17 - Finishing the Isomorphism Theorems

Theorem

Suppose $K ⊴ G$ (see Normal Subgroup) with $π : G \to \frac{G}{K}$ be the cannonical projection. Then:

$A \leq B$ in $G$ iff $\overset{―}{A} \leq \overset{―}{B}$ in $\frac{G}{K} = \overset{―}{G}$
$A ⊴ G$ implies $\overset{―}{A} ⊴ \overset{―}{G}$ . This becomes an iff when we have subgroups $H \leq G$ that contain $K$ .
Then assuming $H \leq G$ that contain $K$ :
$A \leq B$ then $| B : A | = | \overset{―}{B} : \overset{―}{A} |$ (see Index (Groups))
$\overset{―}{A \cap B} = \overset{―}{A} \cap \overset{―}{B}$
$\overset{―}{⟨ A \cup B ⟩} = ⟨ \overset{―}{A} \cup \overset{―}{B} ⟩$

Example

Consider $D_{8}$ :

If we alternatively look at $K = ⟨ r^{2} ⟩ = {1, r^{2}}$ then the left cosets of $K$ in $D_{8}$ are:

Notice we don't care which representative we use for each subgroup of cosets, so then we can use the bar notation to indicate a coset:

$\overset{―}{1} = \overset{―}{r^{2}}$
$\overset{―}{r} = \overset{―}{r^{3}}$
$\overset{―}{s} = \overset{―}{s r^{2}}$
$\overset{―}{s r} = \overset{―}{s r^{3}}$
which all are elements of $\frac{D_{8}}{K} = {\overset{―}{1}, \overset{―}{r}, \overset{―}{s}, \overset{―}{s r}}$ . But look! We can make some map:
${1, s, s r^{2}, r^{2}} \to {\overset{―}{1}, \overset{―}{s}, \overset{―}{s r^{2}}, \overset{―}{r^{2}}} = {\overset{―}{1}, \overset{―}{s}}$
namely, use $π : D_{8} \to \frac{D_{8}}{K}$ such that $g \mapsto g K = \overset{―}{g}$ . Then for each normal subgroup in the lattice (see the orange boxes in the picture of the lattice):
$H_{1} = {1, s, s r^{2}, r^{2}}$ has $π (H_{1}) = \overset{―}{H_{1}}$
$H_{2} = {1, r, r^{2}, r^{3}} \to {\overset{―}{1}, \overset{―}{r}, \overset{―}{r^{2}}, \overset{―}{r^{3}}} = {\overset{―}{1}, \overset{―}{r^{2}}} = \overset{―}{H_{2}}$ .
... we can repeat with each normal subgroup, and we'll get $π (H_{i}) = \overset{―}{H_{i}}$ .
$H_{3} = {1, s r^{3}, r^{2}, s r} \to {\overset{―}{1}, \overset{―}{s r^{3}}, \overset{―}{r^{2}}, \overset{―}{s r}} = {\overset{―}{1}, \overset{―}{s r}} = \overset{―}{H_{3}}$
But some of them became the other normal subgroups. Namely $π ({1, s}) = \overset{―}{H_{1}}$
$π {1, s r^{2}} = \overset{―}{H_{4}}$
...

We can make a lattice for what becomes what, and what stays the same:

Which is the same lattice picture we had before, but only with the normal subgroups!

Back to Group Actions

We'll be looking at specific examples of Group Actions in, well, action! For each one we'll look at the mechanical actions to see how to get things like:

08 - Centralizers, Normalizers, and Stabilizers
Normal Subgroups of each
...

Recall what a Group Action is:

left action

A (left) action of Group $G$ on set $X$ consists of a set map:
$σ : G \times X \to X$
Where $(g, x) \mapsto σ (g, x) = g \cdot x = g x$ such that:

$g_{1} \cdot (g_{2} \cdot x) = (g_{1} \cdot g_{2}) x$ for all $g_{1}, g_{2} \in G, x \in X$
$1_{G} \cdot x = x$ for all $x \in X$

This definition may seem a bit loose, so mathematicians sometimes use another definition:

left action

A (left) action of $G$ on $X$ is a group Homomorphism (or Group Morphism) (map) $σ : G \to S_{X} = {f X \to X | f is a bijection}$ (see Symmetric Group & Permutations)

We'll later see that these definitions are equivalent. However, let's first look at an example.

Often we'll use $σ_{g}$ to represent how $σ (g) = σ_{g}$ in the alternative definition above. Namely:
$σ_{g} (x) = g \cdot x$

An Example

The trivial action of $G$ on $X$ is $σ : G \to S_{X}$ such that each $g \mapsto 1$ . Namely $g x = x$ for all $g \in G, x \in X$ .
The Dihedral Groups, like $D_{8}$ for the symmetries of a square, acts on the set of vertices of that square ( $X = {v_{1}, v_{2}, v_{3}, v_{4}}$ for simplicity, for example $r v_{1} = v_{2}$ ). Define $σ_{r} = (v_{1} v_{2} v_{3} v_{4}) \equiv (1 2 3 4)$ . Then this is a group action, where $σ : D_{8} \to S_{X} ≅ S_{4}$ where $σ_{r} : r \mapsto (1 2 3 4)$ .

$D_{8}$ can also act on the set $Y = {d_{1}, d_{2}}$ , the diagonals of the square:

Here $σ_{r} = (d_{1} d_{2}) \equiv (1 2)$ . Here $D_{8} \to S_{Y} ≅ S_{2}$ .

One action of the group onto itself is left multiplication by the group operation:
$g ⋆ g^{'} = g g^{'}$

Note

Why left multiplication? Namely, it's because our definition is in terms of terms from the group being on the left.

Is there any group elements that keeps each element fixed? Namely is there some $g \in G$ such that $g ⋆ x = 1_{G}$ for all $x \in X$ ? See the answer on Trivial Group Action.

We can ask two questions, for each $x \in X$ :

We have the Orbit of $x$ , denoted $O_{x} = {g \cdot x | g \in G}$ , which is just the outputs of all $g$ 's applied to some $x$ .
We have the Stabilizers of $x$ denoted $G_{x} = {g \in G | g \cdot x = x}$ .

We can ask ourselves a theorem:

Theorem

When $G$ acts on a set $X$ , for each $x \in X$ there's a bijection between the Orbit of $x$ and the cosets of the Stabilizers of $x$ , denoted $G_{x}$ , in $G$ :
$O_{x} ≅ \frac{G}{G_{x}}$
Thus $| O_{x} | = | G : G_{x} |$ (see Index (Groups)). This map is done by $g \cdot x \mapsto g G_{x}$ .

The idea here is that the 25 The Lattice of Subgroups of a Group are themselves vertices of, say, a Dihedral Groups, which themselves can be analyzed! The orbits here let us get to that point.

Proof

Define the set map $f : O_{x} \to \frac{G}{G_{x}}$ by $f (g \cdot x) \mapsto g G_{x}$ .

First, is $f$ well-defined (since we are not mapping from a quotient group)? Suppose $g_{1} \cdot x = g_{2} \cdot x$ for $g_{1}, g_{2} \in G$ and $x \in X$ . Then:
$\begin{aligned} g_{1} \cdot x = g_{2} \cdot x & ⟺ g_{2}^{- 1} \cdot (g_{1} \cdot x) = x \\ ⟺ (g_{2}^{- 1} g_{1}) \cdot x = x \\ ⟺ g_{2}^{- 1} g_{1} \in G_{x} \\ ⟺ g_{1} G_{x} = g_{2} G_{x} \\ ⟺ f (g_{1} \cdot x) = f (g_{2} \cdot x) \end{aligned}$
Next, is $f$ surjective? Take any $g G_{x}$ in $\frac{G}{G_{x}}$ . Then $g \cdot x \in O_{x}$ and $f (g \cdot x) = g G_{x}$ .

Is $f$ injective? Suppose $g_{1} \cdot x = g_{2} \cdot x \in O_{x}$ . Then:
$\begin{array}{r} f (g_{1} \cdot x) = f (g_{2} \cdot x) \Rightarrow g_{1} \cdot x = g_{2} \cdot x \end{array}$
Thus this is a valid Isomorphism!

☐