hcju/proofwikiqa huggingface dataset:
Let $x \mathrel {\mathcal R_G} y \iff y \in \Orb x$. Checking in turn each of the critera for equivalence: === Reflexivity === $x = e * x \implies x \in \Orb x$ from the definition of group action. Thus $\mathcal R_G$ is reflexive. {{qed|lemma}} === Symmetry === {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \Orb x       | c =  }} {{eqn | ll= \leadsto       | l = \exists g \in G: y       | r = g * x       | c =  }} {{eqn | ll= \leadsto       | l = g^{-1} * \paren {g * x}       | r = g^{-1} * y       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = g^{-1} * y       | c =  }} {{eqn | ll= \leadsto       | l = \exists g^{-1} \in G: x       | r = g^{-1} * y       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \Orb y       | c =  }} {{end-eqn}} Thus $\mathcal R_G$ is symmetric. {{qed|lemma}} === Transitivity === {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \Orb x, z \in \Orb y       | c =  }} {{eqn | ll= \leadsto       | l = \exists g_1 \in G: y       | r = g_1 * x, \exists g_2 \in G: z = g_2 * y       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = g_2 * \paren {g_1 * x}       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = \paren {g_2 g_1} * x       | c =  }} {{eqn | ll= \leadsto       | l = z       | o = \in       | r = \Orb x       | c =  }} {{end-eqn}} Thus $\mathcal R_G$ is transitive. {{qed|lemma}} So $\mathcal R_G$ has been shown to be an equivalence relation.  Hence the result, by definition of an equivalence class. {{Qed}}
Follows trivially from the fact that the Group Action Induces Equivalence Relation. {{explain|cite a result about equiv rels and partitions, and one about cardinality and partitions}} {{qed}} Category:Group Actions Category:Named Theorems pwvgvk2c6p89r84hq9vudcd3u47firc
From the {{GroupActionAxiom|2}}: :$e * x = x \implies e \in \Stab x$ and so $\Stab x$ cannot be empty. Let $g, h \in \Stab x$. {{begin-eqn}} {{eqn | l = g, h       | o = \in       | r = \Stab x       | c =  }} {{eqn | ll= \leadsto       | l = g * x       | r = x       | c = {{Defof|Stabilizer}} of $x$ by $G$ }} {{eqn | lo= \land       | l = h * x       | r = x       | c = {{Defof|Stabilizer}} of $x$ by $G$ }} {{eqn | ll= \leadsto       | l = g * \paren {h * x}       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \paren {g \circ h} * x       | r = x       | c = {{GroupActionAxiom|1}} }} {{eqn | ll= \leadsto       | l = g \circ h       | o = \in       | r = \Stab x       | c = {{Defof|Stabilizer}} of $x$ by $G$ }} {{end-eqn}} Let $g \in \Stab x$. Then: :$x = \paren {g^{-1} \circ g} * x = g^{-1} * \paren {g * x} = g^{-1} * x$ Hence $g^{-1} \in \Stab x$. Thus the conditions for the Two-Step Subgroup Test are fulfilled, and $\Stab x \le G$. {{qed}}
Let us define the mapping: :$\phi: G \to \Orb x$ such that: :$\map \phi g = g * x$ where $*$ denotes the group action. It is clear that $\phi$ is surjective, because from the definition $x$ was acted on by all the elements of $G$. Next, from Stabilizer is Subgroup: Corollary: :$\map \phi g = \map \phi h \iff g^{-1} h \in \Stab x$ This means: :$g \equiv h \pmod {\Stab x}$ Thus there is a well-defined bijection: :$G \mathbin / \Stab x \to \Orb x$ given by: :$g \, \Stab x \mapsto g * x$ So $\Orb x$ has the same number of elements as $G \mathbin / \Stab x$. That is: :$\order {\Orb x} = \index G {\Stab x}$ The result follows. {{qed}}
Let $x \in X$. Let $\phi: \Orb x \to G \mathbin / \Stab x$ be a mapping from the orbit of $x$ to the left coset space of $\Stab x$ defined as: :$\forall g \in G: \map \phi {g * x} = g \, \Stab x$ where $*$ is the group action. Note: this is ''not'' a homomorphism because $\Orb x$ is not a group. Suppose $g * x = h * x$ for some $g, h \in G$. Then: :$h^{-1} g * x = h^{-1} h * x$ and so: :$h^{-1} g * x = x$ Thus: :$h^{-1} g \in \Stab x$ so by Left Coset Space forms Partition: :$g \, \Stab x = h \, \Stab x$ demonstrating that $\phi$ is well-defined. Let $\map \phi {g_1 * x} = \map \phi {g_2 * x}$ for some $g_1, g_2 \in G$. Then: :$g_1 \, \Stab x = g_2 \, \Stab x$ and so by Left Coset Space forms Partition: :$g_2^{-1} g_1 \in \Stab x$ So by definition of $\Stab x$: :$x = g_2^{-1} g_1 * x$ Thus: {{begin-eqn}} {{eqn | l = g_2 * x       | r = g_2 * \paren {g_2^{-1} g_1 * x}       | c = applying $g_2$ to both sides }} {{eqn | r = \paren {g_2 g_2^{-1} g_1} x       | c = {{Defof|Group Action}} }} {{eqn | r = g_1 * x       | c =  }} {{end-eqn}} thus demonstrating that $\phi$ is injective. As the left coset $g \, \Stab x$ is $\map \phi {g * x}$ by definition of $\phi$, it follows that $\phi$ is a surjection. The result follows. {{qed}}
First it is necessary to prove that this is a group action. The action is the same as the one defined in Group Action on Coset Space, but this time we are limiting ourselves to the subsets of $G$ which have the same number of elements. From the result in Group Action on Coset Space, we only have to prove that: :$\card {g S} = \card S$ This follows directly from Order of Subset Product with Singleton. It remains to be shown that: :$\order {\Stab S} \divides \card S$ By definition of stabilizer: :$\Stab S = \set {g \in G: g S = S}$ It follows that: :$\forall s \in S: g s \in S$ Then: :$\Stab S s \subseteq S$ Thus: :$\card {\Stab S s} \le \card S$ We have from corollary $1$ to Stabilizer is Subgroup that the stabilizer contains the identity: :$e \in \Stab S \le G$ It therefore follows that: :$s \in S \implies s \in \Stab S s$ Thus: :$\displaystyle S = \bigcup_{t \mathop \in S} \Stab S t$ By Congruence Class Modulo Subgroup is Coset, distinct right cosets are disjoint. Thus $S$ consists of a union of disjoint right cosets of $\Stab S$. Thus $\displaystyle \bigcup_{t \mathop \in S} \Stab S t$ is a partition of $S$. Therefore: :$\forall s \in S: \card {\Stab S s} \divides \card S$ As we have: :$\forall s \in S: \card {\Stab S s} = \order {\Stab S}$ the result follows: :$\card {\Stab S} \divides \card S$ {{qed}}
The identity permutation takes each element of $X$ to itself, thus fulfilling {{GroupActionAxiom|1}}. The group operation in $S_n$ ensures fulfilment of {{GroupActionAxiom|2}}. {{qed}}
Let $h \in G_0$. {{begin-eqn}}      {{eqn | l = h       | o = \in       | r = G_0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = h       | o = \in       | r = \set {g \in G: \forall x \in X: g \cdot x = x}       | c = {{Defof|Kernel of Group Action}} }} {{eqn | ll= \leadstoandfrom       | lo= \forall x \in X:       | l = h       | o = \in       | r = \set {g \in G: g \cdot x = x}       | c =  }} {{eqn | ll= \leadstoandfrom       | lo= \forall x \in X:       | l = h       | o = \in       | r = \Stab x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = h       | o = \in       | r = \bigcap_{x \mathop \in X} \Stab x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = G_0       | r = \bigcap_{x \mathop \in X} \Stab x       | c =  }} {{end-eqn}} From Stabilizer is Subgroup: :$\Stab x \le G$ Thus $G_0$ is the intersection of subgroups. By Intersection of Subgroups is Subgroup: :$G_0 \le G$ To prove normality it is sufficient to show: :$\forall g \in G: g G_0 g^{-1} = G_0$ Let $h \in G_0, g \in G$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \paren {g h g^{-1} } \cdot x       | r = g \cdot \paren {h \cdot \paren { g^{-1} \cdot x } }       | c = Group Action Associates with Group Operation: {{GroupActionAxiom|1}} }} {{eqn | r = g \cdot \paren {g^{-1} \cdot x}       | c = because $h \in \Stab { g^{-1} \cdot x}$ }} {{eqn | r = \paren {g \cdot  g^{-1} } \cdot x       | c = Group Action Associates with Group Operation: {{GroupActionAxiom|1}} }} {{eqn | r = x       | c = $e \cdot x = x$: {{GroupActionAxiom|2}} }} {{end-eqn}} Therefore: :$g h g^{-1} \in G_0$ so: :$g G_0 g^{-1} \subseteq G_0$ Conversely suppose that $h \in G_0$. Then by the above: :$h' = g^{-1} h g \in G_0$ Therefore: :$h = g h' g^{-1} \in g G_0 g^{-1}$ and so: :$G_0 \subseteq g G_0 g^{-1}$ This concludes the proof. {{qed}}
$G$ acts effectively on the left coset space $G / H$ {{iff}} $a H \mapsto g a H$ is faithful, {{iff}}: {{begin-eqn}} {{eqn | lo= \forall g \in G: \forall a H \in G / H:       | l = g a H = a H       | o = \implies       | r = g = e       | c = {{Defof|Faithful Group Action}} }} {{eqn | ll= \leadstoandfrom       | lo= \forall g \in G: \forall a \in G:       | l = a^{-1} g a \in H       | o = \implies       | r = g = e       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom       | lo= \forall g \in G: \forall a \in G:       | l = g \in a H a^{-1}       | o = \implies       | r = g = e }} {{eqn | ll= \leadstoandfrom       | lo= \forall g \in G: \forall a \in G:       | l = g \in H^a       | o = \implies       | r = g = e       | c = {{Defof|Conjugate of Group Subset}} }} {{eqn | ll= \leadstoandfrom       | l = \bigcap_{a \mathop \in G} H^a       | r  = \set e       | c = {{Defof|Intersection of Family}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = g * e       | r = g e g^{-1}       | c =  }} {{eqn | r = g g^{-1}       | c =  }} {{eqn | r = e       | c =  }} {{end-eqn}} So the only conjugate of $e$ is $e$ itself. Thus: :$\Orb e = \set e$ and so: :$\order {\Orb e} = 1$ From the Orbit-Stabilizer Theorem, it follows immediately that: :$\Stab e = G$ {{qed}}
Let $\order G = n$ such that $p \divides n$. Let: :$X = \set {\tuple {a_1, a_2, \ldots, a_p} \in G^p: a_1 a_2 \cdots a_p = e}$ where $G^p$ is the cartesian product $\underbrace {G \times G \times \cdots \times G}_p$. The first $p - 1$ coordinates of an element of $X$ can be chosen arbitrarily. The last coordinate is determined by the fact that: :$a_1 a_2 \cdots a_{p - 1} = a_p^{-1}$ So from the Product Rule for Counting, it follows that: :$\card X = n^{p - 1}$ Let $C_p$ be a cyclic group of order $p$ generated by the element $c$. Let $C_p$ act on the set $X$ by the rule: :$c * \tuple {a_1, a_2, \ldots, a_p} = \tuple {a_2, a_3, \ldots, a_p, a_1}$ By the Orbit-Stabilizer Theorem, the number of elements in any orbit is a divisor of the order of $C_p$, which is $p$. As $p$ is prime, an orbit has either $p$ elements or $1$ element by definition. Let $r$ be the number of orbits with one element. Let $s$ be the number of orbits with $p$ elements. Then by the Partition Equation: :$r + s p = n^{p - 1} = \card X$ By hypothesis, $p \divides n$, so: :$r + s p = n^{p - 1} \implies p \divides r$ We know that $r \ne 0$ because, for example, the orbit of $\tuple {e, e, \ldots, e} \in X$ has only one element. So there must be at least $p$ orbits with only one element. Each such element has the form $\tuple {a, a, \ldots, a} \in X$ so $a^p = e$. So $G$ contains at least $p$ elements $x$ satisfying $x^p = e$. So $G$ contains an element $a \ne e$ such that $a^p = e$. That is, $a$ must have order $p$. {{qed}}
By the corollary to the First Sylow Theorem, $G$ has subgroups of order $p^r$ for all $r$ such that $p^r \divides \order G$. Thus $G$ has at least one subgroup $H$ of order $p$. As a Prime Group is Cyclic, $H$ is a cyclic group. Thus by definition $H$ has an element of order $p$. Hence the result. {{qed}}
Let $\pi = \begin{bmatrix} k_1 & k_2 \end{bmatrix}$ be a transposition. Writing $\pi \pi$ in cycle notation gives: :$\begin{bmatrix} k_1 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_2 \end{bmatrix}$ from which we see that $k_1 \to k_2 \to k_1$ and $k_2 \to k_1 \to k_2$. The result follows from the definition of self-inverse. {{qed}} Category:Symmetric Groups lehv6tztngn5ci5vftba2lh34nrl8ch
The cycle $\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix}$ has the factorisation: :$\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix} = \begin{pmatrix} 1 & k \end{pmatrix} \ldots \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix}$ Therefore, the general $k$-cycle $\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix}$ has the factorisation: :$\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix} = \begin{pmatrix} i_1 & i_k \end{pmatrix} \ldots \begin{pmatrix} i_1 & i_3 \end{pmatrix} \begin{pmatrix} i_1 & i_2 \end{pmatrix}$ The cycle $\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix}$ also has the factorisation: :$\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix} = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix} \ldots \begin{pmatrix} k - 1 & k \end{pmatrix}$ Therefore, the general $k$-cycle $\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix}$ also has the factorisation: :$\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix} = \begin{pmatrix} i_1 & i_2 \end{pmatrix} \begin{pmatrix} i_2 & i_3 \end{pmatrix} \ldots \begin{pmatrix} i_{k - 1} & i_k \end{pmatrix}$ {{qed}}
If $\exists i, j \in \N_n$ such that $x_i = x_j$, then $\map {\Delta_n} {x_1, x_2, \ldots, x_n} = 0$ and the result follows trivially. So, suppose all the elements $x_k$ are distinct. Let us use $\Delta_n$ to denote $\map {\Delta_n} {x_1, x_2, \ldots, x_n}$. Let $1 \le a < b \le n$. Then $x_a - x_b$ is a divisor of $\Delta_n$. Then $x_{\map \pi a} - x_{\map \pi b}$ is a factor of $\pi \cdot \Delta_n$. There are two possibilities for the ordering of $\map \pi a$ and $\map \pi b$: Either $\map \pi a < \map \pi b$ or $\map \pi a > \map \pi b$. If the former, then $x_{\map \pi a} - x_{\map \pi b}$ is a factor of $\Delta_n$. If the latter, then $-\paren {x_{\map \pi a} - x_{\map \pi b} }$ is a factor of $\Delta_n$. The same applies to all factors of $\Delta_n$. Thus: {{begin-eqn}} {{eqn | l = \pi \cdot \Delta_n       | r = \pi \cdot \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j}       | c =  }} {{eqn | r = \pm \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j}       | c =  }} {{eqn | r = \pm \Delta_n       | c =  }} {{end-eqn}} {{Qed}}
We need to show that: :$\forall \pi, \rho \in S_n: \map \sgn \pi \, \map \sgn \rho = \map \sgn {\pi \rho}$ Let $\Delta_n$ be an arbitrary product of differences. {{begin-eqn}} {{eqn | l = \map \sgn {\pi \rho} \Delta_n       | r = \pi \rho \cdot \Delta_n       | c = {{Defof|Sign of Permutation}} }} {{eqn | r = \pi \cdot \paren {\rho \cdot \Delta_n}       | c = Permutation on Polynomial is Group Action }} {{eqn | r = \pi \cdot \paren {\map \sgn \rho \cdot \Delta_n}       | c = {{Defof|Sign of Permutation}} }} {{eqn | r = \pi \, \map \sgn \rho \cdot \Delta_n       | c = Permutation on Polynomial is Group Action }} {{eqn | r = \map \sgn \pi \, \map \sgn \rho \Delta_n       | c = {{Defof|Sign of Permutation}} }} {{end-eqn}} As $\struct {\set {1, -1}, \times}$ is the parity group, the result follows immediately. {{Qed}}
Let $\pi, \rho \in S_n$. Let $\pi * f$ be the permutation on the polynomial $f$ by $\pi$. Let $e \in S_n$ be the identity of $S_n$. From Symmetric Group is Group: :$e * f = f$ thus fulfilling {{GroupActionAxiom|1}}. Then we have that: {{begin-eqn}} {{eqn | l = \paren {\pi \circ \rho} * f       | r = \map \pi {\rho * f}       | c =  }} {{eqn | r = \pi * \paren {\rho * f}       | c =  }} {{end-eqn}} thus fulfilling {{GroupActionAxiom|2}} {{qed}}
From Parity Function is Homomorphism: :$\map \sgn {I_{S_n} } = 1$ Thus: :$\pi \pi^{-1} = I_{S_n} \implies \map \sgn \pi \, \map \sgn {\pi^{-1} } = 1$ The result follows immediately. {{qed}}
As $\map \sgn \pi = \pm 1$ for any $\pi \in S_n$, we can apply the laws of commutativity and associativity: {{begin-eqn}} {{eqn | l = \map \sgn \pi \, \map \sgn \rho       | r = \map \sgn \rho \, \map \sgn \pi       | c =  }} {{eqn | ll= \leadsto       | l = \map \sgn \pi \, \map \sgn \rho \, \map \sgn {\pi^{-1} }       | r = \map \sgn \rho \, \map \sgn \pi \, \map \sgn {\pi^{-1} }       | c =  }} {{eqn | ll= \leadsto       | l = \map \sgn {\pi \rho \pi^{-1} }       | r = \map \sgn \rho       | c = Parity Function is Homomorphism }} {{end-eqn}} {{Qed}}
From Transposition is of Odd Parity, any transposition is of odd parity. From K-Cycle can be Factored into Transpositions, we see that a $k$-cycle is the product of $k - 1$ transpositions. Thus $\pi$ is even {{iff}} $k$ is odd. {{qed}}
We have that $\map \sgn {S_n}$ is onto $C_2$. Thus from the First Isomorphism Theorem, $A_n$ consists of the set of even permutations of $S_n$. The result follows from Subgroup of Index 2 is Normal. {{Qed}}
From Parity Function is Homomorphism, the mapping: :$\sgn : G \to \Z_2$ is a homomorphism. Then: :$G / \map \ker \sgn \cong \Img \sgn$ from the First Isomorphism Theorem. The only possibilities for $\Img \sgn$ are $\set 0$ or $\Z_2$. So either: :$\order {G / \map \ker \sgn} = 1$ in which case all the permutations of $G$ are even, or: :$\order {G / \map \ker \sgn} = 2$ in which case exactly half of them are. {{qed}} Category:Symmetric Groups jktrlc08nm231gxlx8dba3n056hbeqi
Consider the effect of $\pi \rho \pi^{-1}$ on $\map \pi i$: :$\map {\pi \rho \pi^{-1} } {\map \pi i} = \map \pi {\map \rho i}$ That is: :$\pi \rho \pi^{-1}$ maps $\map \pi i$ to $\map \pi {\map \rho i}$ In the cycle decomposition of $\pi \rho \pi^{-1}$, $\map \pi i$ lies to the left of $\map \pi {\map \rho i}$, whereas in the cycle decomposition of $\rho$, $i$ lies to the left of $\map \rho i$. {{qed}}
Let $\sigma \in G$ have cycle type $\tuple {k_1, k_2, \ldots, k_n}$. From Existence and Uniqueness of Cycle Decomposition, $\sigma$ can be expressed uniquely as the product of disjoint cycles: :$\sigma = \alpha_1 \alpha_2 \dotsm \alpha_l$ where $\alpha_i$ is a $k_i$-cycle. Let $\tau \in G$ such that $\rho = \tau \sigma \tau^{-1}$. Then: {{begin-eqn}} {{eqn | l = \tau \sigma \tau^{-1}       | r = \tau \alpha_1 \alpha_2 \dotsm \alpha_l \tau^{-1}       | c =  }} {{eqn | n = 1       | r = \tau \alpha_1 \tau^{-1} \tau \alpha_2 \tau^{-1} \dotsm \tau \alpha_l \tau^{-1}       | c = Product of Conjugates equals Conjugate of Products }} {{eqn | r = \alpha_{\map \tau 1} \alpha_{\map \tau 2} \dotsm \alpha_{\map \tau l}       | c = Cycle Decomposition of Conjugate }} {{end-eqn}} We have that for $i, j \in \set {1, 2, \ldots, l}$, $\alpha_i$ and $\alpha_j$ are disjoint. We also have that $\tau$ is a bijection, and so an injection. It follows that $\alpha_{\map \tau i}$ and $\alpha_{\map \tau j}$ are also disjoint for all $i, j \in \set {1, 2, \ldots, l}$. Thus: :the product in $(1)$ is $\rho$ written as the product of disjoint cycles where: :$\tau \alpha_i \tau^{-1}$ is a $k_i$-cycle. Thus $\rho$ has the same cycle type as $\sigma$. {{qed}}
Let $\mathbf A$ be a Hermitian matrix. Then, by definition: : $\mathbf A = \mathbf A^*$ where $\mathbf A^*$ denotes the conjugate transpose of $\mathbf A$. Let $\lambda$ be an eigenvalue of $\mathbf A$. Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$. By definition of eigenvector: : $\mathbf{A v} = \lambda \mathbf v$ Left-multiplying both sides by $\mathbf v^*$, we obtain: :$(1):\quad \mathbf v^* \mathbf{A v} = \mathbf v^* \lambda \mathbf v = \lambda \mathbf v^* \mathbf v$ Firstly, note that both $\mathbf v^* \mathbf{A v}$ and $\mathbf v^* \mathbf v$ are $1 \times 1$-matrices. {{explain|Links to the proof of this fact}} Now observe that, using Conjugate Transpose of Matrix Product: General Case: :$\left({\mathbf v^* \mathbf{A v}}\right)^* = \mathbf v^* \mathbf A^* \left({\mathbf v^*}\right)^*$ As $\mathbf A$ is Hermitian, and $\left({\mathbf v^*}\right)^* = \mathbf v$ by Conjugate Transpose is Involution, it follows that: :$\mathbf v^* \mathbf A^* \left({\mathbf v^*}\right)^* = \mathbf v^* \mathbf{A v}$ That is, $\mathbf v^* \mathbf{A v}$ is also Hermitian. By Product with Conjugate Transpose Matrix is Hermitian, $\mathbf v^* \mathbf v$ is Hermitian.  So both $\mathbf v^* \mathbf{A v}$ and $\mathbf v^* \mathbf v$ are Hermitian $1 \times 1$ matrices.  Now suppose that we have for some $a,b \in \C$: :$\mathbf v^* \mathbf{A v} = \begin{bmatrix}a\end{bmatrix}$ :$\mathbf v^* \mathbf v = \begin{bmatrix}b\end{bmatrix}$ Note that $b \ne 0$ as an eigenvector is by definition non-zero. By definition of a Hermitian matrix: : $a = \bar a$ and $b = \bar b$ where $\bar a$ denotes the complex conjugate of $a$. By Complex Number equals Conjugate iff Wholly Real, it follows that $a, b \in \R$, that is, are real. From equation $(1)$, it follows that $\begin{bmatrix}a\end{bmatrix} = \lambda \begin{bmatrix}b\end{bmatrix}$.  Thus, $a = \lambda b$, i.e. $\lambda = \dfrac a b$ (recall that $b \ne 0$). Hence $\lambda$, being a quotient of real numbers, is real. {{qed}}
Let $\mathbf A$ be a Hermitian matrix. Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates the conjugate transpose. Let $\lambda$ be an eigenvalue of $\mathbf A$. Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$ of $\mathbf A$. Denote with $\left\langle{\cdot, \cdot}\right\rangle$ the inner product on $\C$.  {{begin-eqn}} {{eqn | l = \lambda * \left\langle v, v\right\rangle       | r = \left\langle \lambda*v, v\right\rangle       | c = Properties of Complex Inner Product }} {{eqn | r = \left\langle \mathbf A*v, v\right\rangle       | c = {{Defof|Eigenvector}}: $\lambda*v = \mathbf A*v$ }} {{eqn | r = \left\langle v, \mathbf A^\dagger * v\right\rangle       | c = Properties of Adjugate }} {{eqn | r = \left\langle v, \mathbf A*v\right\rangle       | c = $\mathbf A$ is Hermitian, so $\mathbf A^\dagger = \mathbf A$ }} {{eqn | r = \left\langle v, \lambda*v\right\rangle       | c = {{Defof|Eigenvector}}: $\lambda*v = \mathbf A*v$ }} {{eqn | r = \overline{\lambda}*\left\langle v, v\right\rangle       | c = Properties of Complex Inner Product }} {{end-eqn}} We have that $v \ne 0$, and because of the positive definiteness, it must be that: : $\left\langle v, v\right\rangle \ne 0$ {{explain|A link is needed to whatever it is ($\left\langle v, v\right\rangle$, presumably) is positive definite.}} Thus: : $\left\langle v, v\right\rangle \ne 0$ So we can divide both sides by $\left\langle v, v\right\rangle$. Thus $\lambda = \overline{\lambda}$. By Complex Number equals Conjugate iff Wholly Real, $\lambda$ is a real number. $\lambda$ was arbitrary, so it follows that every eigenvalue is a real number. Hence the result. {{qed}}
First we need to show that $\theta_g$ is well-defined and injective. {{begin-eqn}} {{eqn | l = x H       | r = y H       | c =  }} {{eqn | ll= \leadstoandfrom       | l = y^{-1} x       | o = \in       | r = H       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {g y}^{-1} g x       | r = y^{-1} x \in H       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map {\theta_g} {y H}       | r = \map {\theta_g} {x H}       | c =  }} {{end-eqn}} Thus $\theta_g$ is well-defined and injective. Then we see that $\forall x H \in \mathbb S: \map {\theta_g} {g^{-1} x H} = x H$, so $\theta_g$ is surjective. Thus $\theta_g$ is a well-defined bijection on $\mathbb S$, and therefore a permutation on $\mathbb S$. Next we see: {{begin-eqn}} {{eqn | l = \map {\theta_{u v} } {x H}       | r = u v x H       | c =  }} {{eqn | r = \map {\theta_u} {v x H}       | c =  }} {{eqn | r = \map {\theta_u} {\map {\theta_v }{x H} }=       | c =  }} {{end-eqn}} This shows that $\theta_{u v} = \theta_u \theta_v$, and thus: :$\map \theta {u v} = \map \theta u \, \map \theta v$ Thus $\theta$ is a homomorphism. Now to calculate $\map \ker \theta$: {{begin-eqn}} {{eqn | l = \map \ker \theta       | r = \set {g \in G: \theta_g = I_\mathbb S}       | c =  }} {{eqn | r = \set {g \in G: \forall x \in G: \map {\theta_g} {x H} = x H}       | c =  }} {{eqn | r = \set {g \in G: \forall x \in G: g x h = x H}       | c =  }} {{eqn | r = \set {g \in G: \forall x \in G: x^{-1} g x h = H}       | c =  }} {{eqn | r = \set {g \in G: \forall x \in G: x^{-1} g x \in H}       | c =  }} {{eqn | r = \set {g \in G: \forall x \in G: g \in x H x^{-1} }       | c =  }} {{eqn | r = \bigcap_{x \mathop \in G} x H x^{-1}       | c =  }} {{end-eqn}} as required. {{Qed}}
Let $H = \set e$. We can apply Permutation of Cosets to $H$ so that: :$\mathbb S = G$ and: :$\map \ker \theta = \set e$ The result follows by the First Isomorphism Theorem for Groups. {{Qed}}