# 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
- Take a generator, look at all the multiples, let \(m_0\) be the minimum such that \(g^{m_0}\) belongs to the subgroup \(H\)
- Want to show that \(\langle g^{m_0} \rangle = H\)
- It is clear that \(\langle g^{m_0} \rangle \subseteq H\)
- 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.
- \(f(G) = \text{Im}(f)\) is a subgroup of \(H\)
- \(f^{-1} (1_H) = \{ g \in G \vert f(g) = 1_H \}\) is a subgroup of \(G\) – the preimage of the identity is a subgroup in \(G\)
- The identity is preserved: \(f(1_G) = 1_H\)
- 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:
- Let \(k \in H g_1 \cap H g_2\)
- Let \(k \in H g_1\). Then \(k g_1^{-1} \in H\).
- Let \(k \in H g_2\). Then \(k g_2^{-1} \in H\).
- \(x \equiv g_1\) and \(g_1 \equiv k\), meaning \(x \equiv k\)
- \(k \equiv g_2\), so \(x \equiv g_2\).
- 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
- The congruence classes modulo \(H\) form a group \(G / H\)
- 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
- Define the function \(\bar{f}\) : \(\bar{f}(\text{ker}(f) g) = f(g)\)
- Need to check that this function is well-defined. If \(\text{ker}(f) g = \text{ker}(f) g'\), then \(f(g) = f(g')\)
- We need to verify that \(\bar{f}\) is a homomorphism
- 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
- If \(k \subseteq \text{ker}(f)\) with \(k\) a normal subgroup of \(G\), \(f\) induces \(\bar{f} : G / K \to H\)
- If \(k = \text{ker}(f)\), \(\bar{f}\) is injective
- \(f\) 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
- \(\forall g \in G\), there should be \(f_g : X \to X\)
- \(\forall g_1, g_2 \in G\), we have that \(f_{g_1} \circ f_{g_2} = f_{g_1 g_2}\)

- Equivalent condition:
- #1 as before, without bijection
- #2 as before
- \(f_e = \text{id}\) (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 \(X\).
- \(\forall x \in X, \text{Stab}(x) = \{ g \in G : f(g)(x) = x \}\) (stabilizer)
- \(\forall x \in X, \text{Orb}(x) = \{ f(g)(x) : g \in G \}\) (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 \(X\)

- 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 \(G\).