# Lecture Notes

MATH 403

## 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. $$1_H \in \text{Im}(f)$$ (the identity is in the image)
2. If $$h_1, h_2 \in \text{Im}(f)$$, then $$h_1 *_H h_2 \in \text{Im}(f)$$ (closed under the operation)
3. 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 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 $$GL_n / SL_n \cong \mathbb{R} \setminus \{ 0 \}$$

• 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$$ subgroups case. 1. We know that $$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$$. 2. Consider the function $$f : N_1 \times N_2 \to G$$ given by $$f(n_1, n_2) = n_1 n_2$$. This is a surjective function, for free. 3. It is a homomorphism: $$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$$ commute. bitches aint shit frfr ong no cap 4. Injectivity: need to show that the kernel of $$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

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

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

Proof:

1. For all $$p$$ prime, $$p \vert \vert G \vert$$, let $$G(p) = \{ g \in G : \vert g \vert = p^{\alpha} \}$$
2. 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 $$\vert g^{-1} \vert = \vert g \vert$$
3. Now prove that each $$G(p_i)$$ is disjoint except for $$e$$
4. 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 $$G$$.