Title: Part IA / Groups
Tags: Groups_And_Homomorphisms Group_Actions Symmetric_Groups_I Möbius_Groups Matrix_Groups Quotient_Groups Lagrange's_Theorem Symmetric_Groups_II Quaternions

Q: Definition. Binary Operation
A: Formally it is a map, \(*:A \times A \to A\)

Q: Definition. Group
A: A group is a set G with a binary operation \(* \) satisfying the following axioms:
1(Idenitity). There is some \(e \in G\) such that for all a, we have
\[a * e = e * a = a\]
2(Inverse). For all \(a \in G \), there is some \(a^{-1} \in G\) such that \[ a * a^{-1} = a^{-1} * a = e\]3(Associativity). For all \(a,b,c \in G\), we have \[(a*b)*c = a*(b*c)\]

Q: Definition. Order of group
A: The order of the group, denoted by \(|G|\), is the number of elements in G. A group is a finite group if the order is finite.

Q: Definition. Abelian Group
A: A group is abelian if it satisfies commutativity\[(\forall a,b \in G) a * b = b* a\]

Q: Proposition. Let \((G, *)\) be a group. Then
i) The identity is unique
ii) Inverses are unique
A: i) Assume e and e' are both identities. Then \(e = e * e' = e'\) first treating e' as an identity and then e.
ii) Assume \(a^{-1} \text{ and } b\) are both inverses for a. Then \(a = ae = a(a^{-1}b) = (aa^{-1})b = eb = b\)

Q: Proposition. Let \((G, *)\) be a group and \(a,b \in G\). Then
i) \((a^{-1})^{-1} = a\)
ii) \((ab)^{-1} = b^{-1}a^{-1}\)
A: i)Given \(a^{-1}\), both \(a \text{ and } (a^{-1})^{-1}\) satisfy \[xa^{-1} = a^{-1}x = e\]
By uniqueness of inverses they must be the same
ii) We have \[(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aea^{-1} = aa^{-1} = e\]
Repeat with \((b^{-1}a^{-1})(ab)\) and by uniqueness of inverses the statement is true

Q: Definition. Subgroup
A: A H is a subgroup of G, written \(H \leq G\), if \(H \subseteq G\)and H with the restricted operation \(*\) from G is also a group.

Q: Lemma. Subgroup criteria 1. Let \((G, *)\) be a group and \(H \subseteq G\). \(H \leq G\) iff
i) \(e \in H\)
ii) \((\forall a,b \in H)ab \in H\)
iii) \((\forall a \in H) a^{-1} \in H\)
A: 0. Closure: ii)
1. Identity: i) Note that H ang G must have the same identity. Suppose that \(e_H \text{ and } e_G\) are the identities of H and G respectively. Then \(e_He_H = e_H\). Now \(e_H\) has an inverse in G. Thus we have \(e_He_He_H^{-1} = e_He_H^{-1}\). So \(e_He_G = e_G\). Thus \(e_H = e_G\)
2. Inverse: iii)
3. Associativity: inherited from G

Q: Lemma. Subgroup criteria 2. A subset \(H \subseteq G\)  is a subgroup of G iff:
I) H is non-empty
II) \((\forall a,b \in H)ab^{-1} \in H\)
A: I) and II) follow trivially from i), ii), ii) in criteria 1
To prove I) and II) imply i), ii), iii) we do the following:
i) H must contain at least one element a, then \(aa^{-1} = e \in H\): Identity
iii) \(ea^{-1} = a^{-1} \in H\): Inverses
ii)\(a(b^{-1})^{-1} = ab \in H\): Closure
Associativity inherited

Q: Proposition. The subgroups of \((\mathbb{Z}, +)\)are exactly \(n\mathbb{Z}\), for \(n \in \mathbb{N}\)
A: It is trivial to show that for any \(n \in \mathbb{N}\), \(n\mathbb{Z}\) is a subgroup. Now we must show that any subgroup is in this form.
Let \(H \leq \mathbb{Z}\). We know \(0 \in H\). If there are no other elements in H, then \(H = 0\mathbb{Z}\). Otherwise, pick the smallest positive integer n in H. Then \(H = n\mathbb{Z}\).
Otherwise, suppose \((\exists a \in H) n\nmid a\) . Let \(a = pn + q\), where \(0 < q < n\). Since \(a - pn = q \in H\) then \(q<n\) which is a contradicition so every \(a \in H\) is divisible by n. Also by closure all multiples of n must be in H.

Q: Definition. Function
A: Given two sets \(X,Y, f: X \to Y\) sends each \(x \in X\) to a particular \(f(x) \in Y\). \(X\) is called the domain and \(Y\) is the co-domain

Q: Definition. Inclusion Map
A: \(\iota : \mathbb{Z} \to \mathbb{Q}: \iota (n) = n\)
Differs from the identity in that the domain and codomain are different

Q: Definition. Group homomorphism
A: Let \((G, *)\) and \((H, \times)\)be groups. A function \(f: G \to H\) is a group homomorphism iff \[(\forall g_1,g_2 \in G)f(g_1) \times f(g_2) = f(g_1 * g_2)\]

Q: Definition. Group isomorphism
A: Isomorphisms are bijective homomorphisms. We write \(G \cong H\)

Q: Proposition. Suppose that \(f: G \to H\) is a homomorphism. Then
i) \(f(e_G) = e_H\)
ii) \(f(a^{-1}) = f(a)^{-1}\)
A: i) \(f(e_G) = f(e_G^2) = f(e_G)f(e_g) \implies f(e_G)^{-1}f(e_G) = f(e_G)^{-1}f(e_G)^2 \implies f(e_G) = e_H\)
ii) \(e_H = f(e_G) = f(aa^{-1}) = f(a)f(a^{-1})\), by uniqueness of inverses \(f(a)^{-1} = f(a^{-1})\)

Q: Proposition. Suppose that \(f: G \to H\) os a homomorphism.
iii) The composite of 2 group homomorphisms is a group homomorphisms
iv) The inverse of an isomorphism is an isomorphisms
A: iii)Let \(f: G_1 \to G_2\) and \(g: G_2 \to G_3\). Then \(g(f(ab)) = g(f(a)f(b)) = g(f(a))g(f(b))\)
iv) Let \(f: G \to H\) be an isomorphism. \(f^{-1}(ab) = f^{-1} \left\{ f[f^{-1}(a)]f[f^{-1}(b)] \right\} = f^{-1} \left \{f[f^{-1}(a)f^{-1}(b)] \right \} = f^{-1}(a)f^{-1}(b)\)

Q: Definition. Image of homomorphism
A: If \(f: G \to H\) is a homomorphism, then the image of f is \[\text{im} f = f(G) = \{f(g) : g \in G\}\]

Q: Definition. Kernel of homomorphism
A: The kernel of f is \[ \text{ker} f = f^{-1}(\{e_H\}) = \{g \in G: f(g) = e_H\}\]

Q: Proposition. Both the image and the kernel are subgroups of the respective groups, i.e. \(\text{im} f \leq H \text{ and ker} f \leq G\)
A: Use subgroup criteria II.
Since \(e_H \in \text{im}f, e_G \in \text{ker} f\) they are both non-empty.
Moreover suppose \(b_1, b_2 \in \text{im} f\), then \(\exists a_1, a_2 \in G\) such that \(f(a_i) = b_i\), then \(b_1b_2^{-1} = f(a_1)f(a_2^{-1}) = f(a_1a_2^{-1}) \in \text{im}f\)
Then consider \(b_1,b_2 \in \text{ker} f\), Then \(f(b_1b_2^{-1}) = f(b_1)f(b_2^{-1}) = e^2 = e\) so \(b_1b_2^{-1} \in \text{ker} f\)

Q: Proposition. Given any homomorphism \(f: G \to H\) and any \(a \in G\), for all \(k \in \text{ker} f, aka^{-1} \in \text{ker} f\)
A: \(f(aka^{-1}) = f(a)f(k)f(a)^{-1} = f(a)ef(a)^{-1} = e \implies aka^{-1} \in \text{ker} f\)

Q: Proposition. For all homomorphisms \(f: G \to H\), f is
i) surjective iff \(\text{im}f = H\)
ii) injective iff \(\text{ker} f = \{e \}\)
A: i) by definition
ii)\(f(e)= e\) so if f is injective then by definition \(\text{ker} f = \{e\}\). If \(\text{ker} f = \{e\}\) then if \(f(a) = f(b) \implies f(ab^{-1}) = f(b)f(b)^{-1} = e.\) So \(ab^{-1} \in \text{ker} f = \{e\} \implies ab^{-1} = e \implies a=b\)

Q: Definition. Cyclic group \(C_n\)
A: A group G is cyclic if \[(\exists a)(\forall b)(\exists n \in \mathbb{Z})b = a^n\]
We write \(\langle a \rangle\) for the the cyclic group generated by a.

Q: Definition. Order of element
A: The order of an element is the smallest integer n such that \(a^n = e\), if n doesn't exist, a has infinite order

Q: Lemma. For a in g, \(ord (a) = |\langle a \rangle|\)
A: If \(\text{ord} (a) = \infty, a^n \neq a^m \text{ for } n\neq m\), otherwise \(a^{m-n} = e\). Thus \(|\langle a \rangle| = \infty = \text{ord} (a)\).
Otherwise, suppose \(\text{ord} (a) = k\), thus \(a^k = e\).  Now we claim that \(\langle a \rangle = {e,a,a^2, \cdots a^{k-1}}\). Note that \(\langle a \rangle\) does not contain higher powers of a, there are also no repeating elements in the list provided since \(a^m = a^n \implies a^{m-n} = e\)

Q: Definition. Exponent of group
A: The exponent of a group G is the smallest integer n such that \(\forall a \in G, a^n = e\)

Q: Definition. Dihedral groups \(D_{2n}\)
A: The symmetries of a regular n-gon. It contains n rotations and n reflections.

Q: Definition. Direct product of groups
A: Given two groups \((G, \circ)\) and \((H, \bullet)\), we can define a set \(G \times H = \{(g,h): g \in G, h \in H\}\) and an operation \((a_1, a_2)*(b_1,b_2) = (a_1 \circ b_1, a_2 \bullet b_2)\). This forms a group

Q: Proposition. \(C_n \times C_m \cong C_{nm}\) iff \(\text{hcf}(n,m) = 1\)
A: Suppose that \(\text{hcf}(m,n) = 1\). Let \(C_n = \langle a \rangle\) and \(C_m = \langle b \rangle\). Let k be the order of \((a,b)\). Then \((a,b)^k = (a^k, b^k) = e\). This is possible only if \(n|k, m|k\) so since the order is the minimum value of k that satisfies the equation, \(k = \text{lcm}(m,n) = \frac{mn}{\text{hcf}(m,n)}\)
Now consider \(\langle (a,b) \rangle \leq C_n \times C_m\). Since \((a,b)\) has order nm, \(\langle (a,b) \rangle\) must has nm elements and so does \(C_n \times C_m\) so it must be the whole group and \(\langle (a,b) \rangle \cong C_n \times C_m\) so \(C_n \times C_m \cong C_{nm}\).
Suppose that \(\text{hcf}(m,n) \neq 1\), then \(\text{lcm}(m,n) < mn\) so the order of any \((a,b)\) is less than mn so there is no element of order mn so \(C_n \times C_m\) is not a cyclic group of order nm.

Q: Proposition. Direct product theorem. Let \(H_1, H_2 \leq G\). Suppose the following are true:
i) \(H_1 \cap H_2 = \{e\}\)
ii)\((\forall a_i \in H_i)a_1 a_2 = a_2 a_1\)
iii)\((\forall a \in G)(\exists a_i \in H_i)a = a_1 a_2\)
Then \(G \cong H_1 \times H_2\)
A: Define \(f: H_1 \times H_2 \to G\) by \(f(a_1, a_2) = a_1a_2\). This is a homomorphism since:
\(f((a_1,a_2)*(b_1,b_2)) = f(a_1b_1,a_2b_2) = a_1b_1a_2b_2 = a_1a_2b_1b_2 = f(a_1,a_2)f(b_1,b_2)\)
Surjectivity follows from iii), we'll show injectivity by showing that the kernel is {e}. If \(f(a_1,a_2) = e\)  then \(a_1a_2 = e\). Then \(a_1 = a_2^{-1}\). Since \(a_1 \in H\) and \(a_2^{-1} \in H_2\), we have \(a_1 = a_2^{-1} \in H_1 \cap H_2 = {e}\). Thus \(a_1 = a_2 = e\) and \(\ker f = \{e\}\)

Q: Definition. Permutation
A: A permutation of X is a bijection from a set X to X itself. The set of all permutations on X is Sym X.

Q: Theorem. Sym X with composition forms a group
A: 0. If \(\sigma: X \to X\) and \(\tau: R \to R\), then \(\sigma \circ \tau: X \to X\). If they are both bijections then so is the composite,  so if \(\sigma, \tau \in \text{Sym} X\) then \(\sigma \circ \tau \in \text{Sym} X\).
1. The identity \(1_X: X \to X\) is clearly a permutation, and gives the identity.
2. Every bijective function has a bijective inverse. So if \(\sigma \in \text{Sym} X\), then \(\sigma^{-1} \in \text{Sym}X\)
3. Composition of functions is associative

Q: Definition. Symmetric group \(S_n\)
A: If X is finite, say \(|X| = n\), we write \(\text{Sym }X = S_n\). This is the symmetric group of degree n.

Q: Notation. Cycle notation
A: If a map sends \(1 \to 2, 2 \to 3, 3\to 1\) we write it as a cycle \((1 2 3)\). We leave out numbers that don't move, and we can write products of cycle such as \((12)(34)\). The order of each cycle is the length of the cycle and the inverse is the cycle written the other way round.

Q: Definition. k-cycles and transpositions
A: We call \((a_1a_2a_3\cdots a_k)\) a k-cycle. 2-cycles are called transpositions. Two cycles are disjoint if no number appeaers in both cycles

Q: Theorem. Any permutation in \(S_n\) can be written uniquely as a product of disjoint cycles.
A: Let \(\sigma \in S_n\). Start with \((1 \text{ } \sigma(1) \text{ } \sigma^2(1)^ \text{ } \sigma^3(1) \cdots)\). As the set \(\{1,2,3 \cdots n\}\) is finite, for some k we must have \(\sigma^k(1)\) already in the list. If \(\sigma^k(1) = \sigma^l(1)\), with \(l < k\), then \(\sigma^{k-l}(1) = 1\). So all \(\sigma^I(1)\) are distinct until we get back to 1. Thus we have the first cycle \((1 \text{ } \sigma(1) \text{ } \sigma^2(1)^ \text{ } \sigma^3(1) \cdots \sigma^{k-1}(1))\)
Now choose the smallest number that is not yet in a cycle, repeat to obtain a cycle \((j \text{ } \sigma(j) \text{ } \cdots) \sigma^{l-1}(j)\). Since \(\sigma\) is a bijection, nothing in this cycle can be in previous cycles as well. Repeat until all elements are exhausted. This is essentially unique because every number j completely determines the whole cycle it belongs to, and whichever number we start with, we'll end up with the same cycle.

Q: Definition. Cycle type
A: Write a permutation \(\sigma \in S_n\) in disjoint cycle notation. The cycle type is the list of cycle lengths. This is unique up to re-ordering. We often leave out singleton cycles.

Q: Lemma. For \(\sigma \in S_n\), the order of \(\sigma\) is the least common multiple of cycle lengths in the disjoint cycle notation. In particular, a k-cycle has order k.
A: As disjoint cycles commute, we can group together each cycle when we take powers. So if \(\sigma = \tau_1\tau_2\cdots \tau_l\) then \(\sigma^m = \tau_1^m \tau_2^m \cdots \tau_l^m\). Now if cycle \(\tau_i\) has length \(k_i\) then \(\tau_i^{k_i} = e\), so to get \(\sigma^m = e\) we need m to be a common multiple of all \(k_i\). Since the order is the least possiblem m such that \(\sigma^m = e\), the order is the least common multiple of \(k_i\).

Q: Proposition. Every permutation is a product of transpositions
A: As each permutation is a product of disjoint cycles, it suffices to prove that each cycle is a product of transpositions. So we can just write \((a_1 \text{ } a_2 \text{ } a_3 \dots)\) as \((a_1 \text{ } a_2)(a_2 \text{ } a_3)\cdots\). Thus a k-cycle can be written as a product of k-1 transpositions.

Q: Theorem. Writing \(\sigma \in S_n\) as a product of transpositions in different ways, it is either always composed of an even number of transpositions, or always an odd number of transpositions.
A: Write \(\#(\sigma)\) for the number of cycles in disjoint cycle notation, including singleton cycles.
When we multiple \(\sigma\) by a transposition \(\tau = (c \;{ } d)\):
-if \(c,d\) are in the same \(\sigma\)-cycle, say \((c \; a_2 \cdots a_{k-1} \; d \; a_{k+1} \cdots a_{k+l})(c \; d) = (c \; a_{k+1} a_{k+2} \cdots a_{k+l})(d \; a_2 \; a_3 \cdots a_{k-1})\). So \(\# (\sigma \tau) = \# (\sigma) + 1\)
-if \(c,d\) are in different cycles, say \((d \; a_2 \; a_3 \cdots a_{k-1})(c \; a_{k+1} \cdots a_{k+l})(c \; d) = (c \; a_2 \cdots a_{k-1} \; d \; a_{k+1} \cdots a_{k+l})(c \; d)(c \; d) = (c \; a_2 \cdots a_{k-1} \; d \; a_{k+1} \cdots a_{k+l})\) and \(\# (\sigma \tau) = \# (\sigma) - 1\)
There for any transposition the number of cycles stays the same mod 2.
Construct \(\sigma = \tau_1 \cdots \tau_l = \tau_1' \cdots \tau_k'\) by beginning with e and multiplying by transpositions 1 by 1, we can show that \(l \equiv k \mod 2\)using the fact above

Q: Definition. Sign of permutation
A: Viewing \(\sigma \in S_n\) as a product of transpositions, \(\sigma = \tau_1 \cdots \tau_l\), we call \(\text{sgn}(\sigma) = (-1)^l. \) If \(\text{sgn}(\sigma) = 1\) we call it an even permutation, otherwise it is odd.

Q: Theorem. For \(n \geq 2, \text{sgn}:S_n \to \{\pm1\}\) is a surjective group homomorphism
A: Suppose \(\sigma_1 = \tau_1 \cdots \tau_{l_1}\) and \(\sigma_2 = \tau_1 \cdots \tau_{l_2}\), then \(\text{sgn}(\sigma_1 \sigma_2) = (-1)^{l_1 + l_2} = (-1)^{l_1} (-1)^{l_2} = \text{sgn}(\sigma_1)\text{sgn}(\sigma_2)\). So it is a homomorphism.
It is surjective since \(\text{sgn} (e) = 1\) and \(\text{sgn}(1 \; 2) = -1\)

Q: Lemma. \(\sigma\) is an even permutation iff the number of cycles of even length is even.
A: A k-cycle can be written as k-1 transpositions. Thus an even-length cycle is odd, vice versa.
Since sgn is a group homomorphism, writing \(\sigma\) in disjoint cycle notation, \(\sigma = \sigma_1 \sigma_2 \cdots \sigma_l\), we get \(\text{sgn}(\sigma) = \text{sgn}(\sigma_1) \cdots \text{sgn}(\sigma_l)\) . Suppose there are m even length-cycles and n odd-length cycles, then \(\text{sgn}(\sigma) = (-1)^m1^n.\) This is equal to 1 iff m is even.

Q: Definition. Alternating group \(A_n\)
A: The alternating group is the kernel of sgn, i.e. the even permutations. Since \(A_n\) is a kernel of a group homomorphism, \(A_n \leq S_n\)

Q: Proposition. Any subgroup of \(S_n\) contains either no odd permutations or exactly half.
A: If \(S_n\) has at least one odd permutation \(\tau\), then there exists a bijection between the odd and even permuations by \(\sigma \to \sigma \tau\)(bijection since \(\sigma \to \sigma \tau^{-1}\) is a well-defined inverse). So there are as many odd permutations as even permutations.

Q: Definition. Cosets
A: Let \(H \leq G\) and \(a \in G\). Then the set \(aH = \{ah:h \in H\}\) is a left coset of H and \(Ha = \{ha:h \in H\}\) is a right coset of H

Q: Proposition. \(aH = bH \iff b^{-1}a \in H\)
A: \((\Rightarrow)\)Since \(a \in aH, a \in bH\). Then \(a = bh\) for some \(h \in H\). So \(b^{-1}a = h \in H\).
\((\Leftarrow)\)Let \(b^{-1}a = h_0\). Then \(a = bh_0\). Then \(\forall ah \in aH\), we have \(ah = b(h_0h) \in bH\). So \(aH \subseteq bH\). Similarly, \(bH \subseteq aH\). So \(aH = bH\)

Q: Lemma. The left cosets of a subgroup \(H \leq G\) partition G, and every coset has the same size.
A: For each \(a \in G, a \in aH\). Thus the union of all cosets gives all of G. Now we have to show that for all \(a,b \in G\), the cosets aH and bH are either the same or disjoint.
Suppose that aH and bH are not disjoint. Let \(ah_1 = bh_2 \in aH \cap bH\). Then \(b^{-1}a = h_1h_1^{-1} \in H\). So \(aH = bH\).
Note that \(f : H \to aH\) with \(f(h) = ah\) is invertible with inverse \(f^{-1}(h) = a^{-1}h\). Thus there exists a bijection between them and they have the same size.

Q: Definition. Index of a subgroup
A: The index of H in G, written \(|G : H|\), is the number of left cosets of H in G

Q: Theorem. Lagrange's Theorem. If G is a finite group and H is a subgroup of G, then \(|H|\) divides \(|G|\). In particular; \[|H||G : H| = |G|\]
A: Suppose that there are \(|G:H|\) left cosets in total. Since the left cosets partition G, and each coset has size \(|H|\), we have \(|H||G:H| = |G|\).

Q: Corollary: The order of an element divides the order of the group
A: Consider the subgroup generated by a, which has order ord(a). Then by Lagrange's theorem, ord(a) divides |G|

Q: Corollary. The exponent of a group divides the order of the group
A: We know that \(|G| = k \text{ ord}(a)\) for some k. Then \(a^{|G|} = (a^{ord(a)})^k = e^k = e\)

Q: Corollary. Groups of prime order are cyclic and are generated by every non-identity element.
A: Say \(|G| = p\). If \(a \in G\)is not the identity, the subgroup generated by a must have order p since it has to divide p. Thus the subgroup generated by a has the same size as G and they must be equal. Then G must be cuclic since it is equal to the subgroup generated by a.

Q: Definition. Normal subgroups. \(K \unlhd G\)
A: A subgroup K of G is a normal subgroup if \[(\forall a \in G)(\forall k \in K)aka^{-1} \in K\]
Equivalent to \[(\forall a \in G)aKa^{-1} = K\]\[(\forall a \in G)aK = Ka\]

Q: Proposition. Every kernel is a normal subgroup
A: Given homomorphism \(f : G \to H\) and some \(a \in G\), for all \(k \in \ker f\), we have \(f(aka^{-1}) = f(a)f(k)f(a)^{-1} f(a)ef(a)^{-1} = e\). Therefore \(aka^{-1} \in \ker f\) by definition of the kernel

Q: Proposition. A group of order 6 is either cyclic or dihedral
A: Let \(|G| = 6\). By Lagrange the orders of the elements must be 1,2,3 or 6. If there is an element of order 6 then it is the cyclic group.
Otherwise if G only has elements of order 2, the order of the group must be a power of 2 which isn't possible so there has to be an element r of order 3.
So \(\langle r \rangle \unlhd G\) as it has index 2. Now G must also have an element s of order 2.
Since \(\langle r \rangle\) is normal, we know that \(srs^{-1} \in \langle r \rangle\). If \(srs^{-1} = e\) then r = e which is not true. If \(srs^{-1} = r\) then \(sr = rs\) and sr has order 6, which was ruled out above. So \(srs^{-1} = r^2 = r^{-1}\) then G is dihedral by the definition of a dihedral group.

Q: Lemma. Every subgroup of index 2 is normal
A: If \(K \leq G\)  has index 2, then there are only two possible cosets K and G\K. As eK = Ke and cosets partition G, the other left and right coser must be G\K. So all left cosets and right cosets are the same.

Q: Proposition. Let \(K \unlhd G\). Then the set of cosets of K in G is a group under the operation \(aK * bK = (ab)K\)
A: First show the operation is well-defined. If \(aK = a'K\) and \(bK = b'K\) we want to show that \(aK * bK = a'K * b'K\). We know that \(a' = ak_1\) and \(b' = bk_2\) for some \(k_1, k_2 \in K\). Then \(a' b' = ak_1 bk_2\). We know that \(b^{-1}k_1b \in K\). Let \(b^{-1}k_1b = k_3\). Then \(k_1b = bk_3\). So \(a'b' = abk_3k_2 \in (ab)K\). So picking a different representative of the same coset gives the same product.
It is easy to check the group axioms.

Q: Definition. Quotient Group
A: Given a group G and a normal subgroup K, the quotient group or factor group of G by K, written as \(G/K\), is the set of cosets of K in G under the operation \(aK * bK = (ab)K\)

Q: Lemma. Given \(K \unlhd G\), the quotient map \(q : G \to G/K\) with \(g \to gK\) is a surjective group homomorphism
A: \(q(ab) = (ab)K = aKbK = q(a)q(b)\).
Also for all \(aK \in G/K\), \(q(a) = aK\) so it is surjective

Q: Proposition. The quotient of a cyclic group is cyclic.
A: Let \(G = C_n\) with \(H \leq C_n\). We know that H is also cyclic. Say \(C_n = \langle c \rangle\) and \(H = \langle c^k \rangle \cong C_l\), where \(kl = n\). We have \(C_n/H = \{H, cH, c^2H, \cdots c^{k-1}H\} = \langle cH \rangle \cong C_k\)

Q: Theorem. The Isomorphism Theorem. Let \(f : G \to H\) be a group homomorphism with kernel K. Then \(K \unlhd G\) and \(G/K \cong \text{im} f\)
A: We have proved that \(K \unlhd G\) already. We define a group homomorphism \(\theta : G/K \to \text{im}f\) by \(\theta (aK) = f(a)\).
Check that this is well-defined. If \(a_1 K = a_2 K\), then \(a_2^{-1}a_1 \in K\). So \(f(a_2)^{-1}f(a_1) = f(a_2^{-1}a_1) = e\). So \(f(a_1) = f(a_2)\) and \(\theta (a_1 K) = \theta (a_2 K)\). Now we check that it is a group homomorphism: \[\theta (aKbK) = \theta(abK) = f(ab) = f(a)f(b) = \theta(aK) \theta(bK)\]
To show that it is injective, suppose \(\theta (aK) = \theta (bK)\). Then \(f(a) = f(b) \) so \(f(b)^{-1} f(a) = e\). Hence \(b^{-1}a \in K\). So \(aK = bK\). By definition \(\theta\) is surjective since \(\text{im} \theta = \text{im} f\). So \(\theta\) gives an isomorphism \(G/K \cong \text{im }f \leq H\).

Q: Lemma. Any cyclic group is isomorphic to either \(\mathbb{Z}\) or \(\mathbb{Z}/ (n\mathbb{Z})\) for some \(n \in \mathbb{N}\)
A: Let \(G = \langle c \rangle\). Define \(f: \mathbb{Z} \to G\) with \(m \to c^m\). This is a homomorphism since \(c^{m_1 + m_2} = c^{m_1}c^{m_2}\). f is surjective since G is by definition all \(c^m\) for all m. We know that \(\ker f \unlhd \mathbb{Z}\). We have three possibilities, either
i) \(\ker f = \{e\}\), so F is an isomorphism and \(G \cong \mathbb{Z}\)
ii)\(\ker f= \mathbb{Z}\) , then \(G \cong \mathbb{Z} /\mathbb{Z} = \{e\} = C_1\)
iii) \(\ker f = n \mathbb{Z}\)(since these are the only proper subgroups of \(\mathbb{Z}\), then \(G \cong \mathbb{Z} / (n\mathbb{Z})\)

Q: Definition. Simple group
A: No non-trivial proper normal subgroup. Only \(\{e\}\) and G are normal subgroups

Q: Definition. Quaternions
A: \[Q_8 = \langle a,b : a^4 = e, b^2 = a^2, bab^{-1} = a^{-1} \rangle \]

Q: Definition. Group action
A: Let X be a set and G be a group. An action of G on X is a homomorphism \(\varphi : G \to \text{Sym } X\).

Q: Proposition. Let X be a set and G be a group. Then \(\varphi : G \to \text{Sym }X\) is a homomorphism iff \(\theta : G \times X \to X\) defined by \(\theta(g,x) = \varphi(g)(x) \) satisfies:
A: 0. \((\forall g \in G)(x \in X)\theta(g,x) \in X\)
1. \((\forall x \in X) \theta(e,x) = x\)
2. \((\forall g,h \in G)(\forall x \in X)\theta(g, \theta(h,x)) = \theta(gh,x)\)

Q: Definition. Kernel of action
A: The kernel of \(\varphi\), all g such that \(\varphi (g) = 1_X\)

Q: Lemma. Left regular action. Any group G acts on itself by left multiplication. This action is faithful and transitive
A: We have:
\((\forall g \in G)(x \in G)g(x) = g \cdot x \in G\) by definition of a group
\((\forall x \in G)e \cdot x = x\) by definition of a group
\(g(hx) = (gh)x\) by associativity
\([(\forall x \in X)gx = x] \implies g = e\) by uniqueness of identity so it is faithful
\(\forall x, y \in G\) then \((yx^{-1})(x) = y\) so any x can be sent to any y so it is transitive

Q: Lemma. Left coset action. Let \(H \leq G\). Then G acts on the left cosets of H by left multiplication transitively
A: Check axioms trivially.
\((ba^{-1})(aH) = bH\) so any aH can be mapped to bH so it is transitive

Q: Definition. Faithful action
A: An action is faithful if the kernel is just {e}

Q: Definition. Orbit of action
A: Given an action G on X, the orbit of an element \(x \in X\) is \[\text{orb}(x) = G(x) = \{y \in X : (\exists g \in G)g(x) =y\}\]
Intuitively, it is the elements that x can possibly get mapped to

Q: Definition. Stabiliser of action
A: The stabiliser of x is \[\text{stab}(x) = G_x = {g \in G: g(x) = x} \subseteq G\]
Intuitively, it is the elements in G that do not change x

Q: Lemma. stab(x) is a subgroup of G
A: We know that e(x) = x by definition. So stab(x) is not empty. Suppose \(g,h \in \text{stab}(x)\), then \(gh^{-1}(x) = g(h^{-1}(x)) = g(x) = x\). So \(gh^{-1} \in \text{stab}(X)\). So \(\text{stab}(x)\) is a subgroup

Q: Definition. Transitive action
A: An action G on X is transitive if \((\forall x)\text{orb}(x) = X\)
Intuitively you can reach any element from any element.

Q: Lemma. The orbits of an action partition X
A: First, \((\forall x)(x \in \text{orb}(x))\) as \(e(x)= x\). So every x is in some orbit. Then suppose \(z \in \text{orb}(x)\) and \(z \in \text{orb}(y)\), we have to show that \(\text{orb}(x) = \text{orb}(y)\). We know that \(z = g_1(x)\) and \(z = g_2(y)\) for some \(g_1, g_2\). Then \(g_1(x) = g_2(y)\) and \(y = g_2^{-1}g_2(x)\).
For any \(w = g_3(y) \in \text{orb}(y)\), we have \(w = g_3g_2^{-1}g_1(x)\). So \(w \in \text{orb}(x)\). Thus \(\text{orb}(y) \subseteq \text{orb}(x)\) and similarly \(\text{orb}(x) \subseteq \text{orb}(y)\). Therefore \(\text{orb}(x) = \text{orb}(y)\) .

Q: Theorem. Orbit-stabiliser theorem. Let the group G act on X. Then there is a injection between \(\text{orb}(x)\) and cosets of \(\text{stab}(x)\) in G. In particular, if G is finite then \[|\text{orb}(x)||\text{stab}(x)| = |G|\].
A: We biject the cosets of stab(x) with elements in the orbit of x. Recall that \(G : \text{stab}(x)\) is the set of cosets of \(\text{stab}(x)\). We can define \[\theta : (G : \text{stab}(x)) \to \text{orb(x)} \\ g \text{ stab}(x) \mapsto g(x)\].
This is well-defined -- if \(g \text{ stab}(x) = h \text{ stab}(x)\) then \(h = gk\) for some \(k \in \text{ stab}(x)\). So \(h(x) = g(k(x)) = g(x)\).
This map is surjective since for any \(y \in \text{ orb}(x)\), there is some \(g \in G\) such that \(g(x) = y\), by definition. Then \(\theta(g \text{ stab}(x)) = y\). It is injective since if \(g(x) = h(x)\), then \(h^{-1}g(x) = x\). So \(h^{-1}g \in \text{ stab}(x)\). So \(g \text{ stab}(x) = h \text{ stab}(x)\).
Hence the number of closets is \(| \text{ orb}(x)|\). Then the result follows from Lagrange's theorem.

Q: Theorem. Cayley's Theorem. Every group is isomorphic to some subgroup of some symmetric group
A: Take the left regular action of G on itself. This gives a group homomorphism \(\phi: G \to \text{Sym } G\)with \(\text{ker }\phi = \{e\}\) as the actino is faithful. By the isomorphism theorem, \(G \cong \text{im } \phi \leq \text{Sym }G\)

Q: Definition. Conjugation of element
A: Conjugation of \(a \in G\) by \(b \in G\) is given by \(bab^{-1} \in G\). Given any \(a,c\), if there exists some b such that \(c = bab^{-1}\) , then we say that a and c are conjugate

Q: Lemma. Conjugation action. Any group acts on itself by conjugation. \(g(x) = gxg^{-1}\)
A: 0. \(g(x) = gxg^{-1} \in G\) for all \(x,g \in G\)
1. \(e(x) = exe^{-1} = x\)
2. \(g(h(x)) = g(hxh^{-1}) = ghxh^{-1}g^{-1} = (gh)x(gh)^{-1} = (gh)(x)\)

Q: Definition. Conjugacy classes and centralizers
A: The conjugacy classes are the orbits of the conjugacy action. \[\text{ccl}(a) = \{ b \in G : (\exists g \in G)gag^{-1} = b\}\]
The centralizers are the stabilizers of this action. i.e. elements that commute with a \[C_{G}(a) = \{g \in G: gag^{-1} = a\} = \{g\in G: ga = ag\} \]

Q: Definition. Center of group
A: The elements that commute with all other elements. \[Z(G) = \{g\in G:(\forall a)gag^{-1} = a\} = \{ g\in G:(\forall a)ga = ag\}\]

Q: Lemma. Let \(K \unlhd G\), then G acts by conjugation on K
A: We only have to prove closure as other properties follow from conjugation action.
For every \(g \in G, k \in K\) we know \(gkg^{-1} \in K\), so it is closed

Q: Proposition. Normal subgroups are exactly those subgroups which are unions of conjugacy classes
A: Let \(K \unlhd G\). If \( k \in K\), then by definition for every \(g \in G\), we get \(gkg^{-1} \in K\). So \(\text{ccl}(k) \subseteq K\). So K is the union of the conjugacy classes of all its elements.
Conversely, if K is a union of conjugacy classes and a subgroup of G, then for all \(k \in K\), \(g \in G\), we have \(gkg^{-1} \in K\). So K is normal

Q: Lemma. Let X be the set of subgroups of G. Then G acts by conjugation on X
A: 0. If \(H \leq G\), then we have to show that \(gHg^{-1}\) is also a subgroup. We know that \(e \in H\) and thus \(geg^{-1} = e \in gHg^{-1}\), so \(gHg^{-1}\) is non-empty.
For any two elements \(gag^{-1}, gbg^{-1} \in gHg^{-1}\), \((gag^{-1})(gbg^{-1})^{-1} = g(ab^{-1})g^{-1} \in gHg^{-1}\) so it is a subgroup
1.\(eHe^{-1} = H\)
2. \(g_1(g_2Hg_2^{-1})g_1^{-1} = (g_1g_2)H(g_1g_2)^{-1}\)

Q: Definition. Normalizer of subgroup
A: The normalizer of a subgroup is the stabilizer of the group conjudation action. \[N_G(H) = \{ g \in G: gHg^{-1} = H\}\]Can show that \(N_G(H)\)  is the largest subgroup of G in which H is a normal subgroup, hence the name.

Q: Theorem. Cauchy's Theorem. Let G be a finite group and prime p dividing |G|. Then g has an element of order p, in fact there must be at least p-1 elements of order p.
A: Let G and p be fixed. Consider \(G^p = G \times G \times \cdots \times G\), the set of p-tuples of G. Let \(X \subseteq G^p\) be \(X = \{(a_1, a_2, \cdots , a_p) \in G^p : a_1a_2\dots a_p = e\}\).
In particular if an element b has order p then \((b,b,b, \dots , b) \in X\).
Now let \(H = \langle h : h^p = e \rangle \cong C_p\) be a cyclic group of order p with generator h, Let H act on X by 'rotation'. \[h(a_1,a_2,\dots a_p) = (a_2,a_3,\dots , a_p, a_1)\]
This is an action and can be checked.
Orbits partition X so the sum of all orbit sizes must be |X|. We know that \(|X| = |G|^{p-1}\) since we can freely choose the first p-1 entires and the last one must be the inverse of their product. Since p divides |G|, p also divides |X|. We have \(|\text{orb} (a_1, \dots , a_p)||\text{stab}_H (a_1, \dots , a_p)| = |H| = p\). So all orbits have size 1 or p, and they sum to \(|X| = p \times \text{something}\). We know there is one orbit of size 1, namely \((e,e,\dots,e).\) So there must be at least p - 1 other orbits of size 1 for the sum to be divisible by p.
In order to have an orbit of size 1, they must look like \((a,a,\dots ,a)\) for some \(a \in G\), which has order p.

Q: Proposition. If \((a_1 \; a_2 \dots a_k)\) is a k-cycle and \(\rho \in S_n\), then \(\rho (a_1 \dots a_k) \rho^{-1}\) is the k-cycle \((\rho(a_1) \rho(a_2) \dots \rho(a_k))\)
A: Consider any \(\rho(a_1)\) acted on by \(\rho (a_1 \; a_2 \dots a_k)\rho^{-1}\). The three permutations send it to \(\rho(a_1) \mapsto a_1 \mapsto a_2 \mapsto \rho(a_2)\). Since \(\rho\) is bijective we can write every b as \(\rho(a)\). The result is the k-cycle \((\rho(a_1) \rho(a_2) \dots \rho(a_k))\).

Q: Corollary. Two elements in \(S_n\) are conjugate iff they have the same cycle type.
A: Suppose \(\sigma = \sigma_1 \sigma_2 \dots \sigma_l\) in disjoint cycle notation. Then \(\rho \sigma \rho^{-1} = \rho \sigma_1 \rho^{-1} \rho \sigma_2 \rho^{-1} \dots \rho \sigma_l \rho^{-1}\). Since the conjugation of a cycle conserves its length, \(\rho \sigma \rho^{-1}\) has the same cycle type.
Conversely, if \(\sigma, \tau\) have the same cycle type, say \[\sigma = (a_{11} a_{12} \dots a_{1 n_1}) \dots (a_{k1} a_{k2} \dots a_{k n_k}) \;\;\; \tau = (b_{11} \dots b_{1n_1})\dots(b_{k1} b_{k2} \dots b_{k n_k})\]Then let \(\rho(a_{ij}) = b_{ij}\), then \(\rho \sigma \rho^{-1} = \tau\)

Q: Definition. Splitting of conjugacy classes
A: When \(|\text{ccl}_{A_n}(\sigma)| = \frac{1}{2}|\text{ccl}_{S_n}(\sigma)|\), we say that the conjugacy class of \(\sigma\) splits in \(A_n\)

Q: Proposition. For \(\sigma \in A_n\), the conjugacy class of \(\sigma\) splits in \(A_n\) if and only if no odd permutation commutes with \(\sigma\).
A: We have the conjugacy classes splittling if and only if the centralizer does not(by orbit-stabiliser theorem).
Clearly \(C_{A_n}(\sigma) = C_{S_n}(\sigma) \cap A_n\). So splitting of centralizer occurs if and only if an odd permutation commutes with \(\sigma\).
Will split as if it has one odd element, half the elements odd so we lose half the elements.

Q: Theorem. \(A_5\) is simple
A: We know that normal subgroups must be unions of conjugacy classes, must contain e and their order must divide 60 by Lagrange.
The possible orders are 1,2,3,4,5,6,10,12,15,20,30.
The conjugacy classes 1,15,20,12,12 canot add up to any of the possible orders apart from 1 and 60.

Q: Definition. General linear group \(GL_n(F)\)
A: \(n \times n\) matrices with non-zero determinants and real entries

Q: Definition. Special linear group \(SL_n(F)\)
A: \[SL_n(F) = \{ A \in GL_n(F) : \det A = 1\}\]

Q: Proposition. \(GL_n(\mathbb{C})\) acts faithfully on \(\mathbb{C}^n\) by left multiplication to the vector, with two orbits(0 and everything else)
A: Can show it is group action easily.
To prove it is faithful: A linear map is determined by what it does on a basis. Take the standard basis, any matrix which maps each \(\mathbf{e_k} = (1,0,0\dots)\) to itself must be I since the columns of a matrix are the images of the basis vectors.
To show there are two orbits, we know that \(A \mathbf{0 = 0}\) for all A. Also, as A is invertible \(A \bf v = 0 \iff v = 0\). So \(\bf 0\) forms a singleton orbit.
Then given any two vectors \(v \neq w \in \mathbb{C}^n \setminus \{0\}\), there is a matrix A such that \(Av = w\).

Q: Proposition. \(GL_n (\mathbb{C})\) acts on \(M_{n \times n}(\mathbb{C})\) by conjugation.
A: Proof is trivial.
This action can be thought of as a change of basis action. Two matrices are conjugate if they represent the same map but with respect to different bases.

Q: Lemma. Orthogonal matrices are isometries
For any orthogonal A and \(x,y \in \mathbb{R}^n\), we have:
i) \((A\bf{x}) \cdot(Ay) = x \cdot y\)
ii) \(|A\bf x| = x\)
A: Treat the dot product as a matrix multiplication. So \[(Ax)^T(Ay) = x^TA^TAy = x^T Iy = x^Ty\]
Then we have \(|Ax|^2 = (Ax)\cdot(Ax) = x \cdot x = |x|^2\). Since both are positive we can take square root.

Q: Definition. Orthogonal group O(n)
A: \[O(n) = O_n = O_n(\mathbb{R}) =  \{A \in GL_n(\mathbb{R}) : A^TA = I\}\]

Q: Lemma. The orthogonal group is a group
A: Simply show it is a subgroup of \(GL_n(\mathbb{R})\)

Q: Definition. Special orthogonal group SO(n)
A: Kernel of \(\mathbf \det : O(n) \to \{\pm 1\}\)\[SO(n) = SO_n = SO_n(\mathbb{R}) = \{A \in O(n): \det A = 1\}\]

Q: Lemma. SO(2) consists of all rotations of \(\mathbb{R}^2\) around 0
A: Use the properties that \(A^TA = I\) and \(\det A = 1\) to show that the rotation matrix satisfies the equations.

Q: Corollary. Any matrix is O(2) is either a rotation around 0 or a reflection in a line through 0
A: If \(A \in SO(2)\)we have shown its a rotation. Otherwise \(A = \left ( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right)\left ( \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{matrix}\right) = \left ( \begin{matrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{matrix}\right)\)
since \(O(2) = SO(2) \cup \left ( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right)SO(2)\). This has eigenvalues 1,-1. So it is a reflection in the line of the eigenspace \(E_1\). Since the eigenspace is a subspace it must go through 0.

Q: Lemma. Every matrix in \(SO(3)\) is a rotation around some axis
A: We know \(\det A = 1\) and A is an isometry. The eigenvalues must have \(|\lambda|= 1\). They also multiply to get \(\det A = 1\). Since we are in \(\mathbb{R}\) complex eigenvalues come in complex conjugate pairs, and the third one must be 1.
If all are real we have either -1,-1,1 or 1,1,1 all of which contain an eigenvalue of 1.
So pick an eigenvector for our eigenvalue 1 as the third basis vector. Then \[A = \left ( \begin{matrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 1\end{matrix}\right)\]So now let \(A' = \left ( \begin{matrix} a & b \\ c & d \end{matrix}\right )\) with \(\det A' = 1\) and A' is orthogonal so A' must be a rotation and then A must be a rotation through an axis.

Q: Lemma. Every matrix in \(O(3)\) is the product of at most three reflections in planes through 0
A: Recall that \(O(3) = SO(3) \cup \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{matrix}\right)SO(3)\)
If \(A \in SO(3)\) then we know it is a rotation which is the product of 2 reflections, \(A = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{matrix}\right) \left( \begin{matrix} \cos \theta & \sin \theta & 0 \\ \sin \theta & -\cos \theta & 0 \\ 0 & 0 & 1\end{matrix}\right)\)
Then if it isn't in this, it's automatically a product of 3 reflections.

Q: Definition. Möbius map
A: A map from \(\mathbb{C}_{\infty} \to \mathbb{C}_{\infty}\) of the form \[f(z) = \frac{az+b}{cz+d}\]
where \(a,b,c,d \in \mathbb{C}\) and \(ad-bc \neq 0\), with \(f(-\frac{d}{c}) = \infty\) and \(f(\infty) = \frac{a}{c}\)

Q: Proposition. The map \(\theta : GL_2(\mathbb{C}) \to M\)sending \(\left ( \begin{matrix} a & b \\ c & d\end{matrix}\right ) \mapsto \frac{az+b}{cz+d}\) is a surjective group homomorphism. What is the kernel?
A: The determinant ad-bc of any matrix is non-zero so it does map to a Möbius map. This also shows it is surjective.
Easy to check it is a homomorphism.
We can try different values of z to find the kernel.
\(z = \infty \implies c = 0; z=0 \implies b = 0;z = q \implies d = a\)
So \[\text{ker }\theta = Z = \{\lambda I: \lambda \in \mathbb{C}, \lambda \neq 0\}\]

Q: Definition. Fixed point for Möbius maps
A: A z such that \(f(z) = z \)

Q: Proposition. Any Mobius map with at least 3 fixed points must be the identity.
A: Consider \(f(z) = \frac{az + b}{cz+d}\), fixed points satisfy \(\frac{az+b}{cz+d} = z \iff cz^2 + (d-a)z -b = 0\). A quadratic has at most two roots, unless \(c=b=0, d=a\) in which case we have 0=0. But if \(c=b=0, d=a\) then f is just the identity.

Q: Definition. Projective general linear group \(PGL_2(\mathbb{C})\)
A: \(PGL_2(\mathbb{C}) = GL_2(\mathbb{C}) / Z\) where \(Z = \{\lambda I : \lambda \in \mathbb{C}/Z\}\), by the isomorphism theorem taking \(\theta : GL_2(\mathbb{C}) \to M\), sending \(\left ( \begin{matrix} a & b \\ c & d\end{matrix}\right ) \mapsto \frac{az+b}{cz+d}\).
We can restrict \(\theta\) to \(SL_2(\mathbb{C})\), and it will still be surjective as \(\text{ker }\theta = \{\pm I\}\)
So \(M \cong SL_2(\mathbb{C})/\{\pm I\} = PSL_2(\mathbb{C})\)
We also get \(PSL_2(\mathbb{C}) \cong PGL_2(\mathbb{C})\) as both are isomorphic to the Möbius group.

Q: Proposition. Every Möbius map is a composite of maps of the following form:
i)Dilation/rotation \(f(z) = az, a \neq 0\)
ii)Translation \(f(z) = z +b\)
iii)Inversion \(f(z) = \frac{1}{z}\)
A: Let \(\frac{az+b}{cz+d} \in M\).
If \(c = 0\), then \(g(z) = \frac{a}{d}z + \frac{b}{d}\) which is clearly a dilation following by a translation.
If \(c \neq 0\), then we have \(z \mapsto z + \frac{d}{c} \mapsto \frac{1}{z + \frac{d}{c}} \mapsto -\frac{ad+bc}{c^2(z+\frac{d}{c})} \mapsto \frac{a}{c} - \frac{ad+bc}{c^2(z+\frac{d}{c})} = \frac{az+b}{cz+d} \)

Q: Proposition. Any Möbius map is conjugate to \(f(z) = vz, v \neq 0\) or to \(f(z) = z+1\)
A: We have the surjective group homomorphism \(\theta : GL_2(\mathbb{C}) \to M\). The conjugacy classes of \(GL_2(\mathbb{C})\) are of the types;
\[\left ( \begin{matrix} \lambda & 0 \\ 0 & \mu \end{matrix}\right) \mapsto g(z) = \frac{\lambda z + 0}{0 + \mu} = \frac{\lambda}{\mu}z\]\[\left ( \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix}\right) \mapsto g(z) = \frac{\lambda z + 0}{0 + \lambda} =z\]\[\left ( \begin{matrix} \lambda & 1 \\ 0 & \lambda \end{matrix}\right) \mapsto g(z) = \frac{\lambda z + 1}{0 + \lambda} = z + \frac{1}{\lambda}\]We know that we can also represent the last one as \(\left ( \begin{matrix} 1 & \frac{1}{\lambda} \\ 0 & 1 \end{matrix}\right) \), which is conjugate to \(\left ( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right)\) since that is its Jordan-normal form. So it is conjugate to \(z + 1\)

Q: Proposition. Every non-identity Mobius map has exactly 1 or 2 fixed points.
A: \(f(w) = w \iff hf(w) = h(w) \iff hfh^{-1}(h(w)) = h(w)\) so \(h(w)\) is a fixed point of \(hfh^{-1}\).
Since h is a bijection this means f and its conjugate have the same number of fixed points.
Every f is conjugate to either \(vz\) or \(z+1\) so it has either 2 or 1 fixed points respectively

Q: Proposition. Given \(f,g \in M\). If \(\exists z_1, z_2, z_3 \in \mathbb{C}_\infty\)such that \(f(z_i) = g(z_i)\), then \(f=g\). i.e. every Möbius map is uniquely determined by 3 points
A: As the maps are invertible write \(f(z_i) = g(z_i) \implies g^{-1}f(z_i) = z_i\). So \(g^{-1}f\) has 3 fixed points so it must be the identity so \(f = g\)

Q: Proposition. Möbius maps send circles/lines to circles/lines. Alternatively, circles on the Riemann sphere are sent to circles on the Riemann sphere
A: Calculate it directly using \(w = \frac{az +b}{cz+d} \iff z = \frac{dw + b}{-cw +a}\) and substituting z into the circle equation gives \(A'w\overline{w} + \overline{B}'w + B'\overline{w} + C' = 0\)

Q: Theorem. Möbius maps preserve circles
A: Show that \(z \mapsto az, z \mapsto z + b, z \mapsto \frac{1}{z}\) all preserve circles.
Write \(S(A,B,C)\) as the circle that satifies \(Az\overline{z} + \overline{B}z + B\overline{z} + C = 0\).
Then just plug into the formula.

Q: Definition. If \(z_1, z_2, z_3, z_4 \in \mathbb{C}_{\infty}\) are distinct, the their cross ratio \([z_1,z_2,z_3,z_4]\) is defined the be \(f(z_4)\) where \(f \in M\) is the unique map such that \(f(z_1)=0, f(z_2)=1,f(z_2) = \infty\).
A: The formula is given by \[[z_1,z_2,z_3,z_4] = \frac{(z_4 - z_1)(z_2 - z_3)}{(z_2-z_1)(z_4-z_3)}\]

Q: Theorem. Möbius maps preserve the cross ratio \[[g(z_1),g(z_2),g(z_3),g(z_4)] = [z_1,z_2,z_3,z_4]\]
A: Let \(f \in M\)be the unique Mobius map such that \(f(z_1) = 0, f(z_2) = 1, f(z_3) = \infty\). Therefore \(f(z_4) = [z_1,z_2,z_3,z_4]\). Now consider \(f \circ g^{-1}\) \[(f \circ g^{-1})g(z_1) = 0; (f \circ g^{-1})g(z_2) = 1; (f \circ g^{-1})g(z_3) = \infty\]and \(f \circ g^{-1}\) is the unique map with this property so the cross ratio here is \[[g(z_1), g(z_2), g(z_3), g(z_4)] = (f \circ g^{-1})g(z_4) = f(z_4)\]

Q: Corollary. Four distinct points \(z_1,z_2,z_3,z_4 \in \mathbb{C}_{\infty}\) lie on a circle iff their cross-ratio is real.
A: Let f be the unique Möbius map sending \((z_1,z_2,z_3) \mapsto (0,1,\infty)\)so that \(f(z_4)\) is the required cross-ratio. The circle through \(z_1,z_2,z_3\) is sent by f to the unique circle passing through \(0,1,\infty\) which is the real line with the point at infinity. So \(z_4\) lies on C if and only if \(f(z_4)\) lies on the real line with the point at infinity. But \(z_3\) is mapped to infinity so \(z_4\) is not and this means it has to be real.