  • Index
  • Properties of Cosets
  • Lagrange’s Theorem
• Homomorphisms and Normal Subgroups
  • Normal Subgroups
  • The Quotient Group
  • Linear Transformations
  • Homomorphisms and Isomorphisms of Groups
  • Kernel and Image
  • The First Isomorphism Theorem
  • The Sign of a Permutation
  • Classifying Groups
  • Fundamental Theorem of Finite Abelian Groups
• Group Presentations
  • Dihedral Group

Group Cosets & Lagrange’s Theorem

Cosets

Definition: Let $G$ be a group and $H$ a subgroup. Get $g$ be an element of $G$. We call the set
$gH=\{gh:h\in H\}$ a left coset of $H$ in $G$, and the set
$Hg=\{hg:h\in G\}$ a right coset of $H$ in $G$

• In an Abelian group, there is no difference between left and right cosets
• if $G$ is an additive group, we write cosets additively, e.g. $g+H=\{g+h:h\in H\}$
  • As additive groups are always Abelian, $g+H=H+g$

Lemma: Let $G$ be a group and $H$ a subgroup. Let $g_1,g_2\in G$. Then $g_{1}H=g_{2}H$ if and only if $g_1^{-1}{g}_2\in H$
Proof: Suppose $g_{1}H=g_{2}H$. Note that $g_2=g_2\cdot 1_G\in g_{2}H$ as $H$ is a subgroup of $G$. But $g_{2}H=g_{1}H$, and so $g_2\in g_{1}H$, so $g_2=g_{1}h$ for some $h\in H$. Hence $g_1^{-1}{g}_2=h\in H$ as required.
Conversely, suppose $g_1^{-1}{g}_2=h\in H$. We want to show that $g_{1}H=g_{2}H$. Let $k\in g_{1}H$. Then $k=g_1{h}_1$ for some $h_1\in H$. But $h^{-1}=g_2^{-1}{g}_1$, so $g_1=g_2{h}^{-1}$, so $k=g_1{h}_1=g_2{h}^{-1}{h}_1\in g_{2}H$. Therefore every $k\in g_{1}H$ belongs to $g_{2}H$. By a similar argument, every $k\in g_{2}H$ belongs to $g_{1}H$. Thus, $g_{1}H=g_{2}H$

Index

Definition: Let $G$ be a group and $H$ be a subgroup. Define the left index of $H$ in $G$, denoted by $[G:H]$, to be the number of left cosets of $H$ in $G$. Likewise, define the right index of $H$ in $G$ to be the number of right cosets of $H$ in $G$.

Properties of Cosets

Lemma: Let $G$ be a group and $H$ a finite subgroup. If $g\in G$ then $gH$ and $Hg$ have the same number of elements as $H$.
Proof: Let $g\in G$. We want to show that $H$ and $gH$ have the same number of elements. Both $H$ and $gH$ are finite, and the best way to show that two finite sets have the same number of elements is to set up a bijection between them.
Let
$\varphi :H\to gH,h\mapsto gh$From the definition of $gH$ it is clear that $\varphi (h)$ is in the coset $gH$ whenever $h$ is in the subgroup $H$. So the map $\varphi$ makes sense.
To check this map is bijective we must check it is both injective and surjective:

• Injectivity: Suppose two elements $h_1,h_2$ map to the same element in $gH$ ($\varphi (h_1)=\varphi (h_2)$). This means that $g{h}_1=g{h}_2$. If we left multiply both sides by $g^{-1}$, we obtain $g^{-1}(g{h}_1)=g^{-1}(g{h}_2)$, and thus $h_1=h_2$
• Surjectivity: Suppose $k\in gH$, we want to show that $k$ is of the form $\varphi (h)$ for some $h\in H$. By definition, $gH=\{gh:h\in H\}$, so $k=gh=\varphi (h)$ for some $h\in H$

Lemma: Let $G$ be a group and $H$ be a subgroup. Let $g_1,g_2\in G$. Then the cosets $g_{1}H$ and $g_{2}H$ are either equal or disjoint.
Proof: Suppose $g1H$ and $g_{2}H$ are not disjoint. We want to show they are necessarily equal. As $g_{1}H$ and $g_{2}H$ are not disjoint, they must have a common element which we can denote $g_3$. Then $$\begin{gathered} g_3 \\ in{g}_{1}H \end{gathered}$$ and $g_2\in g_{2}H$. Thus $g_3=g_1{h}_1,g_3=g_2{h}_2,h_1,h_2\in H$ Then $g_1^{-1}{g}_2=(g_3{h}_1^{-1}{)}^{-1}{g}_3{h}_2^{-1}=h_1{g}_3^{-1}{g}_3{h}_2^{-1}=h_1{h}_2^{-1}\in H$By lemma we have $g_{1}H=g_{2}H$

Lagrange’s Theorem

Definition: Let $A$ be a set and $B_1,B_2,...,B_n$ be subsets of $A$. We say that $B_1,B_2,...,B_n$ is a partition of $A$ if the following conditions hold:

• $A=B_1\cup B_2\cup ...\cup B_n$
• If $i\ne j$ then $B_i\cap B_j=\varnothing$ (All subsets are pairwise disjoint)
• $B_i\ne \varnothing$ for $i=1,...,n$

Lemma: Let $G$ be a finite group and $H$ a subgroup. The left cosets of $H$ in $G$ form a partition of $G$.
Proof: Let $g_{1}H,g_{2}H,...,g_{m}H$ be the distinct left cosets of $H$. As they are distinct, we know by lemma that they are disjoint. Suppose now that $g\in G$. Then $gH$ must equal one of the $g_{i}H$. But $g\in gH$ as $1\in H$. Hence, the cosets $g_{1}H,g_{2}H,...,g_{m}H$ are not only disjoint, but every element of $G$ belongs to one of them, and so their union is equal to $G$. Finally, $1_G\in H$ so $g_i\in g_{i}H$, so all cosets are non-empty, and so the cosets form a partition of $G$.

Theorem (Lagrange’s Theorem): Let $G$ be a finite group and $H$ a subgroup. Then $\#G=[G:H]\cdot \#H$
Proof: Let $g_{1}H,g_{2}H,...,g_{m}H$ be the distinct left cosets of $H$. By lemma, these form a partition of $G$, and so $\#G=\#g_{1}H+\#g_{2}H+...+\#g_{m}H$ Now, by lemma $\#g_{1}H=\#g_{2}H=...=\#g_{m}H=\#H$ Hence $\#G=m\cdot \#H$ where $m$ is the number of left cosets of $H$ in $G$, which we defined to be the index of $H$ in $G$, so $m=[G:H]$

Corollary: Let $G$ be a group and $H$ a subgroup. Then $\#H\mid \#G$
Proof: This follows from Lagrange’s Theorem as the index $[G:H]$ is an integer.

Corollary: Let $G$ be a finite group. Let $g\in G$ have order $n$. Then $n\mid \#G$
Proof: Let $H=⟨g⟩$ the cyclic group generated by $g$. We know from lemma that $\#H=n$. By corollary we have $n\mid \#G$

Homomorphisms and Normal Subgroups

Normal Subgroups

Let $H$ be a subgroup of $G$. A conjugate of $H$ has the form $gH{g}^{-1}$ for some $g\in G$

We say that a subgroup $H$ of $G$ is normal if and only if $gH{g}^{-1}=H$ for all $g\in G$. That is, $H$ is normal if and only if it is equal to all its conjugates. We write $H⊴G$ to denote that $H$ is a normal subgroup of $G$.

Lemma: Let $H$ be a subgroup of $G$. Then the following are equivalent.

1. $H$ is normal in $G$
2. $gH{g}^{-1}=H$ for all $g\in G$
3. $gH=Hg$ for all $g\in G$
4. $gH{g}^{-1}\subseteq H$ for all $g\in G$
5. $gh{g}^{-1}\in H$ for all $g\in G,h\in H$
   Proof: It is easy to see that $(1)\iff (2)\iff (3),(4)\iff (5)$ and also that $(2)\Rightarrow (4)$. Let’s do $(4)\Rightarrow (2)$. Suppose $gH{g}^{-1}\subseteq H$ for all $g\in G$. Then, since $g^{-}1\in G$ we have $g^{-1}Hg\subseteq H$. Left multiplying by $g$ and right multiplying by $g^{-1}$ gives $H=g(g^{-1}Hg)g^{-1}\subseteq gH{g}^{-1}$ As $gH{g}^{-1}\subseteq H$ and $H\subseteq gH{g}^{-1}$ we have $gH{g}^{-1}=H$

Lemma: Let $G$ be a finite group and let $H$ be a subgroup of $G$ of index 2. Then $H$ is normal in $G$.
Proof: We want to show that $gH=Hg$ for all $g\in G$. By lemma, $gH=H$ if and only if $g\in H$.

• Suppose first that $g\in H$. Then $gH=H$ and $Hg=H$, and so $gH=Hg$
• Suppose instead that $g\notin H$. Then $gH\ne H$. But $H$ has index 2 in $G$ and so has exactly 2 left cosets, which must be $H$ and $gH$. Therefore, $G=H\cup gH$ and $H\cap gH=\varnothing$, since cosets form a partition. Thus, $gH=G\backslash H$. Similarly, $Hg=G\backslash H$. Hence, $gH=Hg$.

The Quotient Group

Let $G$ be a group and $H$ a subgroup. We write $G/H$ for the set of left cosets of $H$ in $G$: $G/H=\{gH:g\in H\}$
If $N$ is normal subgroup of $G$, we define multiplication on $G/N$ by $(gN)(g^{\prime }N)=g{g}^{\prime }N$
Lemma: Suppose $N$ is normal in $G$. Then $(gN)(g^{\prime }N)=g{g}^{\prime }N$ is well-defined.
Proof: Suppose $xN=aN$ and $yN=bN$. For the definition to be sensible we want $xyN=abN$. By lemma we know that $x^{-1}a\in N$ and $y^{-1}b\in N$. We want to prove that $(xy{)}^{-1}ab\in N$. However, $(xy{)}^{-1}ab=y^{-1}{x}^{-1}ab=y^{-1}\cdot (x^{-1}a)\cdot y\cdot y^{-1}b$
We know that as $N$ is normal and $x^{-1}a\in N$, we have $y^{-1}\cdot (x^{-1}a)\cdot y\in N$. Therefore, $(xy{)}^{-1}ab\in N$.

Lemma: Let $G$ be a group and $N$ a normal subgroup. Then $G/N$ with $(gN)(g^{\prime }N)=g{g}^{\prime }N$ is a group with identity element $N=1_{G}N$ and inverses given by $(gN{)}^{-1}=g^{-1}N$. Moreover, if $G$ is finite then $\#(G/N)=[G:N]=\frac{\#G}{\#N}$ We call $G/N$ the quotient group of $G$ over $N$.
Proof: The fact that $G/N$ is a group easily follows from the fact $G$ is a group. For the last part, note that $G/N$ is the set of left cosets of $N$, so $\#(G/N)=[G:N]$ by definition of the index. By Lagrange’s Theorem, $[G:N]=\frac{\#G}{\#N}$

Linear Transformations

Let $F$ be a field. Let $V$ and $W$ be $F$-vector spaces. A linear transformation is a map $\varphi :V\to W$ satisfying:

• $\varphi ({\mathbf{v}}_1+{\mathbf{v}}_2)=\varphi ({\mathbf{v}}_1)+\varphi ({\mathbf{v}}_2)$ for all ${\mathbf{v}}_1,{\mathbf{v}}_2\in V$
• $\varphi (\lambda \mathbf{v})=\lambda \varphi (\mathbf{v})$ for all $\lambda \in F$ and $\mathbf{v}\in V$
  We define the kernel and image of the linear transformation $\varphi$ by $\text{Ker}(\varphi )=\{\mathbf{v}\in V:\varphi (\mathbf{v})=0\},\text{Im}(\varphi )=\{\varphi (\mathbf{v}):\mathbf{v}\in V\}$
  We recall that $\text{Ker}(\varphi )$ is a subspace of $V$ and $\text{Im}(\varphi )$ is a subspace of $W$.

Homomorphisms and Isomorphisms of Groups

Definition: Let $G$, $H$ be groups and let $\varphi :G\to H$ be a map. We say that $\varphi$ is a homomorphism of groups if $\varphi (gh)=\varphi (g)\varphi (h)$ for all $g,h\in G$

Definition: Let $G$ and $H$ be groups. A map $\varphi :G\to H$ is an isomorphism if it is a bijective homomorphism. If $G$ and $H$ are isomorphic we write $G\cong H$.

Kernel and Image

Associated to any homomorphism $\varphi :G\to H$ are its kernel and image:
$\text{Ker}(\varphi )=\{g\in G:\varphi (g)=1_H\},\text{Im}(\varphi )=\{\varphi (g):g\in G\}$
Theorem: Let $\varphi :G\to H$ be a homomorphism of groups. Then

• $\text{Ker}(\varphi )$ is a normal subgroup of $G$
• $\text{Im}(\varphi )$ is a subgroup of $H$

Lemma: Let $\varphi :G\to H$ be a homomorphism of groups. Then $\varphi$ is injective if and only if $\text{Ker}(\varphi )=\{1_G\}$
Proof: Suppose $\text{Ker}(\varphi )=\{1_G\}$. Let $g_1,g_2\in G$ and suppose $\varphi (g_1)=\varphi (g_2)$. Then $\varphi (g_1^{-1}{g}_2)=\varphi (g_1{)}^{-1}\varphi (g_2)=1_H$. Thus $g_1^{-1}{g}_2\in \text{Ker}(\varphi )=\{1_G\}$ and so $g_1^{-1}{g}_2=1_G$ and hence $g_1=g_2$. Therefore $\varphi$ is injective.
Conversely, suppose $\varphi$ is injective. Let $g\in \text{Ker}(\varphi )$. Thus $\varphi (g)=1_H=\varphi (1_G)$. As $\varphi$ is injective, $g=1_G$. Hence $\text{Ker}(\varphi )=\{1_G\}$

The First Isomorphism Theorem

Theorem (The First Isomorphism Theorem): Let $\varphi :G\to H$ be a homomorphism of groups. Let $\hat{\varphi }:G/\text{Ker}(\varphi )\to \text{Im}(\varphi ),\hat{\varphi }(g\text{Ker}(\varphi ))=\varphi (g)$Then $\hat{\varphi }$ is a well-defined group isomorphism.
Proof: Lets show first that $\hat{\varphi }$ is well-defined. Let $g_1,g_2\in G$ and suppose $g_1\text{Ker}(\varphi )=g_2\text{Ker}(\varphi )$. $\hat{\varphi }$ sends $g_1\text{Ker}(\varphi )$ to $\varphi (g_1)$ and $g_2\text{Ker}(\varphi )$ to $\varphi (g_2)$. For $\hat{\varphi }$ to be well-defined we want $\varphi (g_1)=\varphi (g_2)$. $g_1\text{Ker}(\varphi )=g_2\text{Ker}(\varphi )$ tells us that $g_1^{-1}{g}_2\in \text{Ker}(\varphi )$ and so by definition of kernel, $\varphi (g_1^{-1}{g}_2)=1_H$. As $\varphi$ is a homomorphism, we have $\varphi (g_1{)}^{-1}\varphi (g_2)=1_H$, so $\varphi (g_1)=\varphi (g_2)$. Thus, $\hat{\varphi }$ is well-defined.
To show that $\hat{\varphi }$ is a homomorphism note that
$\hat{\varphi }(g_1\text{Ker}(\varphi )\cdot g_2\text{Ker}(\varphi ))=\hat{\varphi }(g_1{g}_2\text{Ker}(\varphi ))=\varphi (g_1{g}_2)=\varphi (g_1)\varphi (g_2)=\hat{\varphi }(g_1\text{Ker}(\varphi ))\cdot \hat{\varphi }(g_2\text{Ker}(\varphi ))$
To show that $\hat{\varphi }$ is injective we apply lemma. Let $g\text{Ker}(\varphi )\in \text{Ker}(\hat{\varphi })$. Then $\varphi (g)=\hat{\varphi }(g\text{Ker}(\varphi ))=1_H$. Thus $g\in \text{Ker}(\varphi )$. Therefore, $g\text{Ker}(\varphi )=1_G\text{Ker}(\varphi )$. It is clear from the definition of $\hat{\varphi }$ that it is surjective, and so $\hat{\varphi }$ is an isomorphism.

The Sign of a Permutation

Let $n\ge 2$. An important homomorphism to know is the sign of a permutation. Recall that a permutation $\sigma \in S_n$ is even if it can be written as a product of an even number of transpositions, and odd if it can be written as a product of an odd number of permutations. A permutation cannot be both even and odd.

We define the sign of a permutation $\sigma \in S_n$ as follows: $\text{sign}(\sigma )=\{\begin{array}{ll}1& \text{if }\sigma \text{ is even}\\ -1& \text{if }\sigma \text{ is odd}\end{array}$ We obtain a homomorphism $\text{sign}:S_n\to \{\pm 1\},\sigma \mapsto \text{sign}(\sigma )$ How do we know that sign is a homomorphism? We need to check that $\text{sign}(\sigma \tau )=\text{sign}(\sigma )\text{sign}(\tau )$. Write $\sigma ={ϵ}_1...{ϵ}_m,\tau ={\delta }_1...{\delta }_n$ as products of transpositions. Then $\text{sign}(\sigma )=(-1{)}^m,\text{sign}(\tau )=(-1{)}^n$
However, $\sigma \tau ={ϵ}_1...{ϵ}_m{\delta }_1...{\delta }_n$ which is a product of $m+n$ transpositions. Thus $\text{sign}(\sigma \tau )=(-1{)}^{m+n}=\text{sign}(\sigma )\text{sign}(\tau )$ Hence sign is a homomorphism

Theorem: Let $n\ge 2$. Then $A_n$ is a normal subgroup of $S_n$. Moreover, $[S_n:A_n]=2,\#A_n=\frac{\#S_n}{2}=\frac{n!}{2}$ Proof: Now that we’ve checked that sign is a homomorphism, we can observe that the kernel is defined to be $A_n=\{\sigma \in S_n:\sigma \text{ is even}\}$ So we know that $A_n$ is a normal subgroup of $S_n$.
It is easy to see that sign is surjective, since $\text{sign}(\text{id})=1$ and $\text{sign}((1,2))=-1$. So $\text{Im}(\text{sign})=\{1,-1\}$
Applying the first isomorphism theorem to $\text{sign}:S_n\to \{1,-1\}$ gives an isomorphism $\widehat{\text{sign}}:S_n/A_n\to \{1,-1\},\widehat{\text{sign}}(\sigma A_n)=\text{sign}(\sigma )$
Thus $[S_n:A_n]=\#(s_n/A_n)=\#\{1,-1\}=2$. Therefore $\#A_n=\frac{\#S_n}{2}=\frac{n!}{2}$

Classifying Groups

Lemma: Let $G$ and $H$ be cyclic groups of order $n$. Then $G$ and $H$ are isomorphic.
Proof: Let $G=⟨g⟩$ and $H=⟨h⟩$, where $g$ and $h$ have order $n$. Define $\varphi :G\to H,\varphi (g^i)=h^i$
Is this well-defined? Suppose there are two integers $i$ and $j$ such that $g^i=g^j$. The map $\varphi$ sends $g^i$ to $h^i$ and $g^j$ to $h^j$. For the definition to be sensible, $h^i$ must be equal to $h^j$. $g^i=g^j$ tells me that $g^{i-j}=1_G$ and since $g$ has order $n$ we have $n\mid (i-j)$. But $h$ also has order $n$, so $h^{i-j}=1_h$ and hence $h^i=h^j$. It follows that $\varphi$ is well-defined.
We now need to check that $\varphi$ is a homomorphism. We want $\varphi (g^i\cdot g^j)=\varphi (g^i)\cdot \varphi (g^j)$ which is the same as $h^{i+j}=h^i\cdot h^j$, which is true. Finally, we need to show that $\varphi$ is a bijection. Let $\psi :H\to G,\psi (h^i)=g^i$Then $\psi$ is well-defined as before and $\psi ,\varphi$ are mutual inverses. Since $\varphi$ has an inverse it must be a bijection.

Lemma: Let $p$ be a prime. Any group of order $p$ is isomorphic to $C_p$
Proof:

Definition: Let $G$ and $H$ be groups. We define the direct product of $G$ and $H$ to be $G\times H=\{(g,h):g\in G,h\in H\}$ i.e. $G\times H$ is the set of ordered pairs $(g,h)$ where $g\in G$ and $h\in H$. The binary operation on $G\times H$ is $(g_1,h_1)\cdot (g_2,h_2)=(g_1{g}_2,h_1{h}_2)$
Lemma: $G\times H$ is a group with identity element $(1_G,1_H)$ and inverse given by $(g,h{)}^{-12}=(g^{-1},h^{-1})$. If $G$ and $H$ are finite then $\#(G\times H)=(\#G)\cdot (\#H)$ Proof:

Theorem: The only groups of order 4 are $C_4$ and $C_2\times C_2$
Proof: What this theorem states is that any group of order 4 is isomorphic to either $C_4$ or $C_2\times C_2$. Lets write $C_2=⟨u⟩$ where $u$ has order 2. Then $C_2=\{1,u\},C_2\times C_2=\{(1,1),(1,u),(u,1),(u,u)\}$ Note that $(1,1)$ has order 1 and all other elements have order 2. In particular, $C_4$ and $C_2\times C_2$ are non-isomorphic as the latter has no elements of order 4.
Let $G$ be a group of order 4. We want to show that $G\cong C_4$ or $G\cong C_2\times C_2$. If $G$ is cyclic, then $G\cong C_4$. Thus we may suppose that $G$ is not cyclic, and so $G$ has no element of order 4. Write $G=\{1,a,b,c\}$. The elements $a,b,c$ have order 2 by Lagrange’s Theorem.

• What is $ab$? It must be one of the 4 elements of $G$. If $ab=1$ then $b=a^{-1}=a$ giving a contradiction. If $ab=a$ then $b=1$ giving a contradiction. Likewise $ab=b$ gives a contradiction. Therefore, $ab=c$. We find that whenever we multiply two distinct elements among $a,b,c$, we obtain the third one.
  Now, let $\varphi :G\to C_2\times C_2=\{\begin{array}{ll}\varphi (1)& (1,1)\\ \varphi (a)& (1,u)\\ \varphi (b)& (u,1)\\ \varphi (c)& (u,u)\end{array}$ and we just check (with the help of a multiplication table) that $\varphi$ is an isomorphism. Thus, $G\cong C_2\times C_2$.

One way of thinking about isomorphisms of groups $\varphi :G\to H$ is that $\varphi$ is a map that sends the multiplication table of $G$ to the multiplication table of $H$.

Fundamental Theorem of Finite Abelian Groups

Theorem (Fundamental Theorem of Finite Abelian Groups): Let $G$ be a non-trivial finite abelian group. Then there are integers $b_1\mid b_2\mid ...\mid b_2,b_1,...,b_n>1$, such that $G\cong C_{b_1}\times C_{b_2}\times ...\times C_{b_n}$
Here, $C_b$ is the cyclic group of order $b$. The integers $b_1,...,b_n$ are called the invariants of the finite abelian group $G$. Note that they are positive integers, that each divides the next one, and that $\#G=b_1{b}_2...b_n$.

Group Presentations

Dihedral Group

A regular $n$-gon has $2n$ symmetries (consisting of $n$ rotations and $n$ reflections). The set of these symmetries forms a group which we will denote $D_{2n}$, which is called the dihedral group of order $2n$.

• Note that some sources will denote this $D_n$, however $D_{2n}$ has been chosen to be consistent with the resources from Algebra 1 (MA151) and Algebra 3 (MA268)

Lemma: Let $a\in D_{2n}$. Suppose $a$ fixes vertices 1 and 2. Then $a=\text{id}$ ^512
Proof: The proof is geometric. Suppose $a$ fixes vertices 1 and 2. Then $a$ fixes the whole line segment joining these two vertices. But $a$ also fixes $O$ (the centre of the $n$-gon). Therefore, $a$ fixes the whole $n$-gon and so $a=\text{id}$

Lemma: Let $b,c\in D_{2n}$. Suppose $b(1)=c(1)$ and $b(2)=c(2)$. Then $b=c$ ^513
Proof: Here we consider $b$ and $c$ as permutations of $\{1,2,...,n\}$. Suppose $b(1)=c(1)=k$ and $b(2)=c(2)=\ell$ where $k$ and $\ell$ are vertex numbers. Let $a=c^{-1}b\in D_{2n}$. Then $a(1)=c^{-1}(b(1))=c^{-1}(k)=1),a(2)b=c^{-1}(b(2))=c^{-1}(\ell )=2$ By lemma we get $a=\text{id}$, so $b=c$

Theorem: Let $n\ge 3$. Then $D_{2n}=\{\text{id},r,r^2,...,r^{n-1}\}\cup \{s,sr,s{r}^2,...,s{r}^{n-1}\}=R\cup sR$ In particular, $\#D_{2n}=2n$ and $R$ is a normal subgroup of index 2.
Proof: Let $a\in D_{2n}$. Let $a(1)=k$. Note that $a$ must map the vertices adjacent to the vertex 1 (2 and $n$) to the vertices adjacent to the vertex $k$ ($k-1$ and $k+1$. Therefore, we have 2 cases, either:

\begin{cases} n \mapsto k −1 & \\ 1 \mapsto k & \\ 2 \mapsto k+1 & \end{cases}$$ or $$a: \begin{cases} n & \mapsto k + 1 \\ 1 & \mapsto k \\ 2 &\mapsto k—1 \end{cases}$$ In the first case, let $b = r^{k-1}$. Note that $$a(1) = k = b(1), \quad a(2) = k + 1 = b(2)$$ so $a = b = r^k$ by [[#ae07c0|lemma]]. Now we take the second case. Note that $$s : \begin{cases} n & \mapsto 2 \\ 1 & \mapsto 1 \\ 2 & \mapsto n \end{cases}$$ So $$as : \begin{cases} n & \mapsto k—1 \\ 1 & \mapsto k \\ 2 & \mapsto k + 1 \end{cases}$$ Hence $as = r^{k-1}$ so $a = r^{k-1}s^{-1} = r^{k-1}s$. We conclude that any $a \in D{2n}$ belongs to either $R$ or $Rs$. Thus $D{2n} = R \cup Rs$. Note that any two cosets $R$ and $Rs$ are distinct as $s \not \in R$. Therefore $[D{2n} : R] = 2$ and so $R$ is normal in $D{2n}$ by [[#6f1afc|lemma]]. Therefore $Rs = sR$ and so $D_{2n} = R \cup RS$ as required. Lemma: With $r, s$ as above, $$r^n = \text{id}, \quad s^2 = \text{id}, \quad srs = r^{-1}$$ *Proof:* The first two relations are trivial. For the third, note that $s^{-1} = s$. Therefore, $srs = srs^{-1}$. As $R$ is normal in $D_{2n}$ and $r \in R$ we have $srs \in R$. So $srs = r^k$ for some $k$. We compute $$(srs)(1) = (sr)(s(1)) = (sr)(1) = s(r(1)) = s(2) = n$$ Hence $srs = r^{n-1} = r^{-1}$ ## Generators ## Group Presentations ## Homomorphisms From Groups Defined By Presentations ## The Quaternion Group $Q_8$ ## A Presentation for $D_{2n}$ # Classification of Groups ## Groups with Exponent 2 ## Groups of Order 6 ## Groups of Order 8 # Group Actions ## Group Actions ## Cauchy’s Theorem ## Conjugation ## Conjugacy Classes in $S_n$