04 - Families of Groups

We have a few families of groups so far (just to review):

$(\frac{Z}{n Z})$ (see Z over nZ - The Integers Modulo n)
1. Finite of order $n$
2. Cyclic, ie: generated by a signle element usually $\overset{―}{1}$ , so it's abelian
$D_{2 n}$ the Dihedral Groups of order $2 n$
1. Finite (order $2 n$ )
2. Not cyclic, but it can be generated by two elements namely $D_{2 n} = ⟨ r, s ⟩$
3. It's not abelian ( $n \geq 3$ ), ex: $s r \neq r s$
$S_{X}$ is the symmetric group on the set $X$
1. Elements are permutations of $X$
2. The notation: $σ = (1 3 2) (4 5)$
3. Usually $X = {1, \dots, n}$ (we can always number the elements) and then write $S_{n}$ .
4. $| S_{n} | = n!$
5. I'ts not cyclic when $n \geq 3$ , but it has some nice sets of generators
6. Not abelian for $n \geq 3$ .

Let's talk about Matrix Groups! There's another one which we won't cover in too much detail but are used a lot outside:

Fields

We haven't really defined Field's yet, but this will be a lot like that definition.

Suppose $F$ is a Field like $Q$ , $R$ , $C$ , or $\frac{Z}{p Z}$ (where $p$ is prime). Then:
$Mat (m, n; F) = {m \times n matrices with entries in F}$
where $m$ is the number of rows, and $n$ is the number of columns. We have matrix addition (by just adding corresponding entries), have the zero matrix $0$ as the identity, and the inverse is just the matrix with negative entries. This satisfies the Group definition, so then this is a valid group. Notably, this is an infinite Abelian Group.

If you drop it down to square matrices:
$Mat (n, n; F)$
where we consider matrix multiplication, then:

It's still associative (annoying but possible to verify)
The identity matrix $I$ is the identity
However, not all matrices have an inverse!
So this is almost a group, but not quite. You can extend the idea to throw away the duds that don't have inverses, using the following notation:
$
(\text{Mat}(n,n; \mathbb{F}))^{\times} := { \text{ invertible $n \times n$ matrices with entries in $F$ } }
$
We call this the general linear group of degree $n$ . Instead of using the above we use:
$G L (n, F)$
You can drop the $F$ if the field is known from context.

Some considerations:

It's not an Abelian Group (for $n \geq 2$ ) since matrix multiplication isn't commutative (ex: $(A + B)^{2} = A^{2} + B^{2} + A B + B A \neq A^{2} + B^{2} + 2 A B$ )
It's infinite if $F$ is infinite.
Finite if $F$ is finite.

You may recall that the determinant is a function:
$det : Mat (n, n; F) \to F$
Where if the determinant is 0 then there's no inverse, while if it isn't then there is an inverse:
$det (M) \neq 0 ⟺ M invertible$
So you could rewrite:
$G L (n, F) := {M \in Mat (n, n; F) | det (M) \neq 0}$

There's a motivation which helps setup the definition.

Consider the set:
$G := {z \in C | z^{4} = 1}$

Here $G = {1, - 1, i, - i} = ⟨ i ⟩$ . This is a Group under complex multiplication. Furthermore it's an Abelian Group.

We can generalize this process for various powers of $z$ . Namely the quaternions is generated from the set:
$Q_{8} = {1, - 1, i, - i, j, - j, k, - k}$
where the key relations are:

$i^{2} = j^{2} = k^{2} = - 1$ (notice that then they are all order 4 elements)
$i j k = - 1$ (notice that implies that $i = - k j$ , and other formulas)
To go in more detail, let's find more products that are formed from these relations.

i j

Let's find $i j$ :
$\begin{aligned} i j k & = - 1 \\ (i j k) k & = - 1 k \\ (i j) k^{2} & = - k \\ i j (- 1) & = - 1 k \\ i j (- 1) & = k (- 1) \\ i j & = k \end{aligned}$

j i

Let's find $j i$ :
$\begin{aligned} i j k & = - 1 \\ i^{2} j k & = - i \\ - 1 j k & = - i \\ j k & = i \\ j^{2} k & = j i \\ - k & = j i \end{aligned}$

Product	Result
$i j$	$k$
$i k$	$- j$
$j k$	$i$
$j i$	$- k$
$k i$	$j$
$k j$	$- i$
Some takeaways:

Not Abelian Group
Order (Groups) of 8