Title: Groups
Tags: G07 G09 G02 G11 G14 G41 G48 G51 G72 G13 G22 G27 G29 G32 G39 G83 G00 G01 G06 G08 G16 G19 G21 G23 G33 G37 G42 G47 G52 G56

Q: Group axioms
A: A \emph{group} is a set $G$ with a binary operation $*$ satisfying the following axioms:
  \begin{enumerate}
      \setcounter{enumi}{-1}
    \item $\forall a, b :\, a*b\in G$ \hfill (closure)
    \item $\exists e:\forall a: \, a*e = e*a = a$ \hfill (identity)
    \item $\forall a: \exists a^{-1}: \, a*a^{-1} = a^{-1}*a = e$ \hfill (inverse)
    \item $\forall a, b, c: \, (a*b)*c = a*(b*c)$ \hfill (associativity)
  \end{enumerate}

Q: Abelian group
A: A group which also satisfies
\begin{enumerate}
      \setcounter{enumi}{+3}
    \item $\forall a, b :\, a*b=b*a$ \hfill (commutativity)
  \end{enumerate}

Q: Prove the identity is unique
A: Suppose $e$ and $e'$ are identities in $G$. Therefore
\begin{align*}
e*e'&=e' \quad\textrm{as $e$ is an identity} \\
&=e\quad\textrm{as $e'$ is an identity} \\
\implies e&=e' \qed
\end{align*}

Q: Prove that inverses are unique
A: Suppose $b$ and $a^{-1}$ are inverses of $a$. Therefore
\begin{align*}
b &=be \quad &\textrm{Axiom 1} \\
&=b (aa^{-1}) \quad &\textrm{Axiom 2}\\
&=(ba)a^{-1} \quad &\textrm{Axiom 3}\\
&=ea^{-1} \quad &\textrm{Axiom 2}\\
&=a^{-1} \quad &\textrm{Axiom 1}
\end{align*}\qed

Q: Prove $(a^{-1})^{-1}=a$
A: Observe that $a$ and $(a^{-1})^{-1}$ are both inverses of $a^{-1}$. So by uniqueness of inverses: \[
(a^{-1})^{-1}=a \qed
\]

Q: Socks and Shoes Theorem
A: We have
      \begin{align*}
        (ab)(b^{-1}a^{-1}) &= a(bb^{-1})a^{-1} \\
        &= aea^{-1}\\
        &= aa^{-1}\\
        &= e
      \end{align*}
and
\begin{align*}
        (b^{-1}a^{-1})ab &= b^{-1} (a^{-1}ab) \\
        &= b^{-1} eb \\
        &= b^{-1}b\\
        &= e
      \end{align*}
So $b^{-1}a^{-1}$ is an inverse of $ab$. By the uniqueness of inverses: \[
 (ab)^{-1} = b^{-1}a^{-1} \qed
\]

Q: Order of a group
A: The number of elements in a finite group $G$, written $|G|$.

Q: Subgroup
A: $H$ is a \emph{subgroup} of $G$, written $H\leq G$, if $H\subseteq G$ and $H$, with the restricted operation $*$ from $G$, is also a group.

Q: Proper Subgroup
A: If $H\leqslant G$ and $H\neq G$ we call $H$ a \emph{proper} subgroup of $G$.

Q: Usual subgroup criterion
A: 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) $(\forall a \in H): \,a^{-1} \in H$
\begin{proof}
First note that associativity is inherited from $G$ so we don't have to check it. Also, clearly (ii) iff. Axiom 0 and (iii) iff Axiom 2 by uniqueness of inverses. \\
\\
Clearly, (i)$\implies$ Axiom 1 but the other direction is trickier. We need to show that $H$ and $G$ must have the same identity. Suppose that $e_H$ 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 $e_H = e_G$.
\end{proof}

Q: Super-efficient subgroup criterion
A: A subset $H\subseteq G$ is a subgroup of $G$ iff: \\
(I) $H$ is non-empty \\
(II) $(\forall a, b\in H):ab^{-1}\in H$
\begin{proof}
``$\implies$'' If $H$ is a subgroup, then $e\in H \implies $ (I). Closure under inverses and closure under multiplication $\implies$ (II). \\
``$\impliedby$'' We need to show (I) and (II) imply the usual subgroup criterion. By (I), $\exists h\in H$ so by (II) on $h,h$: $hh^{-1}\in H$ so $e\in H$. \\
Using (II) on $e,a$ we get $\forall a\in H : ea^{-1}\in H$ so $\forall a: a^{-1}\in H$. \\
Therefore, using (II) on $a,b^{-1}$: $\forall a,b: a(b^{-1})^{-1}=ab \in H$.
\end{proof}

Q: Prove that the subgroups of $\left( \mathbb{Z},+\right)$ are exactly $n\mathbb{Z}$ for all $n\in\mathbb{Z}$
A: Suppose $n\mathbb{Z}$ is a subgroup. $0\in n\mathbb{Z}$ and $a+b^{-1}=na'-nb'=n(a'-b')\in n\mathbb{Z}$. So, by the super-efficient subgroup criterion, $n\mathbb{Z}\leqslant\mathbb{Z}$. \\
\\
Conversely, suppose $H\leqslant\mathbb{Z}$. By the usual subgroup criterion, $0\in H$. If $H=\{ 0\}$, then it is $0\mathbb{Z}$. Otherwise, let $n$ be the smallest positive integer in $H$. We show $H=n\mathbb{Z}$. Suppose $a\in H: n\nmid a$ . By the division algorithm, $a=nk+a' (0<a'<n)$. But as $H$ is a subgroup, $nk=n+n+\ldots n\in H$ so $a-nk=a' \in H$, contradicting that $n$ is the smallest positive integer. Therefore, $\forall a\in H: n\mid a$. So $H=n\mathbb{Z}$. \qed

Q: Group Homomorphism
A: Let $(G, *_G)$ and $(H,*_H)$ be groups. A function $f:G\rightarrow H$ is a \emph{group homomorphism} iff
  \[
   ( \forall g_1, g_2 \in G): f(g_1)*_H f(g_2) = f(g_1 *_G g_2),
  \]

Q: Isomorphism
A: A homomorphism that is also a bijective function. If two groups $G,H$ are isomorphic we write $G\cong H$.

Q: Prove that group homomorphisms send identity to identity
A: Suppose $f:G\rightarrow H$ is a group homomorphism. Therefore
\begin{align*}
f(e_G)=f(e_Ge_G)&=f(e_G)f(e_G) \\
\implies f(e_G)^{-1}f(e_G)&=f(e_G)^{-1}f(e_G)f(e_G) \\
\implies e_H&=f(e_G) \qed
\end{align*}

Q: Prove that group homomorphisms send inverses to inverses
A: Suppose $f:G\rightarrow H$ is a group homomorphism. Therefore, for any $a\in G$:
\begin{align*}
f(a^{-1})f(a)=f(a^{-1}a)=f(e_G)=e_H \\
f(a)f(a^{-1})=f(aa^{-1})=f(e_G)=e_H
\end{align*}
as we have already shown that group homomorphisms send identity to identity. Therefore, $f(a^{-1})$ is the inverse of $f(a)$ (by uniqueness of inverses) in $H$. So \[
f(a^{-1})=f(a)^{-1}
\]

Q: Prove that the composite of two group homomorphisms is a group homomorphism
A: Let $f:G_1 \rightarrow G_2$ and $g:G_2 \rightarrow G_3$. Then $\forall a,b \in G_1$:
\begin{align*}
g\circ  f (a*_1b) &=g(f(a*_1b)) \\
&= g(f(a)*_2f(b))\\
&= g(f(a))*_3g(f(b))\\
&= g\circ f(a)*_3g\circ f(b)
\end{align*}
So $g\circ  f: G_1\rightarrow G_3$ is a group homomorphism.

Q: Prove that the inverse of an isomorphism is an isomorphism
A: Let $f:G \rightarrow H$ be an isomorphism. So $f^{-1}$ exists. Then all members of $H$ can be written as $f(g)$ for some $g$. Thus
      \begin{align*}
        f^{-1}\left[f(a)f(b)\right]&=f^{-1}\left[f(ab)\right] \\
&=ab \\
&=f^{-1}[f(a)]f^{-1}[f(b)]
      \end{align*}
      So $f^{-1}$ is a homomorphism. Since it is bijective, $f^{-1}$ is an isomorphism.

Q: Image of a group homomorphism
A: If $f:G\rightarrow H$ is a group homomorphism then
\begin{align*}
\mathrm{Im}f&=\{ b\in H \ \vert \ \exists a \in G: f(a)=b \} \\
&=``f(G)'' \\
&=\{ f(a) \ \vert \ a\in G \}
\end{align*}

Q: Kernel (or preimage) of a group homomorphism
A: If $f:G\rightarrow H$ is a group homomorphism then
\begin{align*}
\mathrm{Ker}f&=\{ a\in G \ \vert \ f(a)=e_H \} \\
&=``f^{-1}(e_H)'' \\
\end{align*}

Q: Prove that kernels are subgroups
A: Let $f:G\to H$ be a group homomorphism. \\
\\
We know that group homomorphisms map identity to identity, so $e_G\in$ Ker$f$. Thus Ker$f$ is not empty. \\
\\
Given $a,b\in$ Ker$f$:
\begin{align*}
f(ab^{-1})&=f(a)f(b) \\
&=e_He_H \\
&=e_H \\
\implies ab^{-1}\in\textrm{Ker}f
\end{align*}
Therefore, by the super-efficient subgroup criterion: Ker$f\leq G$.

Q: Prove that images are subgroups
A: Let $f:G\to H$ be a group homomorphism. \\
\\
We know that group homomorphisms map identity to identity, so $f(e_G)=e_H$ i.e. $e_H\in$ Im$f$. Thus Im$f$ is not empty. \\
\\
Given $h_1,h_2\in$ Im$f$, there exist $g_1,g_2\in G$ such that $f(g_1)=h_1$,$f(g_2)=h_2$. So
\begin{align*}
f(g_1g_2^{-1})&=f(g_1)f(g_2^{-1}) \\
&=f(g_1)f(g_2)^{-1}\\
&=h_1h_2^{-1} \\
&\in\textrm{Im}f
\end{align*}
Therefore, by the super-efficient subgroup criterion: Im$f\leq H$.

Q: Relationship between surjectivity and image of a group homomorphism
A: A group homomorphism $f:G\to H$ is surjective iff. Im$f=H$.

Q: Relationship between injectivity and kernel of a group homomorphism
A: A group homomorphism $f:G\to H$ is injective iff. Ker$f=\{e_G\}$.
\begin{proof}
Suppose $f:G\to H$ is an injective group homomorphism. So given $k\in$ Ker$f$ then
\begin{align*}
f(k)&=e_H \\
&=f(e_G) \\
\implies k&=e_G \quad\textrm{(by injectivity)}
\end{align*}
Thus Ker$f=\{e_G\}$. \\
\\
Suppose  Ker$f=\{e_G\}$. If there exist $a,b$ such that $f(a)=f(b)$ then
\begin{align*}
f(ab^{-1})&=f(a)f(b)^{-1} \\
&=f(a)f(a)^{-1} \\
&=e_H \\
\implies ab^{-1} &\in\textrm{ Ker}f \\
\implies ab^{-1} &=e_G \quad(\textrm{only element of Ker}f)\\
\implies a&=b
\end{align*}
So $f$ is injective.
\end{proof}

Q: Cyclic Group
A: A group $(G,*)$ is \emph{cyclic} if there is an $a\in G$ (called the generator) such that ell elements of $G$ are powers of $a$. \\
\\
If $|G|=n$, then we write $G\cong C_n$.

Q: $\langle a \rangle$
A: The cyclic subgroup generated by $a$. Defined as the smallest subgroup containing $a$.

Q: Prove that $\langle a \rangle $ is exactly the set of all powers of $a$
A: Consider the set of all powers of $a$. $e=a^0$ so the set is non-empty. Given $a^m$ and $a^n$, $a^m(a^n)^{-1}=a^{m-n}$ is also a power of $a$. So by the super-efficient subgroup criterion, the set of all powers of $a$ is a subgroup. \\
\\
By closure under multiplication and inverses, we can see by induction that the set of all powers of $a$ is a subgroup of any subgroup containing $a$. Therefore, set of all powers of $a$ is the smallest subgroup containing $a$. \qed

Q: Order of an element
A: The order of an element $a\in G$ is the smallest $k$ such that \[
a^k=e.
\]
If no such $k$ exists, then $a$ has infinite order.

Q: Prove ord$(a)=|\langle a\rangle |$
A: Infinite case:
\begin{align*}
&\mathrm{ord}(a)=\infty \\
\iff &\forall n: a^n\neq e \\
\iff  &\forall n\neq m: a^m\neq a^n \\
\iff  &|\langle a \rangle | =\infty
\end{align*}
Finite case:
\begin{align*}
&\mathrm{ord}(a)=k \in\mathbb{Z}^+\\
\iff & a^n=e \textrm{ has least positive solution }k  \\
\iff  &\langle a \rangle  =\lbrace e,a,a^2\ldots a^{k-1}\rbrace \\
\iff  &|\langle a \rangle | =k
\end{align*}

Q: Prove that cyclic groups are abelian
A: Given $a^m,a^n \in C_k$:
\[ a^ma^n=a^{m+n}=a^na^m \]
by associativity. So $C_k$ is abelian. \qed

Q: Prove that any subgroup of a cyclic group is cyclic
A: We are given $G\leqslant C_n$. If $G=\lbrace e \rbrace$ then $G\cong C_1$. \\
\\
Otherwise, $G$ has an element with smallest positive power $a^x$. We show that $a^x$ generates $G$. Suppose we are given $a^p \in G$. By the division algorithm, $p=kx+r$ where $0\leqslant r < x$. Then $a^pa^{-kx}=a^r\in G$ (by closure under inverses and multiplication). So as $a^x$ is the smallest positive integer power, $r=0$ and $a^p=(a^x)^k$. Therefore, $G=\langle a^x \rangle$. \qed

Q: Dihedral group
A: \begin{align*}
D_{2n}&=\langle r,s \vert r^n=e=s^2,srs^-1=r\rangle \\
&=\lbrace e,r,r^2,\ldots , r^{n-1},s,rs,r^2s,\ldots , r^{n-1}s\rbrace
\end{align*}
Note that $sr^k=r^{-k}s=r^{n-k}s$.

Q: ord($r^ks$)
A: \begin{align*}
(r^ks)(r^ks) &=r^ksr^ks \\
&=r^kr^{-1}sr^{k-1}s \\
&=r^{k-1}sr^{k-1}s \\
\ldots &=r^0sr^0 s\\
&=s^2 \\
&=e
\end{align*}
Thus ord($r^ks$)=2.

Q: Cartesian product of groups
A: Given two groups $(G, *_G)$ and $(H, *_H)$, we can define a set \[
G\times H = \{(g, h): g\in G, h\in H\}
\]
 and an operation \[
(a_1, a_2)*(b_1, b_2) = (a_1*_Gb_1, a_2*_H b_2).
\]

Q: When is $C_m\times C_n$ cyclic?
A: \[ C_m\times C_n =C_{mn} \iff \mathrm{hcf}(m,n)=1 \]
\begin{proof}
Let $C_n = \langle a \ \vert \ a^n=e\rangle$ and $C_m = \langle b \ \vert \ b^m=e \rangle$. Note that \[
(a,b)^k=(a^k,b^k)
\]
Therefore
\begin{align*}
(a,b)^k=e &\iff a^k=e,b^k=e \\
 &\iff k=\mathrm{lcm}(m,n)
\end{align*}
Thus the greatest order element of $C_m\times C_n$ has order lcm($m,n$). Given that $|C_m\times C_n|=mn$, $C_m\times C_n$ is cylic iff.
\begin{align*}
& \mathrm{lcm}(m,n)=mn \\
\iff &\mathrm{hcf}(m,n)=1 \\
\iff &C_m\times C_n=C_{mn}
\end{align*}
\end{proof}

Q: Direct Product Theorem
A: Let $H_1, H_2\leq G$. If
  \begin{enumerate}
    \item $H_1\cap H_2 = \{e\}$
    \item $(\forall a_i\in H_i): a_1a_2=a_2a_1$
    \item $(\forall a\in G)(\exists a_i\in H_i) : a = a_1a_2$. We also write this as $G=H_1H_2$.
  \end{enumerate}
  Then $G\cong H_1\times H_2$.
\begin{proof}
  We will prove there is an isomorphism constructively, defining $f:H_1\times H_2\rightarrow G$ by $f(a_1, a_2) = a_1a_2$. This is a homomorphism since
  \begin{align*}
    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).
  \end{align*}
By (3), $f$ is surjective. We'll show injectivity by showing that the kernel is $\{e\}$.
\begin{align*}
\textrm{Given } &f(a_1, a_2)=e \\
\implies &a_1a_2 = e\\
\implies &a_1=a_2^{-1} \\
\implies &a_1=a_2=e \textrm{ by (1)} \\
\implies &\mathrm{Ker}f=\lbrace e \rbrace
\end{align*}
\end{proof}

Q: Permutation
A: A bijection from a set to itself.

Q: Sym$X$
A: The set of all permutations of $X$.

Q: Prove that Sym$X$ forms a group under composition.
A: Axiom 0. The composite of bijections is bijective. Thus if $\sigma: X\to X$, $\tau: X\to X$, then $\sigma\circ\tau:X\to X \in$ Sym$X$. \\
Axiom 1. The identity $1_X:X\to X$ is clearly a permutation which acts as the identity. \\
Axiom 2. Every bijection has an inverse which is also bijective. So given $\sigma\in$ Sym$X$, there exists $\sigma^{-1}:X\to X$. \\
Axiom 3. Composition of functions is always associative.

Q: Symmetric group
A: $S_n=$ Sym$X$ where $X=\lbrace 1,2,3,\ldots  ,n \rbrace$.

Q: Two row notation
A: If $\sigma\in S_n$, then we write \[
\begin{bmatrix}
1 & 2 & \ldots & n \\
\sigma(1) & \sigma(2) & \ldots & \sigma(n)
\end{bmatrix}
\]

Q: $|S_n|$
A: Imagine a permutation in two row notation. Consider the second row. There are $n$ possibilities for the first element, leaving $n-1$ for the second etc. Therefore
\[ |S_n|=n! \]

Q: Cycle notation
A: We call \[
(a_1a_2\ldots a_k)
\]
a $k$-cycle. It means \[
\begin{bmatrix}
a_1 & a_2 &\ldots & a_k \\
a_2 & a_3 &\ldots & a_1
\end{bmatrix}.
\]
Note 2-cycles are called transpositions. 1-cycles are implied.

Q: Prove that disjoint cycles commute
A: Let $\sigma, \tau \in S_n$ be disjoint cycles. Let $a \in \lbrace 1,2,\ldots n \rbrace$. As $\sigma,\tau$ are disjoint, $a$ cannot appear in both cycles. So wlog let $a \notin \tau$. Therefore\[
\sigma(\tau(a))=\sigma(a)=\tau(\sigma(a))
\]
\qed

Q: Prove disjoint cycle notation works
A: For $n\geqslant 2$, every permutation in $S_n$ can be written (essentially uniquely i.e. excluding moving and rotating cycles) as a product of disjoint cycles.
\begin{proof}
Let $\sigma\in S_n$. Start by enumerating forever the cycle containing 1 \[
(1\; \sigma(1)\; \sigma^2(1)\; \sigma^3(1)\ldots).
\]
As the set $\{1, 2, 3\ldots n\}$ is finite, for some $k$, we must have $\sigma^k(1)$ is a repeat of an element already in the list. Say wlog $\sigma^k(1) = \sigma^l(1)$, with $l < k$, so $\sigma^{k-l}(1) = 1$. Thus all $\sigma^i(1)$ are distinct until we get back to $1$. Thus we have the first cycle $(1\; \sigma(1)\; \sigma^2(1)\; \sigma^3(1)\;\cdots\;\sigma^{k-1}(1))$. \\
  Now choose the smallest number that is not yet in a cycle, say $j$. Repeat to obtain a cycle \[
(j\; \sigma(j)\; \sigma^2(j)\;\cdots\; \sigma^{l - 1}(j)).
\]
Since $\sigma$ is a bijection, nothing in this cycle can be in previous cycles as well.\\
\\
 Repeat until all $\{1, 2, 3\cdots n\}$ 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.
\end{proof}

Q: Cycle types
A: The list of cycle lengths when written in disjoint cycle notations.

Q: Relationship between cycle type and order of the permutation
A: The order of a permutation is the least common multiple of the cycle lengths.
\begin{proof}
As disjoint cycles commute, we can group together each cycle when we take powers, ie. if $\sigma = \tau_1\tau_2\cdots\tau_l$ with $\tau_i$ all disjoint cycles, then $\sigma^m = \tau_1^m\tau_2^m\cdots\tau_l^m$.
  Now if cycle $\tau_i$ has length $k_i$, then $\tau_i^{k_i} = e$, and $\tau_i^m = e$ iff $k_i \mid m$. To get an $m$ such that $\sigma^m = e$, we need all $k_i$ to divide $m$, ie. $m$ is a common multiple of $k_i$. Therefore the order (i.e. the smallest such $m$) is the least common multiple of $k_i$.
\end{proof}

Q: Prove that $S_n$ is generated by transpositions
A: By disjoint cycle notation, it is sufficient to show that a cycle is a product of transpositions. \[
(a_1a_2\ldots a_k)=(a_1a_2)(a_2a_3)\ldots(a_{k-1}a_k)
\]
which is a product of $k-1$ transpositions. \qed

Q: Prove that a permutation is either always the product of an even number of transpositions or always the product of an odd number of transpositions
A: Write $\#(\sigma)$ for the number of cycles in disjoint cycle notation, including singleton cycles. We multiply $\sigma$ by a transposition $\tau = (cd)$ [wlog assume $c < d$] \\
    Case 1: $c, d$ are in the same cycle (in $\sigma$), say
\begin{align*}
(c a_2 \cdots  a_{k - 1}d a_{k + 1} \ldots a_{k + l}) (cd) \\
= (c a_{k+1} a_{k+2}\ldots a_{k + l})(da_2a_3\ldots a_{k - 1}) \\
\implies \#(\sigma\tau) = \#(\sigma) + 1
\end{align*}
    Case 2: $c, d$ are in different $\sigma$-cycles, say
\begin{align*}
(d, a_2\; a_3,\ldots\;a_{k - 1})(c, a_{k + 1}\; a_{k + 2}\;\ldots\; a_{k + l})(c\; d)  \\
=(c, a_2, \ldots \; a_{k - 1}, d, a_{k + 1}, \ldots a_{k + l}) \\
\implies\#(\sigma\tau) = \#(\sigma) - 1
\end{align*}
Therefore (in either case) for any transposition $\tau$, $\#(\sigma\tau) \equiv \#(\sigma) + 1 \mod 2$.
  Now suppose $\sigma = \tau_1\ldots\tau_l = \tau_1'\ldots\tau_{l'}'$. Since disjoint cycle notation is essentially unique, $\#(\sigma)$ is uniquely determined by $\sigma$. But $\sigma=e\tau_1\ldots\tau_l$ so
\begin{align*}
\#(\sigma) &\equiv \#(e) + l\pmod 2 \\
\#(\sigma) &\equiv \#(e) + k\pmod 2
\end{align*}
So $l \equiv k \pmod 2$. \qed

Q: Sign of a permutation
A: Given $\sigma\in S_n$ which can be written as a product of transpositions \[
\sigma=\tau_1\tau_2\ldots\tau_k,
\]
then sgn($\sigma)=(-1)^k$ and we say \[
\begin{cases}
\textrm{even if sgn}(\sigma)=1 \\
\textrm{odd if sgn}(\sigma)=-1
\end{cases}
\]

Q: sgn is a surjective group homomorphism ($n\geqslant 2$)
A: Given $\sigma_1=\tau_1\ldots\tau_l$ and $\sigma_2=\tau_1'\ldots\tau_{l'}'$
\begin{align*}
\mathrm{sgn}(\sigma_1\sigma_2)=(-1)^{l+l'} \\
&=(-1)^l(-1)^{l'} \\
&=\mathrm{sgn}(\sigma_1)\mathrm{sgn}(\sigma_2)
\end{align*}
So sgn is a homomorphism. \\
\\
sgn$(e)=1$, sgn$(12)=-1$. Thus sgn is surjective for $n\geqslant 2$.

Q: A permutation is even 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, $\sigma = \sigma_1\sigma_2\ldots\sigma_l$, we get sgn$(\sigma) = $sgn$(\sigma_1)\ldots $sgn$(\sigma_l)$. Suppose there are $m$ even-length cycles and $n$ odd-length cycles, then sgn$(\sigma) = (-1)^m 1^n$. This is equal to $1$ iff $(-1)^m = 1$, ie. $m$ is even. \qed

Q: Alternating group
A: $A_n=$Ker(sgn) i.e. the set of even permutations

Q: Prove that 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 permutations by $\sigma \mapsto \sigma\tau$ (bijection since $\sigma \mapsto \sigma \tau^{-1}$ is a well-defined inverse). So there are as many odd permutations as even permutations.

Q: Coset
A: Let $H\leqslant G$ and $a\in G$. Then the set \[
aH=\lbrace ah \ \vert \ h \in H \rbrace
\]
is the left coset of $H$ and \[
Ha=\lbrace ha \ \vert \ h \in H \rbrace
\]
is the right coset of $H$.

Q: Partition
A: Let $X$ be a set, and $X_1, \ldots X_n$ be subsets of $X$. The $X_i$ are called a \emph{partition} of $X$ if \[
\bigcup_{i=1}^n X_i = X \]
and \[
X_i\cap X_j = \emptyset \]
 for $i\neq j$.  \\
In other words, the $X_i$ span the whole of $X$ and are pairwise disjoint. \\
\\
Alternatively: every element of $X$ is in exactly one of the $X_i$.

Q: Lagrange's Theorem
A: Given finite groups $G,H$ such that $H\leqslant G$, then $|H|$ divides $|G|$.
\begin{proof}
Step 1. We show that the left cosets of $H$ partition $G$.
\begin{itemize}
\item[Span] $\forall a \in G: a\in a H$ so every element of $G$ is in at least one of the left cosets. \\
\item[Disjoint] Given $aH,bH$ which coincide, then there exists $h_1, h_2 \in H$ such that
\begin{align*}
ah_1&=bh_2 \\
\implies a&=bh_2h_1^{-1} \\
\implies \forall h \in H: ah&=bh_2h_1^{-1} \\
&\in bH \\
\implies aH&\subseteq bH \\
\implies bH&\subseteq aH \textrm{ (by symmetry)} \\
\implies aH&=bH
\end{align*}
Therefore, distinct cosets are disjoint.
\end{itemize}
Step 2. We prove that any left coset has the same size as H. Given $a\in G$, we construct a bijection $H \longleftrightarrow aH$ by
\begin{align*}
h &\longmapsto ah \\
b &\longmapsto a^{-1}b
\end{align*}
for all $h\in H$ and $b\in aH$. These are well-defined functions, and mutually inverse. \\
Step 3. We have shown that $G$ is the disjoint union of distinct cosets, each with size $|H|$. Therefore, $|G|=$number of cosets $\times |H|$.
\end{proof}

Q: Index
A: The index of $H$ in $G$, written $|G:H|$ is the number of distinct cosets of $H$ in $G$. Therefore \[
|G|=|G:H||H|
\]

Q: How do you determine that two cosets are the same?
A: \[
aH=bH\iff b^{-1}a\in H(\iff a^{-1}b\in H)
\]
Suppose $aH=bH$. So
\begin{align*}
ah_1&=bh_2 \\
\implies b^{-1}a&=h_2h_1^{-1} \\
&\in H
\end{align*}
Conversely, suppose $b^{-1}a\in H$. So
\begin{align*}
b^{-1}a&=h\in H \\
\implies a&=bh \\
\implies ae&=bh \\
&\in bH
\end{align*}
Therefore $aH$ and $bH$ coincide, so $aH=bH$ (as cosets partition). \\
\\
Finally, note that by symmetry (or by closure under inverses with shoes and socks) we also have \[
aH=bH\iff a^{-1}b\in H  \qed
\]

Q: Show that element order divides group order
A: Let $G$ be a finite group with $a\in G$. Then ord$(a)$ divides $|G|$.
\begin{proof}
$\langle a \rangle \leq G$ with $|\langle a \rangle|=$ ord($a$). So by Lagrange \[
\mathrm{ord}(a) \textrm{ divides } |G|
\]
\end{proof}

Q: Show that exponent divides group order
A: Let $G$ be a finite group with $a\in G$ and exponent $p$, then $a^{|G|}=e$.
Let $|G|=k \ \cdot \ $ord($a$). So
\begin{align*}
a^{|G|} &=a^{k \ \cdot \ \mathrm{ord}(a)} \\
&=\left(a^{\mathrm{ord}(a)}\right)^k \\
&=e^k \\
&=e
\end{align*}
Therefore, letting $|G|=pq+r$ ($0\leqslant r \leqslant n-1$)
\begin{align*}
a^{|G|}&=a^{pq+r} \\
&=\left(a^p\right)^qa^r \\
&=e^qa^r \\
&=a^r \\
&=e
\end{align*}
But $p$ is the least positive power for which all elements are the identity, so $r=0$. Therefore \[
p \textrm{ divides } |G| \qed
\]

Q: Show that groups of prime order are cyclic and are generated by any of their non-identity elements.
A: Say $|G|=p$. Let $a\in G: a\neq e$. \\
\\
So $|\langle a \rangle |>1$ and $|\langle a \rangle |$ divides $p$ by Lagrange. Thus $|\langle a \rangle|=p$ and $\langle a \rangle=G$. \\
\\
Therefore, $G\cong C_p$ and is generated by any non-identity element.

Q: Equivalence relation
A: An equivalence relation on a set $X$ is a relation $\sim$ with the properties:
\begin{enumerate}
\item $\forall x \in X: x\sim x$ (reflexive)
\item $\forall x,y \in X: x\sim y \implies y\sim x$ (symmetric)
\item $\forall x,y,z \in X: x\sim y, y\sim z \implies x\sim z$ (transitive)
\end{enumerate}

Q: Equivalence class
A: Given an equivalence relation $\sim$ on $X$ \[
[x]_\sim=[x]=\lbrace y\in x \ | \ x\sim y \rbrace
\]

Q: Prove that equivalence classes partion
A: Let $\sim$ be an equivalence relation on $X$. \\
\\
The equivelance classes span $X$ as $\forall x \in X: x\in [x]$ by reflexivity. \\
\\
Suppose $z\in [x]$ and $z\in[y]$. Then
\begin{align*}
&z\sim x, z\sim y \\
\implies& x\sim z, z\sim y \textrm{ (by symmetry)} \\
\implies& x\sim y \textrm{ (by transitivity)} \\
\therefore \forall w: \ & y\sim w \implies  x\sim w\textrm{ (by transitivity)} \\
\implies& [y]\subseteq[x] \\
\implies& [x]\subseteq[y] \\
\implies& [x]=[y]
\end{align*}
Thus the distinct equivalence classes are pairwise disjoint. \qed

Q: Prove that cosets are equivalence classes
A: Let $H\leqslant G$ and $a\sim b \iff b^{-1}a\in H$ (i.e. $a,b$ are in the same coset). \\
\\
Reflexivity: $\forall a: a^{-1}a=e \in H$ so $a\sim a$. \\
Symmetry: If $b^{-1}a\in H$, then by closure under inverses and shoes \& socks: $(b^{-1}a)^{-1}=a^{-1}b \in H$. \\
Transitivity: If $a\sim b$, $b\sim c$, then
\begin{align*}
&b^{-1}a\in H, c^{-1}b\in H \\
\implies& c^{-1}bb^{-1}a \in H \textrm{ (by closure under multiplication)} \\
\implies& c^{-1}a\in H \\
\implies& a\sim c
\end{align*}

Q: Representatives
A: Given cosets $a_1H,a_2H,\ldots a_nH$ we call $a_1,a_2,\ldots a_n$ the representatives of the cosets.

Q: Defining modular arithmetic as a group under addition
A: Consider the cosets of  $(n\mathbb{Z},+)\leqslant (\mathbb{Z},+)$ as equivalence classes. Then modular arithmetic is defined by \[
[a]+[b]=[a+b]\]
We just need to check that regardless of which representative you choose, you will get the same answer for $[a]+[b]$. If $[a_1]=[a_2]$, $[b_1]=[b_2]$, then
\begin{align*}
&a_1=a_2+kn, \\
&b_1=b_2+ln \\
\implies &a_1+b_1=a_2+b_2+n(k+l) \\
\implies  &[a_1+b_1]=[a_2+b_2]
\end{align*}

Q: Defining modular arithmetic as a group under multiplication
A: Let $U_n=\lbrace [a] \ | \ (a,n)=1\rbrace$. We define \[
[a]\cdot[b]=[ab]\]
We first need to check that regardless of which representative you choose, you will get the same answer. If $[a_1]=[a_2]$, $[b_1]=[b_2]$, then
\begin{align*}
&a_1=a_2+kn, \\
&b_1=b_2+ln \\
\implies &a_1\cdot b_1=a_2b_2+n(kb_2+la_2+kln) \\
\implies  &[a_1b_1]=[a_2b_2]
\end{align*}
Now we show that $(U_n,\cdot)$ is a group. \\
\textbf{Closure}: $(a,n)=1,(b,n)=1 \implies (ab,n)=1 $ so $[a],[b] \in U_n \implies [a]\cdot[b]=[ab]\in U_n$. \\
\textbf{Identity}: $[a][1]=[a]=[1][a]$ \\
\textbf{Inverses}: Let $[a]\in U_n$. We show that the map `multiplication by a' ($U_n\to U_n$ by $[c]\longmapsto [ac]$) is surjective, so hits $[1]$. First, it is injective:
\begin{align*}
&[ac_1]=[ac_2] \\
\implies& n | a(c_1-c_2) \\
\implies& n| (c_1-c_2) \ (\because(a,n)=1) \\
\implies& [c_1]=[c_2]
\end{align*}
$U_n$ is finite, so this is also surjective. \\
\textbf{Associativity} (and commutativity): inherited \qed

Q: Fermat-Euler Theorem via groups
A: Let $n\in \mathbb{N}$ and $a\in \mathbb{Z}$:  $(a,n)=1$. Then
  \[
    a^{\varphi(n)} \equiv 1\pmod n.
  \]
\begin{proof}
  As $a$ is coprime with $n$, $[a]\in U_n$. Then, as exponent divides group order: \[
[a]^{|U_n|} = [1],
\]
i.e. \[
a^{\varphi(n)} \equiv 1\pmod n.
\]
\end{proof}

Q: Unique groups of order 4
A: A group of order 4 is isomorphic to either $C_4$ or $C_2\times C_2$.
\begin{proof}
  Let $|G| = 4$. By Lagrange theorem, possible element orders are $1$ ($e$ only), $2$ and $4$. If there is an element $a\in G$ of order $4$, then $G = \langle a\rangle \cong C_4$. \\
\\
  Otherwise all non-identity elements have order 2. Then $G$ must be abelian.
  Pick $2$ elements of order 2, say $b, c\in G$, then $\langle b\rangle=\{e, b\}$ and $\langle c\rangle= \{e, c\}$. So $\langle b\rangle\cap \langle c\rangle= \{e\}$. As $G$ is abelian, $\langle b\rangle$ and $\langle c\rangle$ commute. We know that $bc = cb$ has order 2 as well, and is the only element of $G$ left. So $G \cong \langle b\rangle\times \langle c\rangle \cong C_2\times C_2$ by the direct product theorem.
\end{proof}

Q: Normal subgroup
A: A subgroup $K$ of $G$ is a \emph{normal} subgroup, written $H\trianglelefteq G$, if $\forall a\in G, k\in K$:
\begin{align*}
&aka^{-1}\in K \\
\iff &aK=Ka \\
\iff &aKa^{-1}=K
\end{align*}

Q: Show that Kernels are normal subgroups
A: Let $f:G\to H$ be a group homomorphism with Ker$f=K$. So for $k\in K$ and $a\in G$:
\begin{align*}
f(aka^{-1})&=f(a)f(k)f(a)^{-1} \\
&=f(a)f(a)^{-1} \\
&=e \\
\implies &aka^{-1}\in K \\
\implies &K\trianglelefteq G
\end{align*}

Q: Groups of order 6
A: Every group of order 6 is either cyclic or dihedral
\begin{proof}
  Let $|G| = 6$. By Lagrange's theorem, possible element orders are $1, 2, 3$ and $6$. If there is an $a\in G$ of order $6$, then $G = \langle a\rangle \cong C_6$.  \\
\\
Otherwise, we can only have non-identity elements of orders $2$ and $3$. If $G$ only has elements of order 2, the order must be a power of 2, which is not the case. So there must 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 $2$ as $6$ is even. Since $\langle r\rangle$ is normal, we know that $srs^{-1}\in \langle r\rangle$. \\
\\
Case 1. If $srs^{-1} = e$, then $r = e$, which is not true.  \\
Case 2. If $srs^{-1} = r$, then $sr = rs$ and $sr$ has order $6$ (lcm of the orders of $s$ and $r$), which was ruled out above.  \\
Case 3. If $srs^{-1} = r^2 = r^{-1}$, then $G\cong D_6$.
\end{proof}

Q: Show that any subgroup of index $2$ is normal
A: If $K\leq G$ has index $2$, then there are only two possible cosets $K$ and $G\setminus K$. As $eK = Ke$ and cosets partition $G$, the other left coset and right coset must be $G\setminus K$. Note that $aK=K=Ka\iff a\in K$. So all left cosets and right cosets are the same (either both $K$ or both $G\setminus K$).

Q: Show that any subgroup of an abelian group is normal
A: If $a,\in G$ and $k\in K\leqslant G$, where $G$ is abelian, then  \[
ak=ka
\]
so \[
aK=Ka
\]
i.e. $K\unlhd G$.

Q: Show that if $G$ is a finite group with all non-identity elements order 2, then $G$ is abelian and $|G|=2^n$ for some $n$.
A: First note that order 2 elements are exactly the self-inverse elements: \[
a^2=e\iff a=a^{-1}
\]
Given $a,b \in G$, $ab\in G$ by closure so
\begin{align*}
ab&=(ab)^{-1} \\
&=b^{-1}a^{-1} \\
&=ba
\end{align*}
Let $S=\lbrace a_1,a_2\ldots a_n \rbrace$ be the minimal generating set of $G$. We show that there is a bijection between the subsets of $S$ and $G$ by simply multiplying the elements of the subset. It is clearly surjective by as all the $a_i$ are order 2 (so higher powers not needed) and form a minimal generating set by assumption. Now we show it is injective. Suppose \[
a_1^{p_1}a_2^{p_2}\ldots a_n^{p_n}=a_1^{q_1}a_2^{q_2}\ldots a_n^{q_n}
\]
where wlog $p_1=1,q_1=0$. Then (as $G$ is abelian) $a_1=a_2^{q_2-p_2}\ldots a_n^{q_n-p_n}$ so $\lbrace a_2 \ldots a_n \rbrace$ form a generating set. Contradiction.

Q: Quotient (factor) group
A: Let $K\unlhd G$. The set $G:K=G/K$ of (left) cosets of $K$ in $G$ is a group under the operation $aK*bK=abK$.
\begin{proof}
Well-defined: Let $aK=a'K$ and $bK=b'K$. Then for some $k_1,k_2,k_3 \in K$:
\begin{align*}
a'&=ak_1,b'=bk_2 \\
\implies a'b'&=ak_1bk_2 \\
&=abk_3k_2 \ (\because K\unlhd G) \\
&\in abK
\end{align*}
Cosets partition so $a'b'K=abK$. \\
Closure: if $aK, bK$ are cosets, then clearly so is $abK$. \\
Identity: $aK*eK=aK=eK*aK$ \\
Inverses: $aK*a^{-1}K=eK=a^{-1}K*aK$ \\
Associativity: follows from associativity in $G$
\end{proof}

Q: Quotient map
A: Given $K\unlhd G$, the map $q:G\to G/K$ by \[
a\mapsto aK
\]
is the quotient map. It is obviously surjective. It is also a group homomorphism:
\begin{align*}
q(ab)&=abK \\
&=aK*bK \\
&=q(a)*q(b)
\end{align*}

Q: Why are normal subgroups exactly the kernels of group homomorphisms?
A: Kernels are normal subgroups \\
\\
Normal subgroups are the kernels of the quotient map (which is a group homomorphism)

Q: The Isomorphism Theorem
A: Let $f:G\to H$ be a group homomorphism with Ker$f=K$. Then $K\unlhd G$ and $G/K\cong$ Im$f$.
\begin{proof}
\textsc{Part 1} For $k\in K$ and $a\in G$:
\begin{align*}
f(aka^{-1})&=f(a)f(k)f(a)^{-1} \\
&=f(a)f(a)^{-1} \\
&=e \\
\implies &aka^{-1}\in K \\
\implies &K\trianglelefteq G
\end{align*}
\textsc{Part 2}
We define $\theta: G/K\to$ Im$f$ by $aK\mapsto f(a)$. We are required to prove $\theta$ is an isomorphism.
\textbf{Well-defined}
\begin{align*}
&a_1K = a_2K \\
\implies &a_2^{-1}a_1\in K \\
\implies &f(a_2)^{-1}f(a_1) = f(a_2^{-1}a_1) = e \\
\implies &f(a_1) = f(a_2) \\
\implies &\theta(a_1K) = \theta(a_2 K)
\end{align*}
\textbf{Group homomorphism}
\begin{align*}
\theta(aKbK) &= \theta(abK) \\
&= f(ab) \\
&= f(a)f(b)  \\
&= \theta(aK) \theta(bK)
\end{align*}
\textbf{Injective}
\begin{align*}
&\theta(aK) = \theta(bK) \\
\implies &f(a) = f(b)\\
\implies &f(b)^{-1}f(a)=e \\
\implies &f(b^{-1}a)=e \\
\implies &b^{-1}a\in K  \\
\implies &aK = bK
\end{align*}
\textbf{Surjective} by definition, since Im$\theta =$ Im$f$.
\end{proof}

Q: Uniqueness of cyclic groups
A: If $G=\langle c \rangle$, then either $G\cong(\mathbb{Z},+)$ or $G\cong(\mathbb{Z}/n\mathbb{Z},+_n)$.
\begin{proof}
Define $f: \mathbb{Z} \to G$ with $m\mapsto c^m$. $f$ is a group homomorphism since $c^{m_1 + m_2} = c^{m_1}c^{m_2}$ and is surjective since $G$ is by definition all $c^m$ for all $m$. We know that Ker$f\unlhd \mathbb{Z}$. But the only subgroups of $\mathbb{Z}$ are $n\mathbb{Z}$ by closure under repeated addition. We have three possibilities.
  \begin{enumerate}
    \item Ker$f = 0\mathbb{Z}=\{e\}$, so $f$ is injective and thus an isomorphism. Then $G\cong \mathbb{Z}$.
    \item Ker$f = 1\mathbb{Z}=\mathbb{Z}$, then, by the isomorphism theorem, $G\cong \mathbb{Z}/\mathbb{Z} = \{e\} = C_1$.
    \item Ker$f = n\mathbb{Z}$ for $n\in\mathbb{N}$, then, by the isomorphism theorem, $G\cong \mathbb{Z}/(n\mathbb{Z})$.
  \end{enumerate}
\end{proof}

Q: Simple group
A: A group is \emph{simple} if it has no non-trivial proper normal subgroup

Q: Group action
A: Let $X$ be a set and $G$ a group. An \emph{action} of $G$ on $X$ is a function $\theta: G\times X \to X$ satisfying
\begin{enumerate}
\setcounter{enumi}{-1}
\item Closure: $\forall g\in G, x\in X: \theta (g,x) \in X$
\item Identity:  $\forall x\in X: \theta (e,x) =x$
\item $\forall g,h\in G, x\in X: \theta (g,\theta(h,x)) =\theta (gh,x)$
\end{enumerate}
We can also write $\theta(g,x)=g\cdot x=g(x)=\theta_g(x)$.

Q: Show that group actions are permutations
A: \begin{align*}
\theta(g,\theta(g^{-1},x))&= \theta(gg^{-1},x) \\
&=\theta(e,x) \\
&=x \\
&=\theta(e,x) \\
&=\theta(g^{-1}g,x) \\
&=\theta(g^{-1},\theta(g,x))
\end{align*}
Thus $\theta_g:X\to X$ has inverse $\theta_{g^{-1}}:X\to X$. So $\theta_g$ forms a bijection $X\leftrightarrow X$ i.e. a permutation.

Q: Show that $\theta_g$ is an action iff $\varphi: G\to $ Sym$X$ with $\varphi(g)=\theta_g$ is a group homomorphism
A: Suppose $\theta_g$ is an action, so is a permutation. Thus
\begin{align*}
\varphi(gh)&=\theta_{gh} \\
&=\theta_g\circ\theta_h \\
&=\varphi(g)\circ\varphi(h)
\end{align*}
Conversely, suppose $\varphi: G\to $ Sym$X$ is a group homomorphism. We show $\varphi(g)=\theta_g$ is an action.
\begin{enumerate}
\setcounter{enumi}{-1}
\item If $\varphi(g)=\theta_g\in$ Sym$X$, then by definition $\theta_g(x)\in X$
\item $\varphi(g)\varphi(e)=\varphi(g)=\varphi(e)\varphi(g)$ so $\theta_e=1_X$, i.e. $\forall x: \theta_e(x)=x$.
\item $\varphi(gh)=\varphi(g)\circ\varphi(h) \implies \theta_g\circ\theta_h=\theta_{gh}$.
\end{enumerate}

Q: Kernels of an action
A: Ker($\varphi: G\to$ Sym$X$) i.e. the group elements which `act as the identity'

Q: Faithful action
A: Ker($\varphi: G\to$ Sym$X)=\lbrace e \rbrace$ i.e. no `cheating' non-identity elements act as the identity.

Q: Orbit
A: Given an action $G$ on $X$ and $x\in X$ \[
\mathrm{orb}(x)=g(x)=\lbrace y\in X \ | \ \exists g \in G: g(x)=y \rbrace \subseteq X
\]
i.e. the set of elements which can be reached from $x$

Q: Stabiliser
A: Given an action $G$ on $X$ and $x\in X$ \[
\mathrm{Stab}(x)=G_x=\lbrace g\in G \ | \ g(x)=x \rbrace \subseteq X
\]
i.e. the set of group elements which fix $x$

Q: Show that Stab($x$) is a subgroup
A: $e(x)=x$ so Stab($x)$ is non-empty \\
If $g,h \in $ Stab($x$), then $gh^{-1}(x)=x$ so $gh^{-1}\in$ Stab($x$) \\
So Stab($x$) is a subgroup (of $G$) by the super-efficient subgroup criterion.

Q: Transitive action
A: \[
\forall x \in X: orb(x)=X
\]
i.e. you can reach any element from any element

Q: Show that orbits partition
A: The orbits of an action $G$ on $X$ partition $X$
\begin{proof}
$\forall x \in X: e(x)=x$ so $x\in\mathrm{orb}(x)$. Thus the orbits make up the whole set. \\
Suppose $z\in $ orb($x$) and $z\in $ orb($y$). Then
\begin{align*}
\exists g_1,g_2\in G: g_1(x)=z=g_2(y) \\
\implies x=g_1^{-1}[g_2(y)]
\end{align*}
Let $w\in \mathrm{orb}(x)$ such that $g(x)=w$. Therefore
\begin{align*}
&g[g_1^{-1}[g_2(y)]]=w \\
\implies &w\in \mathrm{orb}(y) \\
\implies &\mathrm{orb}(x)\subseteq\mathrm{orb}(y) \\
\implies &\mathrm{orb}(y)\subseteq\mathrm{orb}(x) \ \textrm{(by symmetry)}\\
\implies &\mathrm{orb}(x)=\mathrm{orb}(y)
\end{align*}
So distinct orbits do not coincide.
\end{proof}

Q: Orbit-Stabiliser Theorem
A: Let the finite group $G$ act on the set $X$. Then for any $x\in X$: \[
|G|=|\mathrm{orb}(x)||\mathrm{Stab}(x)|
\]
\begin{proof}
Let $y\in$ orb($x$). Consider $g,h$ which both send $x$ to $y$.
\begin{align*}
&h(x)=y=g(x) \\
\iff &g^{-1}h(x)=x \\
\iff &g^{-1}h \in \mathrm{Stab}(x) \\
\iff &g\mathrm{Stab}(x)=h\mathrm{Stab}(x)
\end{align*}
So the left cosets of Stab($x$) are distinct iff they send $x$ to different points ($y$) in orb($x$). Therefore, there are $|\mathrm{orb}(x)|$ distinct left cosets of Stab($x$), each with size $|\mathrm{Stab}(x)|$. The result follows by Lagrange's Theorem.
\end{proof}

Q: Left regular action
A: Every group $G$ acts on itself by left multiplication. The action is faithful and transitive.
\begin{proof}
\textbf{Action}
  \begin{enumerate}
\setcounter{enumi}{-1}
    \item $\forall g,x \in G: g(x) = g\cdot x \in G$ by closure under group multiplication.
    \item $\forall x\in G\,e(x)=e\cdot x = x$ by group identity axiom.
    \item $g(hx) = (gh)x$ by associativity.
  \end{enumerate}
  So it is an action.
\textbf{Faithful} $[\forall x\in X : g(x)=gx = x]\Rightarrow g = e$ by the uniqueness of identity.
\textbf{Transitive} We can send any $x$ to any $y$ by $yx^{-1}(x) = y$.
\end{proof}

Q: Cayley's Theorem
A: Every group is isomorphic to a subgroup of some symmetric group.
\begin{proof}
  Take the left regular action of $G$ on itself. This gives a group homomorphism $\varphi: G\to $ Sym$G$ with Ker$\varphi = \{e\}$ as the action is faithful. By the isomorphism theorem, $G \cong $ Im$ \varphi \leq $ Sym$G$.
\end{proof}

Q: Conjugate
A: The conjugate of $a$ by $b$ is \[
bab^{-1}
\]

Q: Left coset action
A: Let $H\leq G$. Then $G$ acts on the left cosets of $H$ by left multiplication. The action is transitive.
\begin{proof}
\textbf{Action}
  \begin{enumerate}
      \setcounter{enumi}{-1}
    \item $g(aH) = (ga)H$ is a coset of $H$.
    \item $e(aH) = (ea)H = aH$.
    \item $g_1(g_2(aH)) = g_1((g_2a)H) = (g_1g_2a)H = (g_1g_2)(aH)$.
  \end{enumerate}
\textbf{Transitive} Any $aH$ can be mapped to any $bH$ by $(ba^{-1})(aH) = bH$.
\end{proof}

Q: Conjugation action on groups
A: Any group $G$ acts on itself by conjugation.
\begin{proof}
\textbf{Action}
  \begin{enumerate}
      \setcounter{enumi}{-1}
    \item $\forall g, x\in G: g(x) = gxg^{-1} \in G$
    \item $e(x) = exe^{-1} = x$
    \item $g(h(x)) = g(hxh^{-1}) = ghxh^{-1}g^{-1} = (gh)x(gh)^{-1} = (gh)(x)$
  \end{enumerate}
\end{proof}

Q: Centre
A: The Kernel of the conjugation action i.e. the elements of a group that commute with all other elements
  \[
    Z(G) =\{g\in G \ | \ \forall a:gag^{-1} = a\} = \{g\in G \ | \ \forall a:,ga = ag\}.
  \]
  It is sometimes written as $C(G)$ instead of $Z(G)$.

Q: Conjugacy Classes
A: The orbits of the conjugacy action.
  \[
    \mathrm{ccl}(a) = \{b\in G \ | \ \exists g\in g:gag^{-1} = b\}.
  \]

Q: Centralisers
A: The stabilisers of the conjugacy action, ie. elements that commute with $a$.
  \[
    C_G(a) = \{g\in G: gag^{-1} = a\} = \{g\in G: ga = ag\}.
  \]

Q: Conjugation action on normal subgroups
A: Let $K\unlhd G$. Then $G$ acts by conjugation on $K$.
\begin{proof}
  \textbf{Action}
  \begin{enumerate}
      \setcounter{enumi}{-1}
    \item $\forall g\in G,k\in K: g(k) = gkg^{-1} \in K$ as $gK=Kg$  for normal subgroups
    \item $e(x) = exe^{-1} = x$
    \item $g(h(x)) = g(hxh^{-1}) = ghxh^{-1}g^{-1} = (gh)x(gh)^{-1} = (gh)(x)$
  \end{enumerate}
\end{proof}

Q: Show that normal subgroups are exactly those subgroups which are unions of conjugacy classes
A: \begin{proof}
Suppose $K\unlhd G$ with $k\in K$ (say).
\begin{align*}
&gK=Kg \\
\implies &gkg^{-1} \in K \\
\implies &\mathrm{ccl}(k)\subseteq K \\
\implies &\bigcup_{k\in K} \mathrm{ccl} (k) \subseteq K \\
& \forall k\in K: k\in \mathrm{ccl}(k) \\
\implies &K\subseteq \bigcup_{k\in K} \mathrm{ccl} (k) \\
\therefore \ &K=\bigcup_{k\in K} \mathrm{ccl} (k)
\end{align*}
Conversely, if $K$ is a union of conjugacy classes and a subgroup of $G$, then \[
\forall k\in K, g\in G: gkg^{-1}\in K\]
So $K$ is a normal subrgoup.
\end{proof}

Q: Conjugation action on subgroups
A: $G$ acts by conjugation on the set of its subgroups
\begin{proof}
  \textbf{Action}
  \begin{enumerate}
      \setcounter{enumi}{-1}
    \item If $H\leq G$, then we show that $gHg^{-1}$ is also a subgroup by the super-efficient subgroup criterion.
             \subitem $geg^{-1} = e\in gHg^{-1}$, so it is non-empty
             \subitem $gag^{-1}$ and $gbg^{-1}\in gHg^{-1} \implies$ $(gag^{-1})(gbg^{-1})^{-1} = g(ab^{-1})g^{-1}\in gHg^{-1}$.
    \item $eHe^{-1} = H$.
    \item $g_1(g_2Hg_2^{-1})g_1^{-1} = (g_1g_2)H(g_1g_2)^{-1}$.
  \end{enumerate}
\end{proof}

Q: What subgroups are normal subgroups conjugate to?
A: Let $a\in G$, $K\leq G$.
\begin{align*}
&K\unlhd G \\
\iff &aK=Ka \\
\iff &aKa^{-1}=K \\
\iff &\mathrm{ccl}(K)=\lbrace K \rbrace
\end{align*}
So normal subgroups are exactly those with singleton ccls (i.e. only conjugate to themselves)

Q: Show that conjugate subgroups are isomorphic
A: Given $G,H$ conjugate subgroups such that \[
aGa^{-1}=H
\]
let $f:G\to H$ by \[
g\mapsto aga^{-1}
\]
\textbf{Homomorphism}
\begin{align*}
f(g_1)f(g_2)&=ag_1a^{-1}ag_2a^{-1}  \\
&=ag_1g_2a^{-1} \\
&=f(g_1g_2)
\end{align*}
\textbf{Bijective}
Suppose $f(g_1)=f(g_2)$. Then
\begin{align*}
ag_1a^{-1}&=ag_2a^{-1} \\
\implies g_1&=g_2
\end{align*}
Thus $f$ is injective. As $|aGa^{-1}|=|G|=|H|$, $f$ is thus bijective.

Q: Normalisers
A: The stabiliser of the (group) conjugation action.
  \[
    N_G(H) = \{g\in G: gHg^{-1} = H\}.
  \]
In fact, $N_G(H)$ is the largest subgroup of $G$ in which $H$ is a normal subgroup.
\begin{proof}
\begin{align*}
&\forall a \in N: aHa^{-1} = H \\
\iff &\forall a \in N: aH=Ha \\
\iff &H\unlhd N
\end{align*}
Thus $N_G(H)$ is exactly the largest such $N\leq G$.
\end{proof}

Q: Show that stabilisers of elements in the same oribit are conjugate
A: Let $G$ act on $X$ and let $g\in G, x\in X$. Then \[
\mathrm{Stab}(g(x)) = g\mathrm{Stab}(x)g^{-1}
\]
\begin{proof}
Suppose $a\in\mathrm{Stab}(g(x))$. Then
\begin{align*}
g^{-1}ag(x) &=g^{-1}g(x) \\
&= x \\
\implies g^{-1}ag &\in \mathrm{Stab}(x) \\
\implies a&\in g\mathrm{Stab}(x)g^{-1} \\
\therefore \mathrm{Stab}(g(x))&\leqslant g\mathrm{Stab}(x)g^{-1}
\end{align*}
Conversely, $b\in\mathrm{Stab}(x)$. Then
\begin{align*}
gbg^{-1}(g(x))&=gb(x) \\
&= g(x) \\
\implies  gbg^{-1}&\in \mathrm{Stab}(g(x)) \\
\therefore g\mathrm{Stab}(x)g^{-1}&\leqslant \mathrm{Stab}(g(x))
\end{align*}
\end{proof}

Q: Bound on size of finite simple groups in terms of subgroup index
A: Let $G$ be a finite simple group with $|G|>2$ and and a proper subgroup $H$ with $|G:H|=n$. Then $|G| \leq \frac{n!}{2}$.
\begin{proof}
  Consider the left coset action of $G$ on $H$. We get a group homomorphism $\varphi: G\to S_n$ since there are $n$ cosets of $H$. Since $H\not= G$, $\varphi$ is non-trivial and Ker$\varphi \not=G$. Now Ker$\varphi \unlhd G$. Since $G$ is simple, Ker$\varphi = \{e\}$. So $G\cong$ Im$\varphi \subseteq S_n$ by the isomorphism theorem. So $|G| \leq |S_n| = n!$. \\
  We can further refine this by considering sgn$\circ \varphi: G \to \{\pm 1\}$. Ker(sgn$\circ \varphi)\unlhd G$, so $K = $Ker(sgn$\circ\phi) = \{e\}$ or $G$. Since $G/K\cong$ Im (sgn$\circ\phi)$, we know that $|G|/|K| = 1$ or $2$ since Im(sgn$\circ \phi)$ has at most two elements. Hence for $|G|> 2$, we cannot have $K = \{e\}$, or else $|G|/|K| > 2$. So we must have $K = G$, so sgn($\varphi(g)) = 1$ for all $g$ and Im$\varphi \leq A_n$. So $|G|\leq n!/2$
\end{proof}

Q: Cauchy's Theorem
A: Let $G$ be a finite group and prime $p$ dividing $|G|$. Then $G$ has an element of order $p$ (in fact at least $p - 1$ of them).
\begin{proof}
  Let $G$ and $p$ be fixed. Consider $G^p = G\times G\times \ldots \times G$, the set of $p$-tuples of $G$. Let $X \subseteq G^p$ be $X = \{(a_1, a_2, \ldots, a_p)\in G^p \ | \ a_1a_2\ldots a_p = e\}$.
  In particular, if an element $b$ has order $p$, then $(b, b, \ldots, b)\in X$. In fact, if $(b, b, \ldots, b)\in X$ and $b\not= e$, then $b$ has order $p$, since $p$ is prime.
  Now let $H = \langle h \ | \ h^p = e\rangle\cong C_p$ be a cyclic group of order $p$ with generator $h$ (This $h$ is not related to $G$ in any way). Let $H$ act on $X$ by ``rotation'':
  \[
    h(a_1, a_2, \ldots, a_p) = (a_2, a_3, \ldots, a_p, a_1)
  \]
  This is an \textbf{action}:
  \begin{enumerate}
      \setcounter{enumi}{-1}
    \item If $a_1\ldots a_p = e$, then $a_1^{-1} = a_2\cdots a_p$. So $a_2\ldots a_pa_1 = a_1^{-1}a_1 = e$. So $(a_2, a_3, \ldots, a_p, a_1)\in X$.
    \item $e$ acts as an identity by construction
    \item $h^l(a_1,\ldots a_p)=(a_{l+1}, \ldots a_1,\ldots a_l )=h\circ h \circ \ldots \circ h (a_1,\ldots a_p)$ by construction
  \end{enumerate}
  As orbits partition $X$, the sum of all orbit sizes must be $|X|$. We know that $|X| = |G|^{p - 1}$ since we can freely choose the first $p - 1$ entries and the last one must be the inverse of their product. Since $p$ divides $|G|$, $p$ also divides $|X|$. We have $|\mathrm{orb}(a_1, \ldots , a_p)||\mathrm{Stab}_H(a_1, \ldots, a_p)| = |H| = p$. So all orbits have size 1 or $p$, and they sum to $|X| = p \ \times$ something. We know that there is one orbit of size 1, namely $(e, e, \ldots, e)$. 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 ${(a, a, \ldots, a)}$ for some $a(\neq e)\in G$, which has order $p$.
\end{proof}

Q: Group of rotational symmetries of a cube
A: Let $G^+$ be the group of rotational symmetries of a cube acting on the four diagonals. This gives a group homomorphism $\varphi:G^+\to S_4$. \\
We have $(1234)\in$Im$\varphi$ by rotation around axis through top and bottom face.  \\
We also have $(12)\in$Im$\varphi$ by rotation around axis through midpoint of edge containing 1 and 2. \\
These two generate the whole of $S_4$, so $\varphi$ is surjective.  \\
Now let $G^+$ act on the 8 vertices. The action is clearly transitive. The stabiliser of vertex $1$ is exactly the rotations through axis $1$ and the diagonally opposite vertex. There are three of these (including $e$). So by the Orbit-Stabiliser Theorem, $|G^+|=8\times 3=24.$ \\
Thus $|G^+|=|S_4|$ so $\varphi$ is bijective. Therefore, $\varphi$ is an isomorphism $G^+\leftrightarrow S_4$.

Q: The group of all symmetries of the cube
A: Let $G$ be the group of all symmetries of the cube. Now let $G$ act on the 8 vertices of the cube. The action is transitive. The stabiliser is
\begin{align*}
\lbrace &e,\textrm{$2$ more rotations in axis through $1$,} \\
&\textrm{$3$ reflections in planes through $1$ and an edge coming out of $1$}  \rbrace
\end{align*}
i.e. 3 for each $120^\circ$ rotation and $1$ for each edge coming out of $1$. This has $6$ elements. So by the Orbit-Stabiliser Theorem, $|G|=8\times 6=48$. \\
Now consider the ``reflection in midpoint'', i.e. sending each point to its opposite on the cube. View this as the transformation $\tau$ as $-I\in\mathbb{R}^3$. Now we see that  \\
$G^+\cap \langle\tau\rangle=\lbrace e \rbrace$ \\
$\tau$ commutes with all other symmetries (because multiplication by $-1$ is commutative) \\
$|G^+\times \langle \tau \rangle|=24=|G|$ so $G^+\langle \tau \rangle=G$ \\
\\
Therefore, by the Direct Product Theorem, $G\cong G^+\times \langle \tau \rangle$ \\
Note we can put each vertex of an octahedron on the centre of the face of a cube, so they are dual and thus have the same group of symmetries.

Q: How to determine ccls in $S_n$
A: Elements of $S_n$ are conjugate iff they have the same cycle type
\begin{proof}
If $(a_1\ldots a_k)$ is a $k$-cycle and $\rho\in S_n$, then \[
\rho(a_1\ldots a_k)\rho^{-1}=(\rho(a_1)\rho(a_2)\ldots\rho(a_k))
\]
because
\begin{align*}
&\rho(a_1)\mapsto a_1 \mapsto a_2 \mapsto \rho(a_2) \\
&\rho(a_2)\mapsto a_2 \mapsto a_3 \mapsto \rho(a_3) \\
&\ldots\\
&\rho(a_k)\mapsto a_k\mapsto a_1\mapsto\rho(a_1).
\end{align*}
So if $\sigma=\sigma_1\sigma_2\ldots\sigma_l$ is a product of disjoint cycles, then
\begin{align*}
\rho\sigma\rho^{-1}=\rho\sigma_1\rho^{-1}\rho\sigma_2\rho^{-1}\ldots\rho\sigma_l\rho^{-1}
\end{align*}
so $\sigma$ and its conjugate have the same cycle type. \\
Conversely, if $\sigma,\tau$ have the same cycle type, say
\begin{align*}
\sigma&=(a_1a_2\ldots a_k)(a_{k+1}\ldots a_{k+l}) \ldots \\
\tau&=(b_1b_2\ldots b_k) \ (b_{k+1}\ldots b_{k+l}) \ldots
\end{align*}
then use \[
\rho=
\begin{pmatrix}
a_1 &\ldots &a_{k+l} &\ldots &a_n \\
b_1 &\ldots &b_{k+l} &\ldots &b_n
\end{pmatrix}
\]
such that $\rho\sigma\rho^{-1}=\tau$. So $\sigma$ and $\tau$ are conjugate.
\end{proof}

Q: How to find normal subgroups
A: They must contain $e$, divide the group size and be a union of the ccls.

Q: Conjugacy classes in $A_n$
A: \begin{align*}
|S_n|&=|\mathrm{ccl}_{S_n}(\sigma)||C_{S_n}(\sigma)| \\
|A_n|&=|\mathrm{ccl}_{A_n}(\sigma)||C_{A_n}(\sigma)| \\
|A_n|&=\frac{1}{2}|S_n| \\
\end{align*}
We know that a subgroup of $S_n$ (e.g. centralisers) has either none or half of its elements odd. This leaves two options
\begin{align*}
\mathrm{ccl}_{S_n}(\sigma)=\mathrm{ccl}_{A_n}(\sigma),|C_{A_n}(\sigma)|=\frac{1}{2}|C_{S_n}|(\sigma) \\
|\mathrm{ccl}_{A_n}(\sigma)|=\frac{1}{2}|\mathrm{ccl}_{S_n}(\sigma)|, C_{A_n}(\sigma)=C_{S_n}(\sigma)
\end{align*}
Thus the ccl of $\sigma$ splits in $A_n$ iff
\begin{align*}
&C_{A_n}(\sigma)=C_{S_n}(\sigma) \\
\iff &\forall \tau \in C_{S_n}(\sigma): \mathrm{sgn}(\tau)=1 \\
\iff &\forall \tau: \tau\sigma\tau^{-1}=\sigma \implies \mathrm{sgn}(\tau)=1 \\
\iff &\forall \tau: \tau\sigma=\sigma\tau \implies \mathrm{sgn}(\tau)=1 \\
\iff &\neg\exists \tau: \tau\sigma=\sigma\tau, \ \mathrm{sgn}(\tau)=-1 \\
\iff &\sigma \textrm{ commutes with no odd permutation}
\end{align*}

Q: Groups of order 8
A: If $G$ has order 8, then either \\
 $G$ is abelian and $G\cong C_8, C_4\times C_2$ or $C_2\times C_2\times C_2$ \\
or $G$ is not abelian and isomorphic to $D_8$ or $Q_8$.
\begin{proof}
\textbf{Case 1 } If $G$ contains an element of order 8, then $G\cong C_8$. \\
\textbf{Case 2 } If all non-identity elements have order 2, then $G$ is abelian. Let $a\not= b\in G\setminus\{e\}$. By the direct product theorem, $\langle a, b\rangle = \langle a\rangle \times\langle b\rangle$. Then take $c\in G\setminus\langle a, b\rangle$. By the direct product theorem, we obtain $\langle a, b, c\rangle= \langle a\rangle\times\langle b\rangle \times\langle c\rangle = C_2\times C_2\times C_2$. Since $\langle a, b, c\rangle\subseteq G$ and $|\langle a, b, c\rangle | = |G|$, $G = \langle a, b, c\rangle \cong C_2\times C_2\times C_2$. \\
    \textbf{Case 3 } $G$ has no element of order 8 but has an order 4 element $a\in G$. Let $H = \langle a\rangle$. Since $H$ has index 2, it is normal in $G$. So $G/H \cong C_2$ since $|G/H| = 2$. This means that for any $b\not\in H$, $bH$ generates $G/H$. Then $(bH)^2 = b^2H = H$. So $b^2\in H$. Since $b^2\in \langle a\rangle$ and $\langle a\rangle$ is a cyclic group, $b^2$ commutes with $a$.
      If $b^2 = a$ or $a^3$, then $b$ has order 8. Contradiction. So $b^2 = e$ or $a^2$.
      We also know that $H$ is normal, so $bab^{-1}\in H$. Let $bab^{-1} = a^\ell$. Since $a$ and $b^2$ commute, we know that $a = b^2 ab^{-2} = b(bab^{-1})b^{-1} = ba^\ell b^{-1} = (bab^{-1})^{\ell} = a^{\ell^2}$. So $\ell^2 \equiv 1\pmod 4$. So $\ell \equiv \pm 1 \pmod 4$. \\
\textbf{Case 3.1 }If $l\equiv 1\pmod 4$, $bab^{-1} = a$, ie. $ba = ab$. So $G$ is abelian. \\
\textbf{Case 3.1.1 }If $b^2 = e$, then by the Direct Product Theorem $G = \langle a, b\rangle \cong \langle a\rangle \times \langle b\rangle\cong C_4\times C_2$. \\
\textbf{Case 3.1.2} If $b^2 = a^2$, then $(ba^{-1})^2 = e$. So by the Direct Product Theorem $G = \langle a, ba^{-1}\rangle\cong C_4\times C_2$. \\
  \textbf{Case 3.2 } If $l \equiv -1\pmod 4$, then $bab^{-1} = a^{-1}$.
          \begin{itemize}
            \item If $b^2 = e$, then $G = \langle a, b: a^4 = e = b^2, bab^{-1} = a^{-1}\rangle$. So $G\cong D_8$ by definition.
            \item If $b^2 = a^2$, then we potentially have $G\cong Q_8$ where $Q_8=\langle a,b \ | \ a^4=e,b^2=a^2,bab^{-1}=a^{-1} \rangle$ (called the Quaternions)
          \end{itemize}
\end{proof}
Indeed, $Q_8$ is well-defined and satisfies the group axioms. It can be defined as the set containing
\begin{align*}
&\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix},\begin{pmatrix} i & 0 \\ 0 & -i\end{pmatrix},\begin{pmatrix} -i & 0 \\ 0 & i\end{pmatrix} , \\
&\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}, \begin{pmatrix} 0 & i \\ i & 0\end{pmatrix},\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix},\begin{pmatrix} 0 & -i \\ -i & 0\end{pmatrix}
\end{align*}
under the operation of matrix multiplication. Alternatively, we can define \[
Q_8=\{1,-1,i,-i,j,-j,k,-k\}
\]
with the dot product operation where $i,j,k$ act as right-handed normal vectors with modulus $-1$.

Q: $M_{n\times n} (\mathbb{C})$
A: The set of $n\times n$ matrices. They do not form a group because they do not all have inverses.

Q: $GL_n(\mathbb{C})$
A: \[
GL_n(\mathbb{C}) =\{ A\in M_{n\times n} (\mathbb{C}) \ | \ \mathbb{det}A\neq 0 \}
\]
i.e. the General Linear group is the set of $n\times n$ invertible matrices. These form a group.
\begin{proof}
\textbf{Closure } $\mathrm{det}(AB)=\mathrm{det}(A)\mathrm{det}(B)$ so $\mathrm{det}(A),\mathrm{det}(B)\neq 0 \implies \mathrm{det}(AB)\neq 0$ i.e. the product of invertible matrices is invertible. \\
\textbf{Identity} Clearly $I=\mathrm{diag}(1,1,\ldots 1)$ \\
\textbf{Inverses} Inverses exist for all matrices with non-zero determinant \\
\textbf{Associativity} True for all matrix multiplication
\end{proof}

Q: Show that det is a group homomorphism
A: $\mathrm{det}: GL_n(\mathbb{C}) \to (\mathbb{C}\setminus\{0\},\cdot)$ is a surjective group homomorphism.
\begin{proof}
  $\mathrm{det} AB = \mathrm{det}A\mathrm{det}B$. If $A$ is invertible, it has non-zero determinant and $\mathrm{det}A\in \mathbb{C}\setminus\{0\}$.
  To show it is surjective, for any $x\in \mathbb{C}\setminus\{0\}$, if we take the identity matrix and replace $I_{11}$ with $x$, then the determinant is $x$. So it is surjective.
\end{proof}
  $\mathrm{det}: O_n \to \{\pm 1\}$ is a surjective group homomorphism.
\begin{proof}
\textbf{Homomorphism} $\mathrm{det}(AB) = \mathrm{det} A\mathrm{det} B$. \\
\textbf{Image}  For $A\in O_n$
\begin{align*}
\mathrm{det} A^TA &= \mathrm{det}(I)=1 \\
&=(\mathrm{det} A)(\mathrm{det} A^T)  \\
&=(\mathrm{det} A)(\mathrm{det} A) \\
&=(\mathrm{det} A)^2 \\
\implies \mathrm{det} A&= \pm 1 \\
\end{align*}
\textbf{Surjective}
  \[
    \mathrm{det} I = 1,\;\;\;\mathrm{det}
    \begin{pmatrix}
      -1 & 0 &\cdots & 0\\
      0 & 1 & \cdots & 0\\
      \vdots & \vdots & \ddots & 0\\
      0 & 0 & \cdots & 1
    \end{pmatrix} = -1,
  \]
\end{proof}
  $\mathrm{det}: U_n\to S^1$, where $S^1$ is the unit circle in the complex plane, is a surjective group homomorphism.
\begin{proof}
\textbf{Homomorphism} $\mathrm{det}(AB) = \mathrm{det} A\mathrm{det} B$. \\
\textbf{Image} $1 = \mathrm{det} I = \mathrm{det} A^\dagger A = |\mathrm{det} A|^2$. So $|\mathrm{det} A| = 1$.  \\
\textbf{Surjective}  Given $\lambda\in S_1$, we have
  $\begin{pmatrix}
    \lambda & 0 & \cdots & 0\\
    0 & 1 & \cdots & 0\\
    \vdots & \vdots & \ddots & 0\\
    0 & 0 & 0 & 1
  \end{pmatrix}\in U_n$. So it is surjective.
\end{proof}

Q: Special linear group
A: The \emph{special linear group} $SL_n(\mathbb{C})$ is the kernel of the determinant, ie.
  \[
    SL_n(\mathbb{C}) = \{A\in GL_n(\mathbb{C}): \mathrm{det} A = 1\}.
  \]

Q: $GL_n$ as an action
A: $GL_n(\mathbb{C})$ acts faithfully on $\mathbb{C}^n$ by left multiplication, with two orbits ($\mathbf{0}$ and everything else).
  \begin{proof}
   \textsc{Action} \\
   \textbf{Closure} If $A\in GL_n(\mathbb{C})$ and $\mathbf{v}\in \mathbb{C}^n$, then $A\mathbf{v}\in \mathbb{C}^n$. \\
   \textbf{Identity} $I\mathbf{v} = \mathbf{v}$ for all $\mathbf{v}\in \mathbb{C}^n$. \\
   \textbf{Compatibility} $A(B\mathbf{v}) = (AB)\mathbf{v}$. \\
   \textsc{Faithful} A linear map is determined by what it does on a basis. Take the standard basis $\mathbf{e}_1 = (1, 0, \ldots, 0), \ldots \mathbf{e}_n = (0, \ldots, 1)$. Any matrix which maps each $\mathbf{e}_k$ to itself must be $I$ (since the columns of a matrix are the images of the basis vectors)
   \\
   \textsc{Orbits} We know that $A\mathbf{0} = \mathbf{0}$ for all $A$. Also, as $A$ is invertible, $A\mathbf{v} = \mathbf{0}\Leftrightarrow \mathbf{v} = \mathbf{0}$. So $\mathbf{0}$ forms a singleton orbit. Then given any two vectors $\mathbf{v}\not= \mathbf{w}\in \mathbb{C}^n\setminus\{0\}$, we know there is a matrix $A\in GL_n(\mathbb{C})$ such that $A\mathbf{v} = \mathbf{w}$
  \end{proof}
  Similarly, $GL_n(\mathbb{R})$ acts on $\mathbb{R}^n$. $SL_n(\mathbb{C})$ acts on $\mathbb{C}^n$ also with two orbits. To construct a matrix $A$ which sends $\hat{e}_1$ to $\vec{v}$ with det$A=1$, we create a basis from $\vec{v}$ and construct a matrix with those columns. Then divide any column (apart from the the first) by det$A$. You can now send any vector to $\hat{e}_1$ and $\hat{e}_1$ to any vector, so by matrix multiplication you're done. \\
  Also $GL_n(\mathbb{C})$ acts on $M_{n\times n}(\mathbb{C})$ by conjugation.
  \begin{proof}
   Let $A\in M_{n\times n}(\mathbb{C}),P,Q\in GL_n(\mathbb{C}).$ \\
   \textbf{Closure} $PAP^{-1}\in M_{n\times n}$\\
   \textbf{Identity} $IAI^{-1}=A$\\
   \textbf{Compatibility} $ P(QAQ^{-1})P^{-1}=PQ(A)(PQ)^{-1}$
   \end{proof}
Note that the orbits of this action are the matrices which are the same map with respect to different bases.

Q: Orthogonal group
A: \[
    O_n =O_n(\mathbb{R}) = \{A\in GL_n(\mathbb{R}): A^TA = I\},
  \]
forms a group under matrix multiplication
\begin{proof}
  We show $O_n(\mathbb{R})\leqslant GL_n(\mathbb{R})$ using the super-efficient subgroup criterion: \\
It is non-empty, since $I^TI=I$ so $I\in O_n $. \\
If $A, B\in O_n $, then $(AB^{-1})(AB^{-1})^T = AB^{-1}(B^{-1})^TA^{T} = AB^{-1}BA^{-1} = I$, so $AB^{-1}\in O_n $
\end{proof}

Q: Special orthogonal group
A: The kernel of det: $O_n \to \{\pm 1\}$.
  \[
SO_n = SO_n(\mathbb{R}) = \{A\in O_n: \mathrm{det} A = 1\}.
  \]

Q: Relation between $O_n$ and $SO_n$
A: $O_n = SO_n \cup
  \begin{pmatrix}
    -1 & 0 &\cdots & 0\\
    0 & 1 & \cdots & 0\\
    \vdots & \vdots & \ddots & 0\\
    0 & 0 & \cdots &1
  \end{pmatrix}SO_n$
\begin{proof}
 By the isomorphism theorem,
\begin{align*}
O_n/ \mathrm{Ker}(\mathrm{det}:O_n\to\{\pm 1\})&\cong\mathrm{Im} (\mathrm{det}:O_n\to\{\pm 1\}) \\
\implies O_n/ SO_n&\cong C_2
\end{align*}
The result follows from the fact that the two cosets partition the group.
\end{proof}

Q: Show that orthogonal matrices are isometries
A: For $\mathsf{A}\in O_n$ and $\vec{x},\vec{y}\in\mathbb{R}^n$ \[
(\mathsf{A}\vec{x})^T(\mathsf{A}\vec{y})=\vec{x}^T\vec{y} \]
and \[
|\mathsf{A}\vec{x}|=|\vec{x}| \]
i.e. $O_n$ preserves dot product and length
\begin{proof}
\textbf{Dot product}
\begin{align*}
(\mathsf{A}\vec{x})^T(\mathsf{A}\vec{y})&=\vec{x}^T\mathsf{A}^T\mathsf{A}\vec{y} \\
&= \vec{x}^T\mathsf{I}\vec{y} \\
&=\vec{x}^T\vec{y}
\end{align*}
\textbf{Length}
\begin{align*}
|\mathsf{A}\vec{x}|^2&=(\mathsf{A}\vec{x})^T(\mathsf{A}\vec{x}) \\
&=\vec{x}^T\vec{x} \\
&=|\vec{x}|^2 \\
\implies |\mathsf{A}\vec{x}|&=|\vec{x}|
\end{align*}
\end{proof}

Q: $O_2$
A: $SO_2$ consists of all rotations of $\mathbb{R}^2$ around 0.
\begin{proof}
  Let $A\in SO_2$. So $A^T = A^{-1}$ and $\mathrm{det} A = 1$. Suppose $A =
  \begin{pmatrix}
    a & b\\c & d
  \end{pmatrix}$. Then $A^{-1} =
  \begin{pmatrix}
    d & -b\\-c & a
  \end{pmatrix}.$ So $ad - bc = 1$, $c = -b$, $d = a$. Combining these equations we obtain $a^2 + c^2 = 1$. Set $a = \cos\theta = d$, and $c = \sin\theta = -b$. Then these satisfies all three equations. So
  \[
    A =
    \begin{pmatrix}
      \cos\theta & -\sin\theta\\
      \sin\theta & \cos\theta
    \end{pmatrix}.
  \]
  Note that $A$ maps $(1, 0)$ to $(\cos\theta, \sin \theta)$, and maps $(0, 1)= (-\sin\theta, \cos\theta)$, so $A$ represents a rotation by $\theta$.
\end{proof}
  Any matrix in $O_2$ is either a rotation around $0$ or a reflection in a line through $0$.
\begin{proof}
  If $A\in SO_2$, we've show that it is a rotation. Otherwise,
  \[
    A =
    \begin{pmatrix}
      1 & 0\\
      0 & -1
    \end{pmatrix}
    \begin{pmatrix}
      \cos\theta & -\sin\theta\\
      \sin\theta & \cos\theta
    \end{pmatrix} =
    \begin{pmatrix}
      \cos\theta & -\sin\theta \\
      -\sin\theta & -\cos\theta
    \end{pmatrix}
  \]
  since $O_2=SO_2\cup
  \begin{pmatrix}
    1&0\\0&-1
  \end{pmatrix}SO_2$. This has eigenvalues $1, -1$. So it is a reflection in the line of the $1$ eigenspace. The line goes through $\mathbf{0}$ because it is a linear map.
\end{proof}

Q: $O_3$
A: Every matrix in $SO_3$ is a rotation around some axis.
\begin{proof}
  Let $A\in SO_3$. We know that $\mathrm{det} A = 1$ and $A$ is an isometry. The eigenvalues $\lambda$ must have $|\lambda| = 1$. They also multiply to $\mathrm{det} A = 1$. Since we are in $\mathbb{R}$, complex eigenvalues come in complex conjugate pairs. \\
\textbf{Case 1} If there are complex eigenvalues $\lambda$ and $\lambda^*$, then $\lambda \lambda^* = |\lambda|^2 = 1$. The third eigenvalue must be real and has to be $+1$. \\
\textbf{Case 2} If all eigenvalues are real, then eigenvalues are either $1$ or $-1$, and must multiply to $1$. The possibilities are $1, 1, 1$ and $-1, -1, 1$, all of which contain an eigenvalue of $1$.
  So pick an eigenvector for our eigenvalue $1$ as the third basis vector. Then in some orthonormal basis,
  \[
    A = \begin{pmatrix}
     a & b & 0\\
      c & d &0\\
      0 & 0 & 1
    \end{pmatrix}
  \]
  Since the third column is the image of the third basis vector, and by orthogonality the third row is $0, 0, 1$. Now let
  \[A' = \begin{pmatrix}
      a & b\\
      c & d
    \end{pmatrix} \in GL_2(\mathbb{R})
  \]
  with $\mathrm{det} A' = 1$. $A'$ is still orthogonal, so $A'\in SO_2$. Therefore $A'$ is a rotation and
  \[
    A =
    \begin{pmatrix}
      \cos\theta & -\sin\theta & 0\\
      \sin\theta & \cos\theta & 0\\
      0 & 0 & 1
    \end{pmatrix}
  \]
  in some basis, and this is exactly the rotation through an axis.
\end{proof}
  Every matrix in $O_3$ is the product of at most three reflections (in planes through 0).
Note that a rotation is a product of two reflections so every matrix in $O_3$ is a reflection, a rotation or a product of a reflection and a rotation.
\begin{proof}
  Recall $O_3= SO_3 \cup
  \begin{pmatrix}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & -1
  \end{pmatrix}SO_3$.
  So if $A\in SO_3$, we know that $A =
  \begin{pmatrix}
    \cos\theta & -\sin\theta & 0\\
    \sin\theta & \cos\theta & 0\\
    0 & 0 & 1
  \end{pmatrix}$ in some basis, which is a composite of two reflections:
  \[
    A =
    \begin{pmatrix}
      1 & 0 & 0\\
      0 & -1 & 0\\
      0 & 0 & 1
    \end{pmatrix}
    \begin{pmatrix}
      \cos\theta & \sin\theta & 0\\
      \sin\theta & -\cos\theta & 0\\
      0 & 0 & 1
    \end{pmatrix},
  \]
  Then if $A\in \begin{pmatrix}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & -1
  \end{pmatrix}SO(3)$, then it is automatically a product of three reflections.
\end{proof}

Q: Unitary group
A: \[
    I_n= U_n = \{A\in GL_n(\mathbb{C}) \ | \ A^\dagger A = I\}.
  \]
forms a group under matrix multiplication
\begin{proof}
  We show $U_n(\mathbb{C})\leqslant GL_n(\mathbb{C})$ using the super-efficient subgroup criterion: \\
It is non-empty, since $I^\dagger I=I$ so $I\in U_n $. \\
If $A, B\in U_n $, then $(AB^{-1})(AB^{-1})^\dagger = AB^{-1}(B^{-1})^\dagger A^{\dagger} = AB^{-1}BA^{-1} = I$, so $AB^{-1}\in U_n $
\end{proof}

Q: Special Unitary Group
A: \[
\mathrm{Ker}(\mathrm{det}:U_n\to S^1 )=SU_n
\]

Q: Show that Unitary matrices are isometries
A: For $\mathsf{A}\in U_n$ and $\vec{x},\vec{y}\in\mathbb{C}^n$ \[
(\mathsf{A}\vec{x})^\dagger(\mathsf{A}\vec{y})=\vec{x}^\dagger\vec{y} \]
and \[
|\mathsf{A}\vec{x}|=|\vec{x}| \]
i.e. $U_n$ preserves dot product and length
\begin{proof}
\textbf{Dot product}
\begin{align*}
(\mathsf{A}\vec{x})^\dagger(\mathsf{A}\vec{y})&=\vec{x}^\dagger\mathsf{A}^\dagger\mathsf{A}\vec{y} \\
&= \vec{x}^\dagger\mathsf{I}\vec{y} \\
&=\vec{x}^\dagger\vec{y}
\end{align*}
\textbf{Length}
\begin{align*}
|\mathsf{A}\vec{x}|^2&=(\mathsf{A}\vec{x})^\dagger(\mathsf{A}\vec{x}) \\
&=\vec{x}^\dagger\vec{x} \\
&=|\vec{x}|^2 \\
\implies |\mathsf{A}\vec{x}|&=|\vec{x}|
\end{align*}
\end{proof}

Q: M\"obius maps
A: Maps 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 $ad - bc\not= 0$, with $f(-\frac{d}{c}) = \infty$, $f(\infty) = \frac{a}{c}$ when $c\not= 0$ and $f(\infty)=\infty$ if $c = 0$.
We impose $ad - bc\not= 0$ or else the map will be constant (so not invertible): for any $z, w\in \mathbb{C}$
\begin{align*}
f(z) - f(w) &= \frac{(az + b)(cw + d) - (aw + b)(cz + d)}{(cw + d)(cz + d)} \\
&= \frac{(ad - bc)(z - w)}{(cw + d)(cz + d)}
\end{align*}
If $c\not=0$, then $f(-\frac{d}{c})$ involves division by 0. So we add $\infty$ to $\mathbb{C}$ to form the extended complex plane (Riemann sphere) $\mathbb{C}\cup \{\infty\}= \mathbb{C}_\infty$. Then we define $f(-\frac{d}{c}) = \infty$. We call $\mathbb{C}_\infty$ a one-point compactification of $\mathbb{C}$.

Q: Riemann Sphere / Stereographic projection
A: Take the unit sphere, cut it in the equator with the complex plane. Take a rod attached to the north
pole. Then the intersection points of the rod with the complex plane correspond to intersection
points of the rod with the sphere. The north pole corresponds to $\infty$.

Q: Show that M\"obius maps are bijections
A: The M\"obius maps are bijections $\mathbb{C}_\infty \to \mathbb{C}_\infty$.
\begin{proof}
\begin{align*}
f(z) - f(w) &= \frac{(az + b)(cw + d) - (aw + b)(cz + d)}{(cw + d)(cz + d)} \\
&= \frac{(ad - bc)(z - w)}{(cw + d)(cz + d)}
\end{align*}
So $f$ is injective on $\mathbb{C}\setminus \{-\frac{d}{c}\}$ \\
The inverse of $f(z) = \frac{az + b}{cz+ d}$ is $g(z) = \frac{dz - b}{-cz + a}$, which we can check by composition both ways. So $f$ is surjective on $\mathbb{C}\setminus \{\frac{a}{c},-\frac{d}{c}\}$. \\
Now the special cases. \\
$c\neq 0$: $f(-\frac{d}{c})=\infty$, $f(\infty)=\frac{a}{c}$ \\
$c=0$: $f(\infty)=\infty$
\end{proof}

Q: M\"obius group
A: The M\"obius maps form a group $M$ under function composition. (The M\"obius group)
\begin{proof}
\textbf{Closure} If $f_1(z) = \frac{a_1z + b_1}{c_1z + d_1}$ and $f_2(z) = \frac{a_2z + b_2}{c_2 z + d_2}$, then
\begin{align*}
f_2\circ f_1 (z) &= \frac{a_2\left(\frac{a_1z + b_1}{c_1z + d_1}\right) + b_2}{c_2\left(\frac{a_1z + b_1}{c_1z + d_1}\right) + d_2} \\
&= \frac{(a_1a_2 + b_2c_1)z + (a_2b_1 + b_2d_1)}{(c_2a_1 + d_2c_1)z + (c_2b_1 + d_1d_2)}
\end{align*}
Now we have to check that ``$ad - bc \not = 0$'': we have $(a_1a_2 + b_2c_1)(c_2b_1 + d_1d_2) - (a_2b_1 + b_2d_1)(c_2a_1 + d_2c_1) = (a_1d_1 - b_1c_1)(a_2d_2 - b_2c_2)\not =0 $. \\
\emph{Case} $f_2\circ f_1 (-\frac{d_1}{c_1})=f_2(\infty)=\frac{a_2}{c_2}$. This agrees with the formula
\begin{align*}
f_2\circ f_1\left(-\frac{d_1}{c_1}\right) &= \frac{(a_1a_2 + b_2c_1)(-\frac{d_1}{c_1}) + (a_2b_1 + b_2d_1)}{(c_2a_1 + d_2c_1)(-\frac{d_1}{c_1}) + (c_2b_1 + d_1d_2)} \\
&=\frac{a_1(c_1b_1-a_1d_1)}{c_2(b_1c_1-a_1d_1)}=\frac{a_2}{c_2}
\end{align*}
\emph{Case} $f_2\circ f_1 (\infty)=f_2(\frac{a_1}{c_1})=\frac{a_1a_2+b_2c_1}{c_2a_1d_2c_1}$ which agrees with the formula for $z=\infty$. \\
\emph{Case} $c_1=0$. $f_2\circ f_1 (\infty)=f_2(\infty)=\frac{a_2}{c_2}$ which agrees with the formula for $z=\infty$. \\
\emph{Case} $c_2=0$. $f_2\circ f_1 (-\frac{d_1}{c_1})=f_2(\infty)=\infty$ which agrees with the formula
\begin{align*}
f_2\circ f_1\left(-\frac{d_1}{c_1}\right) &= \frac{(a_1a_2 + b_2c_1)(-\frac{d_1}{c_1}) + (a_2b_1 + b_2d_1)}{(d_2c_1)(-\frac{d_1}{c_1}) +  d_1d_2} \\
&=\frac{d_1(a_1a_2+b_2c_1)-c_1(a_2b_1+b_2d_1)}{0} \\
&=\frac{a_2(a_1d_1-b_1c_1)}{0}=\infty
\end{align*}
because the numerator is non-zero for $c_2=0$. \\
\textbf{Identity} $1(z) = \frac{1z + 0}{0 + 1}$ which satisfies $ad - bc \not= 0$. \\
\textbf{Inverses} We can check that that $f^{-1}(z) = \frac{dz - b}{-cz + a}$ with $da - bc\not= 0$, which are also M\"obius maps \\
\textbf{Associativity} Always true for composition of functions
\end{proof}

Q: Relationship between $M$ and matrices
A: The map $\theta: GL_2(\mathbb{C})\to M$ sending $
  \displaystyle \begin{pmatrix}
    a & b\\
    c & d
  \end{pmatrix} \mapsto \frac{az + b}{cz + d}$ is a surjective group homomorphism.
\begin{proof}
  Firstly, since the determinant $ad - bc$ of any matrix in $GL_2(\mathbb{C})$ is non-zero, it does map to a M\"obius map. This also shows that $\theta$ is surjective.
  We have previously calculated that
  \[
    \theta(A_2)\circ \theta(A_1) = \frac{(a_1a_2 + b_2c_1)z + (a_2b_1 + b_2d_1)}{(c_2a_1 + d_2c_1)z + (c_2b_1 + d_1d_2)} = \theta(A_2A_1)
  \]
  So it is a homomorphism.
\end{proof}
The kernel of $\theta$ is
\[
  \mathrm{Ker}(\theta) = \left\{A\in GL_2(\mathbb{C}):z \equiv \frac{az + b}{cz + d}\right\}
\]
We can try different values of $z$: $z = \infty \Rightarrow c = 0$; $z = 0 \Rightarrow b = 0$; $z = 1\Rightarrow d = a$. So
\[
  \mathrm{Ker}\theta = Z = \{\lambda I: \lambda \in \mathbb{C}, \lambda\not= 0\},
\]
where $I$ is the identity matrix and $Z$ is the centre of $GL_2(\mathbb{C})$.
By the isomorphism theorem, we have
\[
  M \cong GL_2(\mathbb{C})/Z
\]
We call $GL_2(\mathbb{C})/Z=PGL_2(\mathbb{C})$ the projective general linear group.  \\
Since $f_A = f_B$ iff $B = \lambda A$ for some $\lambda\not= 0$, if we restrict $\theta$ to $SL_2(\mathbb{C})$, we have $\left.\theta\right|_{SL_2(\mathbb{C})}: SL_2(\mathbb{C})\to M$ is also surjective (as we can rescale any matrix to have determinant $1$). The kernel is now just $\{\pm I\}$. So
\[
  M \cong SL_2(\mathbb{C})/\{\pm I\} = \mathrm{PSL_2}(\mathbb{C})
\]
where $\mathrm{PSL_2}(\mathbb{C})$ is the projective special linear group. \\
Clearly $\mathrm{PSL}_2(\mathbb{C})\cong \mathrm{PGL}_2(\mathbb{C})$ since both are isomorphic to the M\"obius group.

Q: M\"obius maps in terms of geometric transformations
A: Every M\"obius map is a composite of maps of the following form:
  \begin{enumerate}
    \item Dilation/rotation: $f(z) = az$, $a\not= 0$
    \item Translation: $f(z) = z + b$
    \item Inversion: $f(z) = \frac{1}{z}$
\end{enumerate}
\begin{proof}
  Let $\frac{az + b}{cz + d}\in M$.
\textbf{Case 1}: $c=0$ (i.e. $g(\infty)=\infty$). $g(z) = \frac{a}{d}z + \frac{b}{d}$ \\
\textbf{Case 2}: $c\not= 0$. Let $g(\infty)=z_0$. Let $h(z) = \frac{1}{z - z_0}$. Then $hg(\infty) = \infty$ so $hg$ is of the same type as the first case. $h^{-1}(w) = \frac{1}{w} + z_0$. Thus $g = h^{-1} (hg)$ is a composition of the three maps. \\
\end{proof}
 Alternatively
  \[
    z\mapsto z + \frac{d}{c} \mapsto \frac{1}{z + \frac{d}{c}} \mapsto -\frac{ad + bc}{c^2(z + \frac{d}{c})}\mapsto \frac{a}{c} -\frac{ad + bc}{c^2(z + \frac{d}{c})} = \frac{az + b}{cz + d}
  \]

Q: Fixed points of M\"obius maps
A: Any M\"obius map with at least 3 fixed points must be the identity.
\begin{proof}
  Consider $f(z) = \frac{az + b}{cz + d}$. This has fixed points at those $z$ which satisfy
\begin{align*}
\frac{az + b}{cz + d} = z \Leftrightarrow cz^2 + (d - a)z - b = 0
\end{align*}
A quadratic has at most two roots, unless $c = b = 0$ and $d = a$, in which case the equation just says $0 = 0$.
  However, if $c = b= 0$ and $d = a$, then $f$ is just the identity.
\end{proof}
  Every non-identity M\"obius map has exactly 1 or 2 fixed points.
\begin{proof}
  Given $f\in M$ and $f\not= \mathrm{id}$. So $\exists h\in M$ such that $hfh^{-1}(z) = \nu{z}$. Now
\begin{align*}
f(w) = w &\Leftrightarrow hf(w) = h(w)  \\
&\Leftrightarrow hfh^{-1}(h(w)) = h(w)
\end{align*}
 So $h(w)$ is a fixed point of $hfh^{-1}$. Since $h$ is a bijection, $f$ and $hfh^{-1}$ have the same number of fixed points. \\
Obviously $\nu z$ and $z+1$ fixed points exactly $\{0,\infty \}$ and  $\{\infty \}$ respectively.
  So $f$ has exactly $2$ fixed points if $f$ is conjugate to $\nu z$, and exactly 1 fixed point if $f$ is conjugate to $z + 1$.
\end{proof}
It's intuitive that conjugation preserves fixed points. We move the Riemann sphere around by $h$, apply $f$ (that fixes the fixed points) then restore the Riemann sphere to its original orientation. So we have simply moved the fixed point around by $h$.

Q: Conjugacy classes of M\"obius maps
A: Any M\"obius map is conjugate to $f(z) = \nu z$ for some $\nu\not= 0$ or to $f(z) = z + 1$.
\begin{proof}
  We have the surjective group homomorphism $\theta: GL_2(\mathbb{C}) \to M$. The conjugacy classes of $GL_2(\mathbb{C})$ are of types
  \begin{align*}
    \begin{pmatrix}
      \lambda & 0\\
      0 & \mu
    \end{pmatrix} &\mapsto g(z) = \frac{\lambda z + 0}{0z + \mu} = \frac{\lambda}{\mu}z\\
    \begin{pmatrix}
      \lambda & 0\\
      0 & \lambda
    \end{pmatrix} &\mapsto g(z) = \frac{\lambda z + 0}{0z + \lambda} = 1 z\\
    \begin{pmatrix}
      \lambda & 1\\
      0 & \lambda
    \end{pmatrix} &\mapsto g(z) = \frac{\lambda z + 1}{\lambda} = z + \frac{1}{\lambda}
  \end{align*}
  But the last one is not in the form $z + 1$. We know that the last $g(z)$ can also be represented by $
  \begin{pmatrix}
    1 & \frac{1}{\lambda}\\
    0 & 1
  \end{pmatrix}$, which is conjugate to $
  \begin{pmatrix}
    1 & 1\\
    0 & 1
  \end{pmatrix}$ (since that's its Jordan-normal form). So $z + \frac{1}{\lambda}$ is also conjugate to $z + 1$.
\end{proof}

Q: 3-point determination of M\"obius maps
A: Let $f, g\in M$. If $\exists z_1, z_2, z_3\in \mathbb{C}_{\infty}$ such that $f(z_i) = g(z_i)$, then $f = g$. ie. every M\"obius map is uniquely determined by three points.
\begin{proof}
  As M\"obius maps are invertible, write $f(z_i) = g(z_i)$ as $g^{-1}f(z_i) = z_i$. So $g^{-1}f$ has three fixed points. So $g^{-1}f$ must be the identity. So $f = g$.
\end{proof}

Q: Three-transitive action
A: An action of $G$ on $X$ is called \emph{three-transitive} if the induced action on $\{(x_1, x_2, x_3)\in X^3\ | \ x_i\text{ pairwise disjoint}\}$, given by $g(x_1, x_2, x_3) = (g(x_1), g(x_2), g(x_3))$, is transitive.
  This means that for any two triples $x_1, x_2, x_3$ and $y_1, y_2, y_3$ of distinct elements of $X$, there exists $g\in G$ such that $g(x_i) = y_i$.
  If this $g$ is always unique, then the action is called \emph{sharply three transitive}

Q: Show that $M$ is three-transitive
A: The M\"obius group $M$ acts sharply three-transitively on $\mathbb{C}_\infty$.
\begin{proof}
  We want to show that we can send any three points to any other three points. However, it is easier to show that we can send any three points to $0, 1, \infty$.
  Suppose we want to send $z_1\to \infty, z_2\mapsto 0, z_3 \mapsto 1$. Then the following works:
  \[
    f(z) = \frac{(z - z_2)(z_3 - z_1)}{(z - z_1)(z_3 - z_2)}
  \]
  If any term $z_i$ is $\infty$, we simply remove the terms with $z_i$, eg. if $z_1 = \infty$, we have $f(z) = \frac{z - z_2}{z_3 - z_2}$.
  So given also $w_1, w_2, w_3$ distinct in $\mathbb{C}_\infty$ and $g\in M$ sending $w_1\mapsto \infty, w_2\mapsto 0, w_3\mapsto 1$, then we have $g^{-1}f(z_i) = w_i$.
  The uniqueness of the map $g^{-1}f$ follows from the fact that a M\"obius map is uniquely determined by 3 points.
\end{proof}

Q: The equation of a circle/line in $\mathbb{C}$
A: The general equation of a circle or straight line in $\mathbb{C}$ is
  \[
    Az\bar z + \bar Bz + B\bar z + C = 0,
  \]
  where $A, C\in \mathbb{R}$ and $|B|^2 > AC$.
$A = 0$ gives a straight line. If $A \not= 0, B = 0$, we have a circle centered at the origin. If $C = 0$, the circle passes through 0.
\begin{proof}
 \textbf{Circle}
\begin{align*}
&|z + B| = r \ ( B\in\mathbb{C},r\in\mathbb{R}>0) \\
\implies &|z + B|^2 = r^2 \\
\implies &(z+B) \overline{(z+B)}=r^2 \\
\implies &(z+B)(\bar z + \bar B)=r^2 \\
\implies &z\bar z + \bar B z + B\bar z + B\bar B-r^2 =0
\end{align*}
where $|B|^2>AC=|B|^2-r^2 \iff r^2\in\mathbb{R}>0$.
\textbf{Lines}
\begin{align*}
&|z - a| = |z - b| \ (a\neq b) \\
\implies &|z - a|^2 = |z - b|^2 \\
\implies &z\bar z -a\bar z -\bar a z +a \bar a =z\bar z -b\bar z -\bar b z +b\bar b \\
\implies &(\bar b-\bar a)z+(b-a)\bar z+a\bar a + b\bar b=0  \\
\implies &\bar B z + B \bar z + C=0
\end{align*}
where $|B|^2=|b-a|^2>AC=0 \iff b\neq a$.
\end{proof}

Q: M\"obius maps on circles/lines
A: M\"obius maps send circles/straight lines to circles/straight lines. Note that it can send circles to straight lines and vice versa.
  Alternatively, M\"obius maps send circles on the Riemann sphere to circles on the Riemann sphere.
\begin{proof}
\textbf{Method 1} $w = \frac{az + b}{cz + d}\Leftrightarrow z = \frac{dw - b}{-cw + a}$ so substituting $z$ into the circle equation gives $A' w\bar w + \bar B' w + B'\bar w + C' = 0$ with $A', C'\in \mathbb{R}$.
\textbf{Method 2} We know that each M\"obius map is a composition of translation, dilation/rotation and inversion. We can check for each of the three types. Clearly dilation/rotation and translation maps a circle/line to a circle/line. So we simply do inversion: if $w = z^{-1}$
  \begin{align*}
    &\; Az\bar z + \bar Bz + B\bar z + C = 0\\
    \Leftrightarrow &\; Cw\bar w + Bw + \bar B\bar w + A = 0
  \end{align*}
\end{proof}

Q: Finding loci of M\"obius maps
A: Use three point determination. Usually $0,1,\infty$ are easiest. M\"obius maps are smooth. \\
Direction is preserved as we walk between points.

Q: Cross ratio
A: Given four distinct points $z_1, z_2, z_3, z_4\in \mathbb{C}_\infty,$ their \emph{cross-ratio} is $[z_1, z_2, z_3, z_4] = g(z_4)$, with $g$ being the unique M\"obius map that maps $z_1\mapsto \infty, z_2\mapsto 0, z_3\mapsto 1$. So $[\infty, 0, 1, \lambda] = \lambda$ for any $\lambda\not= \infty, 0, 1$. We have
  \[
    [z_1, z_2, z_3, z_4] = \frac{z_4 - z_2}{z_4 - z_1} \cdot \frac{z_3 - z_1}{z_3 - z_2}
  \]
  (where if $z_i=\infty$ then its two factors are left out).

Q: Relationship between M\"obius maps and cross-ratios
A: If $f\in M$, then $[z_1, z_2, z_3, z_4] = [f(z_1), f(z_2), f(z_3), f(z_4)]$.
\begin{proof}
Let $g$ be the unique M\"obius map such that $[z_1, z_2, z_3, z_4] = g(z_4)$, ie.
  \begin{align*}
    z_1 &\mapsto  \infty \leftarrow f(z_1)\\
    z_2 &\mapsto 0 \leftarrow f(z_2)\\
    z_3 &\mapsto 1 \leftarrow f(z_3)
  \end{align*}
To calculate $[f(z_1), f(z_2), f(z_3), f(z_4)]$ we need to find the (unique) function which gives the 3 left mappings as shown. We can recognise that $gf^{-1}$ works. So
\begin{align*}
[f(z_1), f(z_2), f(z_3), f(z_4)] &= gf^{-1}[f(z_4)] \\
&= g(z_4) \\
&= [z_1, z_2, z_3, z_4]
\end{align*}
\end{proof}
In fact, $\exists f: f(z_i)=w_i \iff [z_1, z_2, z_3, z_4]=[f(z_1), f(z_2), f(z_3), f(z_4)]$
\begin{proof}
If such an $f$ exists, then as M\"obius maps preserve cross-ratio, $[z_1, z_2, z_3, z_4]=[f(z_1), f(z_2), f(z_3), f(z_4)]$. \\
Conversely, if $[z_1, z_2, z_3, z_4]=[f(z_1), f(z_2), f(z_3), f(z_4)]$ such that $g(z_4)=h(z_4)=\lambda$, then
 \begin{align*}
    z_1 &\mapsto  \infty \leftarrow w_1\\
    z_2 &\mapsto 0 \leftarrow w_2\\
    z_3 &\mapsto 1 \leftarrow w_3\\
    z_4 &\mapsto \lambda \leftarrow w_4
  \end{align*}
so we can map $z_i$ to $w_i$ by $f=h^{-1}g$.
\end{proof}

Q: Double transpositions preserve cross ratio
A: For $z_1, z_2, z_3, z_4\in \mathbb{C}_\infty$ all distinct, then
  \[
    [z_1, z_2, z_3, z_4] = [z_2, z_1, z_4, z_3] = [z_3, z_4, z_1, z_2] = [z_4, z_3, z_2, z_1]
  \]
simply by inspection of the symmetry of the formula
 \[
    [z_1, z_2, z_3, z_4] = \frac{z_4 - z_2}{z_4 - z_1} \cdot \frac{z_3 - z_1}{z_3 - z_2}
  \]

Q: Condition for colinear points (or points on the same cirle
A: $z_1, z_2, z_3, z_4$ lie on some circle/straight line iff $[z_1, z_2, z_3, z_4]\in \mathbb{R}$.
\begin{proof}
  Let $C$ be the circle/line through $z_1, z_2, z_3$. Let $g$ be the unique M\"obius map with $g(z_1) = \infty$, $g(z_2) = 0$, $g(z_3) = 1$. Then $g(z_4) = [z_1, z_2, z_3, z_4]$ by definition.
The circle/line defined by $\infty,0,1$ is the real line. So $g(z_1),g(z_2),g(z_3),g(z_4)$ lie on a circle/line iff $g(z_4)\in\mathbb{R}$.  \\
As M\"obius maps preserve circle/lines, $z_1,z_2,z_3,z_4$ lie on a circle/line iff $g(z_1),g(z_2),g(z_3),g(z_4)$ lie on a circle/line. \\
Therefore, $z_1,z_2,z_3,z_4$ lie on a circle/line iff $g(z_4)\in\mathbb{R}$.
\end{proof}