Dihedral Groups

Look at a plane square:

Let $D_{8}$ be the (soon to be) Group of symmetries of this square. What we mean is all the operations that bring the square back to it's original "look" (even thought the points will move around). For this, this only consists of rotations and reflections.

Let's list these operations:

Do nothing: identity transformation $I$
Rotate 90 degrees CW:

to denote this we use $r$ 3. Rotate 180 degrees CW: $r^{2} = rr$ 4. Rotate 270 degrees CW: $r^{3}$ (equivalent to 90 degrees CCW) $\equiv r^{-1}$ 5. Rotate 360 degrees CW: $r^{4} = I$. 6. Flip on the vertical midline:

notice that this can't be created from just the $r$'s so we need to have a new denotion of the operation, which we call $s$ 7. Flip on the horizontal midline:

8. Flip on the diagonal (descending midline)

9. Flip on the ascending diagonal midline

Okay we want to know, what are the needed transformations to get all of these operations? For example:
7. Flip on the horizontal midline is just: $r^{2} s$
8. Flip on the descending diagonal midline is just: $r^{3} s$
9. To get the flip on the ascending diagonal midline we can do: $r s$

Notice that:

Every symmetry just re-orders the vertices. Namely, each symmetry causes a permutation
Not every permutation is physically possible. For example since $1, 3$ are points that are never next to each other, you can't have a square where $1, 3$ share some edge.
$D_{8} = {1, r, r^{2}, r^{3}, s, r s, r^{2} s, r^{3} s}$ is the dihedral group, and lists all unique permutations.
Notice that means that so $| D_{8} | = 8$ (since there's 8 unique elements). Some books call this $D_{4}$ since the Order (Groups) of this group is 4 (looking at $r^{4} = 1$ )
- $r^{4} = 1$ so $| r | = 4$
- $s^{2} = 1$ so $| s | = 2$
- $r^{3} s = s r$ (see the figure below)

## Definition

dihedral group

This Group is the symmetries of the $n$ -gon. It's called the dihedral group of order $2 n$ "generators". Namely as we did above we have:

$r$ is rotation by $\frac{2 π}{n}$ about the center.
$s$ is a reflection across some midline (just one is needed)

D_{2 n} = {1, r, r^{2}, \dots, r^{n - 1}, s, r s, r^{2} s, \dots, r^{n - 1} s}

The "key relations" here are that:

$r^{n} = 1$ (this implies that $r^{- 1} = r^{n - 1}$ )
$s^{2} = 1$ (this implies that $s = s^{- 1}$ )
$s r = r^{n - 1} s$ (this implies that $s r = r^{- 1} s ⟺ s r s^{- 1} = r^{- 1}$ )

Specifically the $s r s^{- 1}$ part is called the conjugate of $r$ by $s$ .

The book will decide to put the

r

's on the right of the

s

's rather than the left here. You can use the

s r = r^{n - 1} s

relation.

Note

Another property is that:

r^{i} s = s r^{- i}

for all $0 \leq i \leq n$ by doing induction on the previous relation.

Mini Example

If we are considering the dihedral group when $n = 4$ then we have:

\begin{aligned} s r^{3} & = s r r r \\ = r^{- 1} s r r \\ = r^{- 1} r^{- 1} r^{- 1} s \\ = r^{- 3} s \\ = (r^{- 1})^{3} s \\ = (r^{3})^{3} s \\ = r^{9} s \\ = r s \end{aligned}

There's an idea from Linear Algebra that is taken where we have a "presentation" for

D_{2 n}

. Namely:

$D_{2 n} = ⟨ r, s | r^{n} = 1, s^{2} = 1, s r = r^{n - 1} s ⟩$

here the $r, s$ part are called the generators.

Danger

Suppose:

G = ⟨ x, y | x^{4} = 1, y^{3} = 1, x y = y^{2} x^{2} ⟩

We could ask what the orders of $x, y$ are, and other things, but actually $x, y = 1$ here (and thus $G = {1}$ as the trivial group). Hence we try to avoid this notation since it may obscure less obvious facts.

When $n \geq 3$ then the group $D_{2 n}$ is not commutative (since $s r \neq r^{n - 1} s$ ), so it's not an Abelian Group.

Let $n$ be a fixed positive integer. Define a relation on $Z$ by:
$a \sim b ⟺ n | (b - a)$
clearly $a \sim a, a \sim b \to b \sim a$ so the relation is trivially reflexive and symmetric. Further work shows it's transitive too, so then $\sim$ is an equivalence relation, so we can write:
$a \sim b \Rightarrow a \equiv b (mod n)$
For any $k \in Z$ we'll denote the equivalence class of $a$ as $\overset{―}{a}$ , named the congruence/residue class of $a$ . It consists of the integers from $a$ that by an integral multiple of $n$ (using the relation above), ie:
$\begin{aligned} \overset{―}{a} & = {a + k n | k \in Z} \\ = {a, a \pm n, a \pm 2 n, \dots} \end{aligned}$
There are precisely $n$ distinct equivalence classes $mod n$ , namely:
$\overset{―}{0}, \overset{―}{1}, \dots, \overset{―}{n - 1}$
determined by the possible remainders after division by $n$ .

\frac{Z}{n Z}

The groups $\frac{Z}{n Z}$ are the sets of equivalence classes $\overset{―}{0}, \overset{―}{1}, \dots, \overset{―}{n - 1}$ such that each one describes the integers with remainder of the equivalence class.

\frac{Z}{6 Z}

There are 6 possible remainders here:
$\overset{―}{0}, \overset{―}{1}, \overset{―}{2}, \overset{―}{3}, \overset{―}{4}, \overset{―}{5}$
Where, for example, $\overset{―}{2} = {\dots, - 10, - 4, 2, 8, \dots}$ .

A cool thing here is that any element of the equivalence class can represent it. Namely:
$\dots = \overset{―}{- 4} = \overset{―}{0} = \overset{―}{6} = \overset{―}{12} = \dots$

Some Properties

For $(\frac{Z}{n Z}, +)$ then $\overset{―}{a} + \overset{―}{b} = \overset{―}{a + b}$ (so it's an Abelian Group of order $n$ )
For $(\frac{Z}{n Z}, \cdot)$ then $\overset{―}{a} \overset{―}{b} = \overset{―}{a b}$ (almost an Abelian Group, but it lacks the inverse for some elements like $\overset{―}{0}$ )
- Here $\overset{―}{a} \in \frac{Z}{n Z}$ has in inverse under $\cdot$ iff $gcd (a, n) = 1$ .

\frac{Z}{6 Z}

Here:
$- \overset{―}{2} = \overset{―}{- 2} = \overset{―}{4}$
$\overset{―}{1} \cdot \overset{―}{1} = \overset{―}{1}$
$\overset{―}{5} \cdot \overset{―}{5} = \overset{―}{25} = \overset{―}{1}$
Notice that $\overset{―}{1}, \overset{―}{5}$ are the only elements with other.

The group ${(\frac{Z}{n Z})}^{\times} = {\overset{―}{a} \in \frac{Z}{n Z} | gcd (a, n) = 1}$ is an Abelian Group under $\cdot$ .
- Back in the day, they used $\overset{―}{a} = \overset{―}{b} ⟺ a \equiv b (mod n)$ . In $\frac{Z}{n Z}$ this is " $a$ congruent to $b$ modulo $n$ ".

Permutation Group

Consider the Permuation Group. Any finite set $X$ will have perumations:

permuation

A permutation of $X$ is a bijection $σ : X \to X$ . It's essentially a shuffling of the elements in $X$ .

X = {1, 2, 3, 4, 5}

Count the number of bijections $σ : X \to X$ . Using principles of proofs, this is the number of injections $σ : X \to X$ (since $X$ is finite). Once we know where $1$ goes we have 5 options, then for 2 we have 4 options, ... so then we have $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5! = 120$ possibilities.

There are some notational options here (which is where the old-timey modulo notation comes into play):

(Functional Notation:) one permuation is: $σ (1) = 2, σ (2) = 3, σ (3) = 5, σ (4) = 1, σ (5) = 4$ . This is obviously super slow so we don't want to do this one.
(Modified Functional Notation:) write it instead with $\to$ : $1 \mapsto 2, 2 \mapsto 3, \dots$ .
(Use A Table:) write where the top row maps to the bottom row:
$(\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 5 & 1 & 4 \end{matrix})$
("Cycle Notation":) write out a visual graph of what number maps to where:

We write this as:
$σ = (\begin{matrix} 1 & 2 & 3 & 5 & 4 \end{matrix})$
Another example would be:
$τ = (\begin{matrix} 1 & 3 & 5 \end{matrix}) (\begin{matrix} 2 & 4 \end{matrix})$
Creates the following:

Here $τ$ has a 3-cycle and a 2-cycle (or a transposition). This $τ$ is a product of disjoint cycles. Notice that since we have a cycle where $n \geq 3$ then the group isn't abelian with operation $τ$ .