09 - Working with the -izers

Last time we talked about:

We talked about the example of Conjugation Subgroup below:

Consider Group $G$ acting on its own elements by conjugation:

$X = G$
$g \cdot x = g x g^{- 1}$

Let's look at Stabilizers:
$\begin{aligned} G_{x} & = {g \in G | g \cdot x = x} \\ = {g \in G | g x g^{- 1} = x} \\ = {g \in G | g x = x g} \end{aligned}$
which are all the elements in $G$ that commute with $x$ . This is the Centralizer of the element $x$ .

For example, for a subset $S \subseteq G$ that acts via conjugation:

g \cdot S = g S g^{- 1} = {g x g^{- 1} | x \in S}

For a single subset $S$ :

G_{S} = {g \in G | g S g^{- 1} = S}

Some More Specific Examples

Let's put all of these into practice. Let $G = D_{8} = ⟨ r, s | r^{4} = s^{2} = 1, r s = s r^{- 1} ⟩ = {1, r, r^{2}, r^{3}, s, s r, s r^{2}, s r^{3}}$ be a Dihedral Groups. We made some Subgroup $S = ⟨ r ⟩ = {1, r, r^{2}, r^{3}}$ .

Subgroup or Subset?

All of these definitions allow for subsets, and don't restrict to subgroups; however, for this class you can expect we'll usually only consider subgroups.

Let's make a conjugation table for $g x g^{- 1}$ :

For the sake of simplicity let $G = {x_{1}, \dots, x_{8}}$ as described in the order above (so $x_{3} = r^{2}$ for example. For each we compute $g x g^{- 1}$ and find the corresponding $x_{i}$ in the table).

$g$ is below, $x$ is right	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$
$1$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$
$r$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$
$r^{2}$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$
$r^{3}$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$
$s$	$x_{1}$	$x_{4}$	$x_{3}$	$x_{2}$
$s r$	$x_{1}$	$x_{4}$	$x_{3}$	$x_{2}$
$s r^{2}$	$x_{1}$	$x_{4}$	$x_{3}$	$x_{2}$
$s r^{3}$	$x_{1}$	$x_{4}$	$x_{3}$	$x_{2}$

For example:
$\underset{g}{\underset{⏟}{s r^{2}}} \cdot \underset{x}{\underset{⏟}{r}} \cdot (s r^{2})^{- 1} = s r^{2} \cdot r \cdot r^{- 2} s^{- 1} = s r s^{- 1} = r^{3} \equiv x_{4}$
Notice that this table alone tells you what happens to $S$ , and helps you identify Stabilizers and Centralizers. Namely:
$C_{D_{8}} (⟨ r ⟩) = {g \in D_{8} | g 1 g^{- 1} = 1, g r g^{- 1} = r, \dots, g r^{3} g^{- 1} = r^{3}}$
Which means look at the rows where all the column entries don't change. We see that only for the first four rows (the $g s), s o t h e n :$
C_{D_{8}}(\langle r \rangle) = { 1, r, r^{2}, r^{3} }
$
which convieniently happened to be the subgroupo $⟨ r ⟩$ itself. For the Normalizer:
$N_{D_{8}} (⟨ r ⟩) = {g \in D_{8} | g 1 g^{- 1} \in ⟨ r ⟩, \dots, g r^{3} g^{- 1} \in ⟨ r ⟩}$
is a weaker condition (one can show that $N \subseteq C$ here). Here everyone is in that set (no one has an $s$ in our table) so then:
$= G = D_{8}$

Knowing the Subgroups

It'd be nice, when computing the Centralizer, if we can make it out of known subgroups. We want a picture of the subgroups of a group. Let's highlight with some tiny examples.

Look at $\frac{Z}{2 Z} = {\overset{―}{0}, \overset{―}{1}}$ . There's really only two subgroups, the trivial one and ${\overset{―}{0}}$ (since they need to have the identity $\overset{―}{0}$ ). Notice that:
${\overset{―}{0}} = ⟨ \overset{―}{0} ⟩$
and
${\overset{―}{0}, \overset{―}{1}} = ⟨ \overset{―}{1} ⟩$
Notice here that the latter includes the former, ie: $⟨ \overset{―}{1} ⟩ \supseteq ⟨ \overset{―}{0} ⟩$ . But this may not always be the case! For $\frac{Z}{3 Z}$ then:
$⟨ \overset{―}{1} ⟩ = ⟨ \overset{―}{2} ⟩ = G$
So again these subgroups include the trivial subgroup $⟨ \overset{―}{0} ⟩$ .

What about $\frac{Z}{4 Z}$ ? Notice that $⟨ \overset{―}{1} ⟩ = ⟨ \overset{―}{3} ⟩ = G$ . But:
${\overset{―}{0}, \overset{―}{2}} = ⟨ \overset{―}{2} ⟩$
So we have some sort of tree of inclusions:

What about $\frac{Z}{5 Z}$ ? Here $1, 2, 3, 4$ generate the whole subgroups (I'm treating the numbers as their modulus equivalent, so $1 \equiv \overset{―}{1}$ ), so there's no intermediate subgroups.

What about $\frac{Z}{6 Z}$ ?

You may notice that any generator's subgroup also is equivalent to the inverse generator's subgroup (ex: $⟨ 2 ⟩ = ⟨ 4 ⟩$ as they are inverses).

How can we generalize this? Here's some facts we can try to prove for $\frac{Z}{n Z}$ (although we can prove this for any Cyclic Group of the same order, then use Isomorphism to show the properties still hold):

For any $a \in Z$ then $| \overset{―}{a} | = \frac{n}{gcd (a, n)}$ (see Order (Groups) for more info).
Every subgroup of $\frac{Z}{n Z}$ is cyclic, so they are of the form $⟨ \overset{―}{a} ⟩$ (so you only need one generator).
$⟨ \overset{―}{a} ⟩ = ⟨ \overset{―}{b} ⟩ ⟺ gcd (a, n) = gcd (b, n) ⟺ | \overset{―}{a} | = | \overset{―}{b} |$ .
There's exactly one unique subgroup for each divisor $d$ of $n$ , and it's $⟨ \frac{n}{d} ⟩$ .
$⟨ \overset{―}{a} ⟩ \leq ⟨ \overset{―}{b} ⟩ ⟺ b | a$ assuming $a, b | n$ and $a > b$ for simplicity.

To highlight these, consider $\frac{Z}{12 Z}$ . The divisors of 12 are: 1,2,3,4,6,12. Then:

$⟨ \overset{―}{1} ⟩ = G$
$⟨ \overset{―}{2} ⟩ = {0, 2, 4, 6, 8, 10}$ where $| ⟨ 2 ⟩ | = \frac{12}{2} = 6$ using (2)
$⟨ 3 ⟩ = {0, 3, 6, 9}$ has $| ⟨ 3 ⟩ | = \frac{12}{3} = 4$ .
...
Here $⟨ 6 ⟩ \leq ⟨ 3 ⟩$ using (5)
(4) says any divisors that aren't dividing 12 thus are co-prime and generate just $G$ itself.

This picture we draw is a lattice of subgroups of our group $\frac{Z}{12 Z}$ .