hcju/proofwikiqa huggingface dataset:
By definition, the order of a group is the cardinality of its underlying set. By definition, the underlying set of $\struct {\Z_m, +_m}$ is the set of residue classes $\Z_m$: :$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$ From Cardinality of Set of Residue Classes, $\Z_m$ has $m$ elements. Hence the result. {{qed}}
By definition, the order of a group is the cardinality of its underlying set. By definition, the underlying set of $\struct {\Z'_m, \times_m}$ is the reduced residue system  $\Z'_m$: :$\Z'_m = \set {\eqclass {a_1} m, \eqclass {a_2} m, \ldots, \eqclass {a_{\map \phi m} } m}$ where: :$\forall k: a_k \perp m$ From Cardinality of Reduced Residue System, $\Z'_m$ has $\map \phi m$ elements. Hence the result. {{qed}}
From Quotient Group of Reals by Integers is Circle Group, $\struct {K, \times}$ is isomorphic to the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$. But $\dfrac {\struct {\R, +} } {\struct {\Z, +} }$ is the half-open interval $\hointr 0 1$. A real interval is uncountable by (some result). Hence the result. {{qed}} {{MissingLinks|results in here}}
We have that $S_{n - 1}$ is the symmetric group on $n - 1$ letters. Let $\pi \in S_{n - 1}$. Then $\pi$ is a permutation on $n - 1$ letters. Hence $\pi$ is also a permutation on $n$ letters which fixes $n$. So $S_{n - 1} \subseteq H$. Now let $\pi \in H$. Then $\pi$ is a permutation on $n$ letters which fixes $n$. That is, $\pi$ is a permutation on $n - 1$ letters. Thus $\pi \in S_{n - 1}$. So we have that $H = S_{n - 1}$. We also have that $S_{n - 1}$ is a group, and: :$\forall \rho \in S_{n - 1}: \rho \in S_n$ So $S_{n - 1}$ is a subset of $S_n$ which is a group. Hence $S_{n - 1}$ is a subgroup of $S_n$ by definition. Then we have: {{begin-eqn}} {{eqn | l = \index {S_n} {S_{n - 1} }       | r = \dfrac {\order {S_n} } {\order {S_{n - 1} } }       | c = {{Defof|Index of Subgroup}} }} {{eqn | r = \dfrac {n!} {\paren {n - 1}!}       | c = Order of Symmetric Group }} {{eqn | r = \dfrac {n \paren {n - 1}!} {\paren {n - 1}!}       | c = {{Defof|Factorial}} }} {{eqn | r = n       | c =  }} {{end-eqn}} {{qed}}
Let $\order a = m, \order b = n$. Let $c = \lcm \set {m, n}$. Then: {{begin-eqn}} {{eqn | l = c       | r = r m       | c = for some $r \in \Z$ }} {{eqn | r = s n       | c = for some $s \in \Z$ }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \paren {a b}^c       | r = a^c b^c       | c = Power of Product of Commuting Elements in Semigroup equals Product of Powers }} {{eqn | r = a^{r m} b^{s n}       | c =  }} {{eqn | r = \paren {a^m}^r \paren {b^n}^s       | c =  }} {{eqn | r = e^r e^s       | c = {{Defof|Order of Group Element}} }} {{eqn | r = e       | c =  }} {{eqn | ll= \leadsto       | l = \order {a b}       | o = \divides       | r = c       | c = Element to Power of Multiple of Order is Identity }} {{end-eqn}} {{qed}}
From Identity is Only Group Element of Order 1: :$\order e = 1$ and so from One Divides all Integers: :$\order e \divides n$ Thus $G_n \ne \O$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = G_n       | c =  }} {{eqn | ll= \leadsto       | l = \order x       | o = \divides       | r = n       | c =  }} {{eqn | ll= \leadsto       | l = \order {x^{-1} }       | o = \divides       | r = n       | c = Order of Group Element equals Order of Inverse }} {{eqn | ll= \leadsto       | l = x^{-1}       | o = \in       | r = G_n       | c =  }} {{end-eqn}} Let $a, b \in G_n$ such that $\order a = r, \order b = s$. Then from Product of Orders of Abelian Group Elements Divides LCM of Order of Product: :$\order {a b} = \divides \lcm \set {a, b}$ But $r \divides n$ and $s \divides n$ by definition of $G_n$. Therefore, by definition of lowest common multiple: :$\order {a b} \divides n$ Thus we have: :$G_n \ne \O$ :$x \in G_n \implies x^{-1} \in G_n$ :$a, b \in G_n \implies a b \in G_n$ and the result follows by the Two-Step Subgroup Test. {{qed}}
From Existence of Cyclic Group of Order n we have that one such group of order $4$ is the cyclic group of order $4$: This is exemplified by the additive group of integers modulo $4$, whose Cayley table can be presented as: {{:Modulo Addition/Cayley Table/Modulo 4}} From Group whose Order equals Order of Element is Cyclic, any group with an element of order $4$ is cyclic. From Cyclic Groups of Same Order are Isomorphic, no other groups of order $4$ which are not isomorphic to $C_4$ can have an element of order $4$. {{qed|lemma}} From Order of Element Divides Order of Finite Group, any other group of order $4$ must have elements of order $2$. We have the Klein $4$-group, whose Cayley table can be presented as: {{:Klein Four-Group/Cayley Table}} and is seen to have that property. {{qed|lemma}} We have that Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic. {{qed|lemma}} It remains to be shown that the Klein $4$-group is the only groups of order $4$ whose elements are all of order $2$ (except the identity). Let the Cayley table be populated as far as can be directly established: :$\begin{array}{c|cccc}   & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e &   &   \\ b & b &   & e &   \\ c & c &   &   & e \\ \end{array}$ Consider $a b$. As $a^2 = e$, $a b \ne e$. As $a e = a$, $a b \ne a$. As $e b = b$, $a b \ne b$. It follows that $a b = c$. Hence we have: :$\begin{array}{c|cccc}   & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c &   \\ b & b &   & e &   \\ c & c &   &   & e \\ \end{array}$ and the rest of the table is completed by following the result that Group has Latin Square Property. {{qed}}
Suppose $G$ is not a $p$-group. Then there exist two distinct primes $p_1, p_2$. By Cauchy's Group Theorem, there exist subgroups $H, K$ such that: :$\order H = p_1$ :$\order K = p_2$ That is: :$H = \gen a$ :$k = \gen b$ for some $a, b \in G: a \ne b$, where: :$\order a = p_1$ :$\order b = p_2$ and so both $H \nsubseteq K$ and $K \nsubseteq H$ Thus, by definition, the subgroups of $G$ do not form a nest. It follows by the Rule of Transposition that $G$ is a $p$-group. It remains to be shown that $G$ is cyclic. Suppose to the contrary that $G$ is not cyclic. Then it is not the case that $G$ is generated by a single element of $G$. Thus there exist $a, b \in G, a \ne b$ such that $H = \gen a$ and $K = \gen b$ are subgroups of $G$. But as $a \in H, b \in K, a \notin K, b \notin H$ it follows that $H \nsubseteq K$ and $K \nsubseteq H$. It follows by the Rule of Transposition that $G$ is cyclic. {{qed}}
By definition of the multiplicative group of complex $n$th roots of unity: :$U_n := \set {z \in \C: z^n = 1}$ By definition of the circle group: :$K = \set {z \in \C: \cmod z = 1}$ By Modulus of Complex Root of Unity equals 1: :$\forall z \in U_n: \cmod z = 1$ Thus: :$U_n \subseteq K$ We further have that the operation $\times$ on both $U_n$ and $K$ is complex multiplication. Finally, from Roots of Unity under Multiplication form Cyclic Group, we have that $\struct {U_n, \times}$ is a group. The result follows by definition of subgroup. {{qed}} Category:Circle Group Category:Multiplicative Groups of Complex Roots of Unity lfvp2f3yr0fc3tttjne1u8897x8k13i
By definition, the normalizer of $\set \alpha$ is: :$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \set \alpha g^{-1} = \set \alpha}$ That is: :$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \alpha = \alpha g}$ First let $g = \alpha^k$ for some $k \in \Z$. Then: :$\alpha \alpha^k = \alpha^k \alpha$ which includes $k = 0$, that is $e$. Thus: :$\forall k \in \Z: \alpha^k \in \map {N_{D_n} } {\set \alpha}$ Now let $g = \alpha^k \beta$. Suppose $g \alpha = \alpha g$. Then: {{begin-eqn}} {{eqn | l = \alpha^k \beta \alpha       | r = \alpha \alpha^k \beta       | c =  }} {{eqn | ll= \leadsto       | l = \beta \alpha       | r = \alpha \beta       | c = Cancellation Laws }} {{end-eqn}} But $\beta \alpha \ne \alpha \beta$ in $D_n$ except for $n < 3$. Hence the result: :$\map {N_{D_n} } {\set \alpha} = \set {e, \alpha, \alpha^2, \ldots, \alpha^{n - 1} }$ Hence the result, by definition of generator of subgroup. {{qed}} Category:Dihedral Groups Category:Normalizers l8t1v1hdfwid63js3d7czck2lhjh8dr
By definition, the normalizer of $\set \beta$ is: :$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \set \beta g^{-1} = \set \beta}$ That is: :$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \beta = \beta g}$ First let $g = \beta^k$ for $k \in \set {0, 1}$. Then: :$\beta \beta^k = \beta^k \beta$ Thus: :$\forall k \in \set {0, 1}: \beta^k \in \map {N_{D_n} } {\set \beta}$ Now let $g = \alpha^j \beta^k$ for $0 < j < n, k \in \set {0, 1}$. Suppose $g \alpha = \alpha g$. Then: {{begin-eqn}} {{eqn | l = \alpha^j \beta^k \beta       | r = \beta \alpha^j \beta^k       | c =  }} {{eqn | ll= \leadsto       | l = \alpha^j \beta \beta^k       | r = \beta \alpha^j \beta^k       | c =  }} {{eqn | ll= \leadsto       | l = \alpha^j \beta       | r = \beta \alpha^j       | c = Cancellation Laws }} {{eqn | ll= \leadsto       | l = \alpha^j \beta       | r = \alpha^{n - j} \beta        | c = Product of Generating Elements of Dihedral Group }} {{eqn | ll= \leadsto       | l = \alpha^j       | r = \alpha^{n - j}       | c =  }} {{eqn | ll= \leadsto       | l = \alpha^2 j       | r = \alpha^n       | c =  }} {{eqn | ll= \leadsto       | l = 2 j       | r = n       | c =  }} {{end-eqn}} Thus when $n$ is odd, there is no such $j$ such that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$. But when $n$ is even such that $n = 2 j$, we have that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$. Hence the result. {{qed}} Category:Dihedral Groups Category:Normalizers or2apavxjj0cj0ux4v60y43iu4pql3h
We have from Subgroup of Symmetric Group that Fixes n that $N = S_{n - 1}$. By definition of normalizer: :$\map {N_{S_n} } {S_{n - 1} } := \set {\rho \in S_n: \rho S_{n - 1} \rho^{-1} = S_{n - 1} }$ We have from Group is Normal in Itself that: :$\forall \rho \in S_{n - 1}: \rho S_{n - 1} \rho^{-1} \in S_{n - 1}$ and so: :$S_{n - 1} \subseteq \map {N_{S_n} } {S_{n - 1} }$ It remains to be shown that $\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$. This will be done by demonstrating that: : $S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$ where $\setminus$ denotes set difference. Let $\rho \in S_n$ such that $\rho \notin S_{n - 1}$. Thus: :$\map \rho n \ne n$ and so: :$\exists a \in S_n, a \ne n: \map \rho a = n$ for some $a \in S_{n - 1}$. Then: :$\map {\rho^{-1} } n = a$ Let $\pi \in S_{n - 1}$ such that: : $\map \pi a = b$ for some $b \ne a$. As $\rho$ is a permutation, $\rho$ is by definition a bijection. Hence: :$\map \rho b \ne n$ We have: {{begin-eqn}} {{eqn | l = \map {\rho \pi \rho^{-1} } n       | r = \map {\rho \pi} a       | c = as $\map {\rho^{-1} } n = a$ }} {{eqn | r = \map \rho b       | c = as $\map \pi a = b$ }} {{eqn | o = \ne       | r = n       | c = as $\map \rho b \ne n$ }} {{end-eqn}} Thus $\rho \pi \rho^{-1}$ does not fix $n$. That is: :$\rho \pi \rho^{-1} \notin S_{n - 1}$ Since this is the case for all arbitrary $\rho \in S_n \setminus S_{n - 1}$, it follows that: :$S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$ So from Intersection with Complement is Empty iff Subset: :$\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$ and so by definition of set equality: :$\map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$ {{qed}}
By definition, the center of $\Dic 2$ is: :$\map Z {\Dic 2} = \set {g \in \Dic 2: g x = x g, \forall x \in \Dic 2}$ We are given that: :$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ We have that $\Dic 2$ is generated by $\alpha$ and $\beta$. Thus: :$x \in \map Z {\Dic 2} \iff x a = a x \land x b = b x$ Let $x \in \map Z {\Dic 2}$. We have that $x$ can be expressed in the form: :$x = a^i b^j$ for $i \in \set {0, 1, 2, 3}$ and $j \in \set {0, 1}$. As $x \in \map Z {\Dic 2}$, we have: {{begin-eqn}} {{eqn | l = x a       | r = a x       | c =  }} {{eqn | ll= \leadsto       | l = a^i b^j a       | r = a^{i + 1} b^j       | c =  }} {{eqn | ll= \leadsto       | l = b^j a       | r = a b^j       | c =  }} {{end-eqn}} For $j = 1$ this means: {{begin-eqn}} {{eqn | l = a b       | r = b a       | c =  }} {{eqn | r = a^{-1} \beta       | c =  }} {{eqn | ll= \leadsto       | l = a       | r = a^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = a^2       | r = e       | c =  }} {{end-eqn}} But the order of $a$ is $4$, not $2$, and hence: :$a^2 \ne e$ So if $x \in \map Z {\Dic 2}$ it follows that $x$ has to be in the form: :$x = a^i$ for some $i \in \Z_{\ge 0}$. Again, as $x \in \map Z {\Dic 2}$, we have: {{begin-eqn}} {{eqn | l = x b       | r = b x       | c =  }} {{eqn | ll= \leadsto       | l = a^i b       | r = b a^i       | c =  }} {{eqn | r = a^{-i} b       | c = Product of Generating Elements of Quaternion Group }} {{eqn | ll= \leadsto       | l = a^i       | r = a^{-i}       | c =  }} {{eqn | ll= \leadsto       | l = a^{2 i}       | r = e       | c =  }} {{eqn | r = a^4       | c = {{Defof|Order of Group Element}} }} {{end-eqn}} So either $i = 0$ or $2 i = 4$, as $0 \le i \le 4$. If $i = 0$ then $x = a^0 = e$. If $2 i = 4$ then: :$x = a^{4 / 2} = a^2$ Hence the result. {{qed}}
The proof proceeds by induction. For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition: :$b a^k = a^{-k} b$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = b a^0       | r = b e       | c =  }} {{eqn | r = e b       | c =  }} {{eqn | r = a^{-0} b       | c =  }} {{end-eqn}} Thus $\map P 0$ is seen to hold. === Basis for the Induction === We have: {{begin-eqn}} {{eqn | l = a b a       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = b a^1       | r = a^{-1} b       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$b a^r = a^{-r} b$ from which it is to be shown that: :$b a^{r + 1} = a^{-r - 1} b$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = b a^{r + 1}       | r = b a^r a       | c =  }} {{eqn | r = a^{-r} b a       | c = Product of Generating Elements of Quaternion Group }} {{eqn | r = a^{-r} \alpha^{-1} b       | c = Basis for the Induction }} {{eqn | r = a^{-r - 1} b       | c =  }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall k \in \Z_{\ge 0}: b a^k = a^{-k} b$ {{qed}} Category:Quaternion Group 4qzdhrhtoodtvkbxkhcfxt7ith3i7ic
From Center of Quaternion Group, we have: :$\map Z {\Dic 2} = \set {e, a^2}$ Thus from Conjugacy Class of Element of Center is Singleton, $\set e$ and $\set {a^2}$ are two of those conjugacy classes. By inspection of the Cayley table: {{:Quaternion Group/Cayley Table}} we investigate the remaining $6$ elements in turn, starting with $a$: {{begin-eqn}} {{eqn | l = a a a^{-1}       | o =        | rr= = a       | c =  }} {{eqn | l = a^2 a \paren {a^2}^{-1}       | r = a^2 a a^2 = a^5       | rr= = a       | c =  }} {{eqn | l = a^3 a \paren {a^3}^{-1}       | r = a^3 a a = a^5       | rr= = a       | c =  }} {{eqn | l = b a b^{-1}       | r = b a \paren {a^2 b} = b a^3 b       | rr= = a^3       | c =  }} {{eqn | l = \paren {a b} a \paren {a b}^{-1}       | r = \paren {a b} a \paren {a^3 b} = b a^3 b       | rr= = a^3       | c =  }} {{eqn | l = \paren {a^2 b} a \paren {a^2 b}^{-1}       | r = \paren {a^2 b} a b       | rr= = a^3       | c =  }} {{eqn | l = \paren {a^3 b} a \paren {a^3 b}^{-1}       | r = \paren {a^3 b} a \paren {a b} = \paren {a^2 b} \paren {a b}       | rr= = a^3       | c =  }} {{end-eqn}} So we have a conjugacy class: :$\set {a, a^3}$ Investigating the remaining $4$ elements in turn, starting with $b$: {{begin-eqn}} {{eqn | l = a b a^{-1}       | r = a b a^3       | rr= = a^2 b       | c =  }} {{eqn | l = a^2 b \paren {a^2}^{-1}       | r = a^2 b a^2       | rr= = b       | c =  }} {{eqn | l = a^3 b \paren {a^3}^{-1}       | r = a^3 b a       | rr= = a^2 b       | c =  }} {{eqn | l = b b b^{-1}       | o =        | rr= = b       | c =  }} {{eqn | l = \paren {a b} b \paren {a b}^{-1}       | r = \paren {a b} b \paren {a^3 b} = \paren {a b} a^3       | rr= = a^2 b       | c =  }} {{eqn | l = \paren {a^2 b} b \paren {a^2 b}^{-1}       | r = \paren {a^2 b} b b = \paren {a^2 b} a^2       | rr= = b       | c =  }} {{eqn | l = \paren {a^3 b} b \paren {a^3 b}^{-1}       | r = \paren {a^3 b} b \paren {a b} = \paren {a^3 b} a       | rr= = a^2 b       | c =  }} {{end-eqn}} So we have a conjugacy class: :$\set {b, a^2 b}$ Investigating the remaining $2$ elements, starting with $a b$: {{begin-eqn}} {{eqn | l = a \paren {a b} a^{-1}       | r = a \paren {a b} a^3 = a^2 b a^3       | rr= = a^3 b       | c =  }} {{end-eqn}} We need go no further: the remaining elements $a b$ and $a^3 b$ are in the same conjugacy class: :$\set {a b, a^3 b}$ Hence the result. {{qed}}
Since $V$ is not a trivial vector space: :$\exists \mathbf v \in V: \mathbf v \ne 0$ By Norm axiom (N1): :$\norm {\mathbf v} > 0$ Let $r, s \in R$: {{begin-eqn}} {{eqn | l = \norm {r s}_R \norm {\mathbf v}       | r = \norm {\paren {r s} \mathbf v}       | c = Norm axiom (N2) }} {{eqn | r = \norm {r \paren {s \mathbf v} }       | c = {{Vector-space-axiom|7}} }} {{eqn | r = \norm r_R \norm {s \mathbf v}       | c = Norm axiom (N2) }} {{eqn | r = \norm r_R \norm s_R \norm {\mathbf v}       | c = Norm axiom (N2) }} {{end-eqn}} By dividing both sides of the equation by $\norm {\mathbf v}$ then: :$\norm {r s}_R = \norm r_R \norm s_R$ The result follows. {{qed}} Category:Normed Division Rings Category:Norm Theory n9j7rsaatesqxu9zjp0c8vke73w4pay
{{AimForCont}}: :$\exists x, y \in {R^*} : x \circ y = 0_R$ By positive definiteness: :$x, y \ne 0_R \iff \norm x, \norm y \ne 0$ Thus: :$\norm x \norm y \ne 0$ But we also have: {{begin-eqn}} {{eqn | l = \norm x \norm y       | r = \norm {x \circ y}       | c = Multiplicativity }} {{eqn | r = \norm {0_R}       | c = by assumption }} {{eqn | r = 0       | c = Positive Definiteness }} {{end-eqn}} a contradiction. {{qed}} Category:Norm Theory Category:Ring Theory azlqrazgbpgo5vzoy5kkmcuepobfreg
From Ring with Multiplicative Norm has No Proper Zero Divisors, $R$ has no proper zero divisors. From Finite Ring with No Proper Zero Divisors is Field, $R$ is a field. {{qed}} Category:Ring Theory Category:Field Theory Category:Norm Theory lwtk5hog282ltnhd0cbmwawzoi1ae1i
An isometry is a distance-preserving mapping which is also a bijection. From Composition of Distance-Preserving Mappings is Distance-Preserving, $\psi \circ \phi$ is a distance-preserving mapping. From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection. {{qed}} Category:Metric Spaces m8cjqegtg9dd5z42hpcuok44xqc5ih0
Let $g \in G$. First we note that since $G$ is closed, and $h \circ g \in G$, it follows that $h * g \in G$. Next we note: :$e * g = e \circ g = g$ and so {{GroupActionAxiom|2}} is satisfied. Now let $h_1, h_2 \in G$. We have: {{begin-eqn}} {{eqn | l = \paren {h_1 \circ h_2} * g       | r = \paren {h_1 \circ h_2} \circ g       | c = Definition of $*$ }} {{eqn | r = h_1 \circ \paren {h_2 \circ g}       | c = {{GroupAxiom|1}} }} {{eqn | r = h_1 * \paren {h_2 * g}       | c = Definition of $*$ }} {{end-eqn}} and so {{GroupActionAxiom|1}} is satisfied. Hence the result. {{qed}}
From Subgroup Action is Group Action we have that $*$ is a group action. Let $x \in G$. Then: {{begin-eqn}} {{eqn | l = \Orb x       | r = \set {g \in G: \exists h \in H: g = h * x}       | c = {{Defof|Orbit (Group Theory)|Orbit}} }} {{eqn | r = \set {g \in G: \exists h \in H: g = h \circ x}       | c = Definition of $*$ }} {{eqn | r = H x       | c =  }} {{end-eqn}} Hence the result, by definition of right coset. {{qed}}
By Riemann Removable Singularities Theorem, as $f$ has a removable singularity at $\infty$, $f$ must be bounded in a neighborhood of $\infty$. That is, there exists a real number $M > 0$ such that $\cmod {\map f z} \le M$ for all $z \in \set{z : \cmod z > r}$ for some real $r \ge 0$. However, by Continuous Function on Compact Space is Bounded, $f$ is also bounded on $\set{z : \cmod z \le r}$.  As $\set{z : \cmod z > r} \cup \set{z : \cmod z \le r} = \C$, $f$ is therefore bounded on $\C$. Therefore by Liouville's Theorem, $f$ is constant. {{qed}} Category:Complex Analysis ft58yz7272i4zngzeagf3r0b1n94if3
From Subgroup Action is Group Action we have that $*$ is a group action. Let $x \in G$. Then: {{begin-eqn}} {{eqn | l = \Stab x       | r = \set {h \in H: h * x = x}       | c = {{Defof|Stabilizer}} }} {{eqn | r = \set {h \in H: h \circ x = x}       | c = Definition of $*$ }} {{eqn | r = \set {h \in H: h = x \circ x^{-1} }       | c =  }} {{eqn | r = \set {h \in H: h = e}       | c =  }} {{eqn | r = \set e       | c =  }} {{end-eqn}} Hence the result, by definition of right coset. {{qed}}
From Stabilizer is Subgroup we have that both $\Stab y$ and $\Stab z$ are subgroups of $G$. From definition of orbits: :$\exists h_1, h_2 \in G: y = h_1 * x, z = h_2 * x$ Then $y = h_1 * \paren {h_2^{-1} * z} = h_1 h_2^{-1} * z$. Thus: {{begin-eqn}} {{eqn | l = \Stab y       | r = \set {g \in G: g * y = y}       | c = {{Defof|Stabilizer}} }} {{eqn | r = \set {g \in G: g * \paren {h_1 h_2^{-1} * z} = h_1 h_2^{-1} * z} }} {{eqn | r = \set {g \in G: h_1^{-1} h_2 * \paren {g h_1 h_2^{-1} * z} = z} }} {{eqn | r = \set {g \in G: \paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } * z = z} }} {{eqn | r = \paren {h_1 h_2^{-1} } \set {\paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } \in G: \paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } * z = z} \paren {h_1 h_2^{-1} }^{-1} }} {{eqn | r = \paren {h_1 h_2^{-1} } \Stab z \paren {h_1 h_2^{-1} }^{-1}       | c = {{Defof|Stabilizer}} }} {{end-eqn}} This shows that $\Stab y$ and $\Stab z$ are conjugate. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \Stab {g H}       | r = \set {k \in K: \paren {k g} H = g H}       | c = {{Defof|Stabilizer}} }} {{eqn | r = \set {k \in K: g^{-1} \paren {k g} \in H}       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | r = \set {k \in K: k \in g H g^{-1} }       | c =  }} {{eqn | r = K \cap H^g       | c = {{Defof|Conjugate (Group Theory)|subdef = Subset|Conjugate of Set}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \card {\Orb {g H} }       | r = \index K {\Stab {g H} }       | c = Orbit-Stabilizer Theorem }} {{eqn | r = \index K {K \cap H^g}       | c = Stabilizer of Subgroup Action on Left Coset Space }} {{end-eqn}} {{qed}}