Lecture Notes

MATH 403

Table of contents

Week 1 Wednesday – Introduction to Groups
Week 2 Monday – Subgroups
Week 2 Wednesday – Subgroups and Homomorphisms
Week 2 Friday – Homomorphisms and Isomorphisms
Week 3 Wednesday – ??
Week 3 Friday – ?? pt 2
Week 4 Monday – Congruence Classes
Week 4 Wednesday – Normal Subgroups
Week 4 Friday – More Normal Subgroups
Week 6 Monday – Isomorphism Theorem for Groups
Week 6 Wednesday – Iso Thm pt 2
Week 6 Friday – Iso Thm pt 3
Week 7 Monday – Third Isomorphism Theorem
Week 7 Wednesday –
Week 7 Friday – Structure of Finite Abelian Groups
Week 9 Monday – More Group Actions
Week 9 Wednesday – Group Actions pt 2
Week 9 Friday – Sylow Theorems 2 and 3
Week 10 Monday – Sylow Theorems

Week 1 Wednesday – Introduction to Groups

A group $(G, *)$ is a pair with a set $G$ and an operation $*: G \times G \to G$ which is associative, has an identity, and has inverses. A group is abelian if the operation is commutative.

The order of a group is its cardinality

Examples of groups:

The reals with addition, $(\mathbb{R}, +)$
Nonzero field with multiplication, $(F - \{0\}, \times)$
Symmetric group on $n$ symbols, $S_n$: $S_n = \{ f \{ 1, ..., n\} \to \{ 1, ..., n \} \}$ with $f$ bijective

Week 2 Monday – Subgroups

Last time:
- A group has a unique identity element and every element has a unique inverse
- Order of an element: minimum integer power to equal the identity
- If $g^l = g^m$ for $m \neq l$, $\vert g \vert$ finite and $l \equiv_{\vert g \vert} m$
The dihedral group $D_n$ is generated by
- $r$, the rotation clockwise
- $s$, symmetry along the $x$-axis
Generated, meaning: every element in $D_n$ can be written as a product of $r$ and $s$
In $D_n$, $r^n = e$ and $s^2 = e$, also $srs = r^{-1}$
Any element in $D_n$ can be written as $r^a s^b$ for $0 \leq a \leq n-1$ and $b \in \{0, 1\}$
Simplify $s r^2 s = sr e r s = srs srs = r^{-1} r^{-1} = r^{-2}$. Generalizing for all $a$, $s r^a s = r^{-a}$
A subgroup of $G$ is a subset $H \subseteq G$ such that $H$ is a group under the same operation as $G$
To check if a subset $k$ is a subgroup of $G$, check
- $k$ is nonempty
- $k$ has the identity
- $k$ is closed under the operation
- $k$ is closed under inverses
You can always construct the trivial subgroup (identity or the group itself)
Or, $\forall g \in G$, the cyclic group generated by $g$ is the subgroup of powers of $g$.
Observation: $\vert \langle g \rangle \vert = \vert g \vert$
$\langle g \rangle$ is abelian

Week 2 Wednesday – Subgroups and Homomorphisms

$D_n = \{ r^a s^b, 0 \le a \le n - 1, 0 \le b \le 1 \}$ with $r^n = 1$, $s^2 = 1$, $srs = r^{-1}$
Cyclic subgroup of $G$: generated by $\langle g \rangle = \{ g^k \vert k \in \mathbb{Z} \}$
The order of $g$ is the order of $\langle g \rangle$
A group is a cyclic group if $\exists g \in G$ such that $\langle g \rangle = G$
$\mathbb{Z}_n$ is a cyclic group under addition
Any subgroup of a cyclic group is cyclic
1. Take a generator, look at all the multiples, let $m_0$ be the minimum such that $g^{m_0}$ belongs to the subgroup $H$
2. Want to show that $\langle g^{m_0} \rangle = H$
3. It is clear that $\langle g^{m_0} \rangle \subseteq H$
4. Argue by contradiction that $\exists h \in H \setminus \langle g^{m_0} \rangle$. This does not hold. So it must be that $H \subseteq \langle g^{m_0} \rangle$. Show this with $h = g^a = g^{k m_0 + r}$ with $0 \le r < h_0$, and show that we can find a smaller element in $H$.
Given a group $G$, $Z(G)$ is the center of $G$, the set of elements in $G$ which commute with everything else
Given two groups $(G, *_G)$ and $(H, *_H)$, a homomorphism $\phi: G \to H$ is a function such that $\phi(a *_G b) = \phi(a) *_H \phi(b)$
- A homomorphism commutes with the group operation
- e.g. $G = (\mathbb{R}^+, \times)$, $H = (\mathbb{R}, +)$, $\phi(x) = \log x$
Let $f : G \to H$ be a group homomorphism.
1. $f(G) = \text{Im}(f)$ is a subgroup of $H$
2. $f^{-1} (1_H) = \{ g \in G \vert f(g) = 1_H \}$ is a subgroup of $G$ – the preimage of the identity is a subgroup in $G$
3. The identity is preserved: $f(1_G) = 1_H$
4. Inverses are preserved: $f(g^{-1}) = f(g)^{-1}$

Week 2 Friday – Homomorphisms and Isomorphisms

If $\text{Im}(f) \subseteq H$, then we need that

$1_H \in \text{Im}(f)$ (the identity is in the image)
If $h_1, h_2 \in \text{Im}(f)$, then $h_1 *_H h_2 \in \text{Im}(f)$ (closed under the operation)
If $h \in \text{Im}(f)$, then $h^{-1} \in \text{Im}(f)$ (closed under inverses)

A homomorphism is an isomorphism if $f$ is injective and surjective.

Cyclic groups are isomorphic to $\mathbb{Z}$ or $\mathbb{Z}_n$
If $G$ is a cyclic group, there is a surjective homomorphism from $\mathbb{Z}$ to $G$, namely $\phi: \mathbb{Z} \to G$ with $\phi(k) = g^k$

Cayley’s Theorem: If $G$ is a group, then $G$ is isomorphic to a subgroup of $A(G)$, where $A(G)$ is the group of bijections from $G$ to $G$ under composition.

Strategy: We will produce an injective homomorphism $\phi: G \to A(G)$ such that $\phi(G)$ is a subgroup of $A(G)$. We will define $\phi(g)$ such that $\phi(g)(h) = gh$. Every function defined by $g$ is basically multiplication by $g$ on the left. Some things to check:

The definition of $\phi(g)$ ensures it is injective
Show that $\phi$ is a homomorphism

Week 3 Wednesday – ??

Isomorphism – “same look”
Automorphism – “self look”
If $f : G \to H$ is an isomorphism, there exists an inverse map $f^{-1} : H \to G$ such that $f^{-1}$ is also an isomorphism
If $f : G \to H$ is an injective homomorphism, then $G \cong \text{Im}(f)$.

Week 3 Friday – ?? pt 2

Week 4 Monday – Congruence Classes

Recap

Every $\sigma \in S_n$ can be written as a product of transpositions and of disjoint cycles.
Disjoint cycles commute

Definition: Given $G$ and a subgroup $H$, $g_1 \equiv_H g_2$ if $g_1 g_2^{-1} \in H$.

Congruence is an equivalence relation – commutative, transitive
Define $Hg$ to be the group of all elements in $G$ which are congruent to $g$, mod $H$
If $H g_1 \cap H g_2 \neq \emptyset$, then $H g_1 = H g_2$. Proof:
1. Let $k \in H g_1 \cap H g_2$
2. Let $k \in H g_1$. Then $k g_1^{-1} \in H$.
3. Let $k \in H g_2$. Then $k g_2^{-1} \in H$.
4. $x \equiv g_1$ and $g_1 \equiv k$, meaning $x \equiv k$
5. $k \equiv g_2$, so $x \equiv g_2$.
6. So $x \equiv g_1 \equiv g_2$
Assume $G$ finite and $H$ subgroup. If $g \in G$, $\vert H e \vert = \vert H g \vert$. All the cosets have the same size.
Lagrange’s theorem: If $G$ is finite and $H$ is a subgroup, then $\vert G \vert = \vert H \vert \cdot \vert H g \vert$.

Week 4 Wednesday – Normal Subgroups

Proposition. If $G$ is a group and $H$ is its subgroup, with $g \in G$, then $\vert H 1 \vert = \vert H g \vert = \{ g' \in G : g' \equiv_H g \}$.

We will construct the function $f : H1 \to Hg$ which is a bijection. We take an element $g \in G$ and $h \in H$, and map $g$ to $hg$.

Lagrange’s theorem

All the cosets are disjoint and have the same number of elements
Subsets cover all of $G$ and these subsets never intersect
We have that $\vert G \vert = \sum \vert Hg \vert = \sum \vert H \cdot 1 \vert = \vert H g \vert \vert H \vert$

Normal subgroups

You can’t add congruence classes in nonabelian groups in the same way you can for abelian groups
A subgroup $H$ of $G$ is normal if $\forall g \in G$, $g H g^{-1} = H$

Week 4 Friday – More Normal Subgroups

Not all groups are normal, e.g. $D_3$ or $\langle (1, 2) \rangle \subseteq S_3$
Criteria is that $g H g^{-1} = H$, which is to say that $g H = H g$
$a \equiv_H b$ and $c \equiv_H d$ implies that $ac \equiv_H bd$
Let $G$ be a group and $H$ its subgroup. Then
1. The congruence classes modulo $H$ form a group $G / H$
2. There is a surjective homomorphism from $G \to G / H$, where each element $g \in G$ is mapped to its congruence class.
We will define $H g_1 \cdot H g_2 = H g_1 g_2$

Week 6 Monday – Isomorphism Theorem for Groups

Let $f : G \to H$ be a group homomorphism. Then the kernel of $f$ is a normal subgroup of $G$
There exists a function $\bar{f} : G / \text{ker}(f) \to H$ such that $\bar{f}(\pi(g)) = f(g)$, where $\pi : G \to G / \text{ker}(f)$ is the natural projection from an element in $G$ to its congruence class
Consider the map from $D_6$ to $\mathbb{Z}_2$ givn by $f(r^a s^b) = b$.
- The kernel of $f$ is $\langle r \rangle$
- The two congruence classes are $\langle r \rangle$ and $\langle r \rangle s$
- The map $\bar{f}$ is given by $\bar{f}(\langle r \rangle) = 0$ and $\bar{f}(\langle r \rangle s) = 1$
Another example: map from $GL_2(\mathbb{R}) \to \mathbb{C}^*$ which sends a matrix to its determinant
- The kernel of this map is the set of all matrices with determinant 1 (the special linear group, $SL_2(\mathbb{R})$)
- $G / \text{ker}(f)$ is the set of congruence classes of matrices modulo the special linear group
Theorem. For $f : G \to H$ homomorphism, there exists an injective homomorphism $\bar{f} : G / \text{ker}(f) \to H$ such that $f = \bar{f} \circ \pi$, where $\pi : G \to G / \text{ker}(f)$ is the natural projection
- Rephrased: for a homomorphism of groups $f : G \to H$, $G / \text{ker}(f) \cong \text{Im}(f)$
Proof of the theorem
1. Define the function $\bar{f}$ : $\bar{f}(\text{ker}(f) g) = f(g)$
2. Need to check that this function is well-defined. If $\text{ker}(f) g = \text{ker}(f) g'$, then $f(g) = f(g')$
3. We need to verify that $\bar{f}$ is a homomorphism
4. We need to check that $\bar{f}$ is injective: namely, check that $\ker(\bar{f}) = \{ 1 \}$
If $f$ is surjective, then $\bar{f}$ is an isomorphism between the quotient group and $H$
- Note that $GL_n / SL_n \cong \mathbb{R} \setminus \{ 0 \}$

Week 6 Wednesday – Iso Thm pt 2

Example: Consider $G = \mathbb{C}$ and $$H = { z \in \mathbb{C}* : \vert z \vert = 1 }
If you work it out, $H$ is closed under multiplication.
Turns out that $(\mathbb{R}^+, *) \cong \mathbb{C}^* / H$
To show that a group is isomorphic to a quotient group $G / H$, you need to find a surjective homomorphism $f : G \to H$ and then show that $\text{ker}(f) = H$
In general: dividing an infinite field by the set of elements with constant norm (say unit norm) gives you $\mathbb{R}^+$?
Suppose $f : G \to H$ is a surjective homomorphism, that $K$ is a normal subgroup of $G$ and contained in the kernel of $f$. Then the map from $G / K \to H$ is injective iff $K = \text{ker}(f)$

Week 6 Friday – Iso Thm pt 3

To prove isomorphisms, show the presence of a homomorphism with a kernel equal to (or contains) the subgroup you want to quotient by, then you know that the quotient group is isomorphic to the image of the homomorphism
For instance, $\mathbb{C}^* / \{ 1, -1 \} \cong \mathbb{C}^*$ using the map $f(z) = z^2$
Proposition: there is a bijection between the subgroups of $G / K$ and the subgroups of $G$ containing $K$.
- The map from subgroups of $G / K$ to the subgroups of $G$ containing $K$: preimage of the subgroup under the natural projection
- The map from subgroups of $G$ containing $K$ to the subgroups of $G / K$: the image of the subgroup under the natural projection
- This bijection sends normal subgroups to normal subgroups

Week 7 Monday – Third Isomorphism Theorem

Two useful definitions of normal subgroups
- For all $g\in G$, $g N g^{-1} \subseteq N$
- For all $g \in G$, for all $n \in N$, exists $n' \in N$ such that $gn = n' g$
Composition of homomorphisms is a homomorphism; you can quotient groups repeatedly
Kernels are normal subgroups. Normal subgroups are the kernels of some homomorphisms
From the quotient process $G \to G / N \to (G / N) / K$, you know that the kernel $\pi^{-1}(K)$ is a normal subgroup of $G$ containing $N$
Restatement of the first isomorphism theorem
1. If $k \subseteq \text{ker}(f)$ with $k$ a normal subgroup of $G$, $f$ induces $\bar{f} : G / K \to H$
2. If $k = \text{ker}(f)$, $\bar{f}$ is injective
3. $f$ surjective iff $\bar{f}$ surjective
Third isomorphism theorem: if $K \lefttriangleeq G / N$, $(G / N) / K \cong G / \pi^{-1}(K)$ for $\pi : G \to G / N$
- Theorem used if you have a quotient and you want to understand what it is
- e.g. $K = \{0, 3, 6, 9 \} \subseteq \mathbb{Z}_{12}$, $\mathbb{Z}_{12} / K = (\mathbb{Z} / 12\mathbb{Z}) / K = \mathbb{Z} / \pi^{-1}(K) = \mathbb{Z} / 3\mathbb{Z}$, where $\pi : \mathbb{Z} \to \mathbb{Z}_{12}$
Can you find a subgroup $H$ of $GL_2$ such that $SL_2 \subseteq H$ and $H$ not normal? Impossible, because there is a bijection from subgroups of $G / N$ and the subgroups of $G$ containing $N$
- Moreover, there is a bijection between normal subgroups of $G / N$ and normal subgroups of $G$ containing $N$
- “Out of the subgroups of $G$ containing $N$, which ones are normal?” becomes “Out of the subgroups of $G / N$, which ones are normal?” Because $GL_2 / SL_2 \cong \mathbb{R}^*$, it is abelian, and therefore every subgroup is normal. This means that every subgroup of $GL_2$ containing $SL_2$ is normal.
A group is simple if every normal subgroup is either the identity or the group itself, which means that if you try to quotient by a normal subgroup, you get either the identity or the group itself
You can decrease complexity by quotienting by normal subgroups
$A_4$ is simple if $n \neq 4$.

Week 7 Wednesday –

Review

The set of all subgroups in $G / N$ is isomorphic to the set of all subgroups in $G$ containing $N$
The set of all normal subgroups in $G / N$ is isomorphic to the set of all normal subgroups in $G$ containing $N$
Then $K \lefttriangleeq G / N$, $(G / N) / K \cong G / \pi^{-1}(K)$, where $\pi : G \to G / N$
Not true in general: if $f$ homomorphism $G \to H$, $N \lefttriangleeq G$, $f(N) \subseteq H$ is not necessarily normal
If $f$ is surjective, then $f(N)$ is normal

Suppose you have a group $G$, with $N_1, N_2, \hdots, N_m \lefttriangleeq G$, satisfying the following properties:

For any $1 \le i \le m$, $N_i \cap N_1 N_2 \hdots N_{i-1} N_{i+1} \hdots N_m = \{ e \}$, where the product is just generation of elements from sets
For all $g \in G$, there exists $n_1, n_2, \hdots, n_m$ such that $g = n_1 \cdot n_2 \cdot \hdots \cdot n_m$, where $n_i \in N_i$ Then $G \cong N_1 \times N_2 \times \hdots \times N_m$. Basically: if you have a bunch of “disjoint” normal subgroups and they generate $G$, then $G$ just is their Cartesian product.

Proof for $2$ subgroups case.

We know that $N_1, N_2 \lefttriangleeq G$, $N_1 \cap N_2 = \{ e \}$, and $N_1 N_2 = G$ We want to prove that $N_1 \times N_2 \cong G$.
Consider the function $f : N_1 \times N_2 \to G$ given by $f(n_1, n_2) = n_1 n_2$. This is a surjective function, for free.
It is a homomorphism: $f((n_1, n_2) \cdot (n_1', n_2')) = f(n_1 n_1', n_2 n_2') = n_1 n_1' n_2 n_2' = f(n_1, n_2) f(n_1', n_2')$. This relies on showing that elements in $N_1, N_2$ commute. bitches aint shit frfr ong no cap
Injectivity: need to show that the kernel of $f$ is one. You can use the fact that the intersection is $e$ to show that the kernel is $e$.

Theorem. $G$ is a finite abelian group. $G \cong \mathbb{Z}_{p_1^{k_1}} \times \hdots \times \mathbb{Z}_{p_m^{k_m}}$, where $p_i$ are primes.

Week 7 Friday – Structure of Finite Abelian Groups

The book’s theorem: If $\forall g \in G, \exists ! n_i \in N_i : g = n_1 \cdot \hdots \cdot n_m$, and $N_i$ cover all of $G$, then $G \cong N_1 \times \hdots \times N_m$

Theorem: If $G$ is a finite abelian group such that $\vert G \vert = p_1^{m_1} \times \hdots \times p_k^{m_k} where$p_i$$ are primes, then

$G \cong G(p_1) \times \hdots \times G(P_k)$ s.t. $\vert G(p_i) \vert = p_i^{m_i}$
\[G(P_i) \cong \mathbb{Z}_{p_i^{m_1}} \times \hdots \times \mathbb{Z}_{p_{i}^{a_n}}\]

That is to say, if $G$ is a finite abelian group, then $G$ is a product of cyclic groups.

Proof:

For all $p$ prime, $p \vert \vert G \vert$, let $G(p) = \{ g \in G : \vert g \vert = p^{\alpha} \}$
Claim: $G(p)$ is a subgroup of $G$.
- \[e \in G(p)\]
- If $g, h \in G(p)$, then $\vert g \vert = p^{\alpha}$ and $\vert h \vert = p^{\beta}$, so $\vert g h \vert = p^{\gamma}$, where $\gamma \le \alpha + \beta$
- If $g \in G(p)$, then $\vert g^{-1} \vert = \vert g \vert$
Now prove that each $G(p_i)$ is disjoint except for $e$
Now prove that $G(p_i)$ covers all of $G$: for every $g \in G$, $\exists g_i \in G(p_i) : g = \sum g_i$. Prove by induction starting from $m = 1$

Important part – if you have a finite abelian group, it is a product of cyclic groups.

Definition of a group action: given a set $X$ and a group $G$, an action of $G$ on $X$ is a group homomorphism $\phi : G \to A(X)$, where $A(X)$ is the set of bijections from $X$ to $X$
Consider $f : G \to H$ a group homomorphism. If we interpret $H$ only as a set, the function $f$ induces an action of $G$ on $H$, given by $a \to F(g)$, with $F(g)(h) = f(g) h$. Indeed, $F$ is a homomorphism.
To give an action of $G$ on $X$, you need
1. $\forall g \in G$, there should be $f_g : X \to X$
2. $\forall g_1, g_2 \in G$, we have that $f_{g_1} \circ f_{g_2} = f_{g_1 g_2}$
Equivalent condition:
1. #1 as before, without bijection
2. #2 as before
3. $f_e = \text{id}$ (identity function)
Example: $D_4$ acts on $X$, the set of vertices of a square, by rotations and reflections
Question: How many 6-faced dices are there up to symmetry?
Let $G$ be a group on a set $X$.
1. $\forall x \in X, \text{Stab}(x) = \{ g \in G : f(g)(x) = x \}$ (stabilizer)
2. $\forall x \in X, \text{Orb}(x) = \{ f(g)(x) : g \in G \}$ (orbit)
Consider $\{ Hg : g \in G \} = X$ for a group $G$ and a subgroup $H$. We can define an action of $G$ on $X$ by $f(g) (H g') = H g' g^{-1}$

Week 9 Monday – More Group Actions

A group action is a homomorphism from $G$ to the group of bijections of $X$
The stabilizer of an element is the set of all elements in $G$ which fix that element
The orbit of an element is the set of all elements in $X$ which can be reached by the action of $G$
We will denote $f_g(x)$ by $g \cdot x$
There is a bijection between the congruence classes of the stabilizer of $x$ and the orbit of $x$
$f : G \to H$ is a group homomorphism. We have a left action of $G$ on $H$ given by $g \cdot h = f(g) h$, and a right action of $H$ on $G$ given by $h \cdot g = g f(h)$, as well as a conjugation action of $G$ on itself given by $g \cdot h = g h g^{-1}$. We assume that $f$ is injective.
- The stabilizer of $h$ is just $\{ e \}$, and the orbit of $h$ is the set of all elements in $G$ which are conjugate to $h$
- We know that $\vert H \vert = \sum_{\text{Orb}} \vert \text{orb}(h _i) \vert = \sum_{\text{Orb}} \frac{ \vert G \vert}{\vert \text{Stab}(h_i) \vert} = \sum_{\text{orb}} \vert G \vert$
  - The cardinality of $H$ is the number of orbits times $G$

Week 9 Wednesday – Group Actions pt 2

There is a bijection between congruence classes of the stabilizer of $x$ and the orbit of $x$
Orbits form a partition of $X$

\[\vert X \ vert = \sum_{\text{orb}} \vert \text{orb} \vert = \sum \frac{\vert G \vert}{\vert \text{Stab}(x) \vert}\]

Prove Lagrange’s Theorem using group actions with $X = K$, $G = G$ with $G < K$, and $g * k = gk$
Prove Giovanni’s Theorem: let $G$ be a group and $p$ divides $\vert G \vert$, $G$ admits a subgroup of order $p$, $p$ is a prime number.
- Prove with $X = G$, $G = G$, and $g * x = gxg^{-1}$
- Given $x \in G$, $\text{Stab}(x) = \{ g \in G : gxg^{-1} = x \}$
- In this case, it ends up being the commutator of $x$ – all the elements in $G$ that commute with $x$
- We have that $\vert G \vert = \sum \vert \text{orb}(x) \vert = \sum_{x_i \in \text{orb}} \frac{\vert G \vert}{\vert C(x_i) \vert}$
- Note that $C(1) = G$, so $C(x_i) = G$ if $x_i \in Z(G)$ (the center of $G$)
- Note that this simplifies as $\sum_{x_i \in \text{orb}, \notin Z(G)} \frac{ \vert G \vert }{ \vert C(x_i) \vert } + \vert Z(G) \vert$
First Sylow Theorem: $G$ group with $p^k \vert G \vert$; $G$ has a subgroup of order $p^k$
- Proof: $X = G$, $G = G$, $g * x = gxg^{-1}$
- Prove by induction on Giovanni’s Theorem ($k = 1$).
- Need to show $k - 1 \implies k$

Week 9 Friday – Sylow Theorems 2 and 3

From last time: $\vert G \vert = \sum_{x_i \in \text{orb}, \notin Z(G)} \frac{ \vert G \vert }{ \vert C(x_i) \vert } + \vert Z(G) \vert$
We can find a normal subgroup $K$ of $Z(G)$, such that $\vert K \vert = p$
Suppose $G$ is a finite group with$p^n \vert \vert G \vert$but$p^{n + 1} \nmid \vert G \vert$with$p$$ prime.
Second Sylow Theorem: If $P_1, P_2$ are subgroups of $G$ of order $p^n$, then $\exists g \in G : g P_1 g^{-1} = P_2$
Third Sylow Theorem: $\vert \{ P < G : \vert P \vert = p^n \} \vert \equiv_p 1$
With the notations from above, $P < G : \vert P \vert = p^n$ is the $p$-Sylow subgroup.
Lemma. For a group $G$ and $P$ sylow in $G$ with $P \lefttriangleeq G$, then $P$ is the unique sylow subgroup of $G$.
To prove this, we need to use the key formula, and find $X$ (the subgroups of $G$ of the form $g P_1 g^{-1}$) and $G$ acting on $X$ by conjugation.
Stabilizer is the set of elements $\{ k \in P_2 : kg P_1 g^{-1} k^{-1} = g P_1 g^{-1} \} = P_2 \cap N_g (g P_1 g^{-1})$
Normalizer: given $H < G$ and $N_G(H) = \{ g \in G : gHg^{-1} \subseteq H \}$ is the normalizer of $H$ in $G$
If $P$ is a $p$-sylow subgroup in $G$, it is the unique sylow subgroup of $N_G(H)$. It is always true that $H \subseteq N_G(H)$ and $H \lefttriangleeq N_G(H)$
To prove theorem II, we have to check that there exists some $g \in G$ such that $P_2 \subseteq N_G(g P_1 g^{-1})$. Indeed that implies $P_2 = g P_1 g^{-1}$ from the previous observation. So $P_2 = g P_1 g^{-1} \iff P_2 \subseteq N_G ( g P_1 g^{-1}) \implies \vert \text{orb}(g P_1 g^{-1}) \vert = 1$.
Sylow 2 and 3 are equivalen tto saying that the number of orbits with 1 element are congruent to 1 mod $p$.

Week 10 Monday – Sylow Theorems

Sylow Theorem 2: let $P_1, P_2$ be two $p$-sylow subgroups of $G$, then $\exists g \in G : g P_1 g^{-1} = P_2$ (all groups of the same order are conjugations of each other)
Sylow Theorem 3: the number of all $p$-sylow groups in $G$ is congruent to 1 mod $p$.
Proof via $X = \{ g P_2 g^{-1} : g \in G \}$ and $$G = P_1$ with $* = \forall k \in P_1, k + g P-2 g^{-1} = kg P_2 g^{-1} k^{-1}$
If $G$ is a group with $P$ a normal $p$-sylow subgroup of $G$, $P$ is the only $p$-sylow subgroup of $G$ (since it conjugates with everything)
Start out with the key formula, you take one of the $p$-sylows and move it around with elements in $G$.