Dihedral Groups

The symmetries of an n-gon form a group (Proved in Prerequisites) We call it Dihedral Group

Theorem 1.1

The dihedral group \(D_n\), has precisely \(\left|D_n\right|=2 n\) elements

Proof

\(D_n\) consists of:

  • The identity \(e \in D_n\) (Trivial)
  • The \(n-1\) rotations (any direction) through angles \(k \cdot \frac{2 \pi}{n}\), \(k \in \mathbb{Z}\) and \(k=1, \ldots, n-1\).
  • The \(n\) reflections (This is left as an exercise for the readers)

Since \(1+(n-1)+n=2n\) we have \(\left|D_n\right|=2 n\) \(\blacksquare\)

Now if we denote \(g\) as rotation and \(h\) as reflection through a particular vertex.

Exercise

\[D_n=\left\{e, g, g^2, \ldots, g^{n-1}, h, g h, g^2 h, \ldots, g^{n-1} h\right\}\]

Hint: What is the physical meaning of \(g^{n-1} h\)

Symmetric Groups

The set \(S_n\) of all symmetries of \(\{1, \ldots, n\}\) is called the symmetric group, where \(\left|S_n\right|=n !\). Consider a graph with \(n\) vertices and no edges connecting them. This is \(S_n\)

Showing \(S_n\) is a group is left as an exercise for the reader \(\blacksquare\)

Matrices

Theorem 1.3

\(\mathrm{GL}(n, \mathbb{R})\) is a group under matrix multiplication.

Proof

G1 (Closure): Claim: Consider two Matrices \(A\) and \(B\). If \(A\) and \(B\) are invertible then \(AB\) is invertible.

Proof: \(A\) and \(B\) invertible \(\Rightarrow\) \({\displaystyle \det(A) \neq 0}\) and \({\displaystyle \det(B) \neq 0}\). From \({\displaystyle \det(AB)=\det(A)\det(B)}\) we can see the above implies \({\displaystyle \det(AB) \neq 0}\). Hence \(AB\) is invertible \(\blacksquare\)

G2 (Associative): \(A(BC)=(AB)(C)\) This is left as an exercise for the reader \(\blacksquare\)

G3(Identity): The Matrix I is the Identity element \(\blacksquare\)

G4(Inverse): For every invertible matrix \(\exists\) a inverse matrix. Let \(A\) be any invertible matrix. Then \(A^{-1}\) is the invertible matrix s.t. \(AA^{-1}=A^{-1}A=I\) \(\blacksquare\)

From \(G1-G4\), \(\mathrm{GL}(n, \mathbb{R})\) is a group under matrix multiplication \(\blacksquare\)

Subgroups

S1. \(H\) is not empty. S2. If \(h, k \in H\) then \(h * k \in H\) S3. If \(h \in H\) then \(h^{-1} \in H\).

A detour on \(\operatorname{SO}(3)\)

\(\operatorname{SO}(3)\) is very important in physics.

Heard the word SO(3)?

\(\operatorname{SO}(n, \mathbb{R}):=\left\{A \in G L(n, \mathbb{R}) \mid \operatorname{det} A=1\right.\) and \(\left.A^T=A^{-1}\right\}\). It is related to Linear Algebra, in particular, Isometry.

Sidenote: \(\operatorname{SO}(n, \mathbb{R})\) is the “combination” of \(\mathrm{SL}(n, \mathbb{R})\) and \(\mathrm{O}(n, \mathbb{R})\)

Exercise

\(\operatorname{SO}(n, \mathbb{R})\) is a subgroup of \(\mathrm{GL}(n, \mathbb{R})\)

Abelian Groups

Suppose that \(G\) is a group. If \(g * h=h * g\) for all \(g, h \in G\), then \(G\) is an abelian group.

Exercise:

Claim: \(D_3\) is not abelian:

Proof:

\[\sigma=\left(\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right), \quad \tau=\left(\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right), \quad \sigma \tau=\left(\begin{array}{lll} 1 & 2 & 3 \\ 1 & 3 & 2 \end{array}\right), \quad \tau \sigma=\left(\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 3 \end{array}\right) .\]

Since \(\sigma \tau \neq \tau \sigma\), \(D_3\) is not abelian \(\blacksquare\)

Normal Subgroups

\(N\) is a normal subgroup and only if \(g n g^{-1} \in N\) for all \(g \in G\) and \(n \in N\). We denote: \(N \triangleleft G\).

Theorem 1.2

Claim: Every Subgroup of an Abelian Group is Normal

Proof

Suppose \(G\) is abelian and let \(H \leq G\). Then \(\forall g\) \(\in\) \(G\), \(\forall h\) \(\in\) \(H\),
\(g h g^{-1}=g g^{-1} h=e h=h \in H\)

Hence \(H \triangleleft G \quad \blacksquare\)

Definition:

Let \(G\) be a group and let \(g \in G\) be an element. We define the subset \(\langle g\rangle:=\left\{g^k \mid k \in \mathbb{Z}\right\}=\left\{\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots\right\}\)

Definition:

A subgroup \(H \leq G\) is cyclic if \(H=\langle h\rangle\) for some \(h \in H\).

Returning to our Dihedral Group \(D_n\), the subgroup \(H\) consisting of the identity and all the rotations, i.e. \(H=\left\{e, g, g^2, \ldots, g^{n-1}\right\}\) is a cyclic subgroup since \(H=\langle g\rangle\)

You should verify that \(H\) is abelian, this is left as an exercise for the reader \(\blacksquare\)

If you have done the exercise above, it should be clear that \(G\) is cyclic \(\Rightarrow\) \(G\) is abelian \(\blacksquare\)

Exercise:

Explain why the converse statement is not true? i.e., \(G\) is abelian \(\nRightarrow\) \(G\) is cyclic

Hint: Consider \(\mathbb{R}\) under addition

First, define \(A_n\), the Alternating Group, as the even permutations of \(S_n\). As \(S_n\) has the same number of odd and even permutations (for any groups this holds!), order of \(A_n=\frac{n!}{2}\). For \(A_4\), order \(\frac{4!}{2}=12\)