hcju/proofwikiqa huggingface dataset:
Let $C_n = \gen g$ be the cyclic group of order $n$. By definition, $g^n = e$. We have that $n - 1$ is coprime to $n$. So it follows from that Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order that: : $C_n = \gen {g^{n - 1} }$ But from Inverse Element is Power of Order Less 1: :$g^{n - 1} = g^{-1}$ {{Qed}}
By definition, the infinite cyclic group with generator $g$ is: :$\gen g = \set {\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots}$ where $e$ denotes the identity $e = g^0$. The fact that $g^{-1}$ generates $G$ is shown by Inverse of Generator of Cyclic Group is Generator. Futhermore: :$\gen e = \set e \ne G$ By definition of infinite cyclic group: :$g^i \ne g^j$ for all $i \ne j$ Let $n \in Z \setminus \set {-1, 0, 1}$. Then: :$\gen {g^n} = \set {\ldots, g^{-2 n}, g^{-n}, e, g^n, g^{2 n}, \ldots}$ But since $\order n > 1$, none of these elements is equal to $g$, because $1 \notin n \Z$. So: :$g \notin \gen {g^n} \implies \gen {g^n} \ne \gen g$ {{qed}}
Let $H$ be a subgroup of $\struct {\Z, +}$. Let $n \in \Z, h \in H$. Then from the definition of cyclic group and Negative Index Law for Monoids: :$n h = n \cdot h \in \gen h \subseteq H$ The result follows. {{Qed}}
From Additive Group of Integers is Subgroup of Rationals, $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$. From Additive Group of Rationals is Subgroup of Reals, $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$. Thus $\struct {\Z, +}$ is a subgroup of $\struct {\R, +}$. {{qed}}
Define $\phi: \R / \Z \to K$ by: :$\map \phi {x + \Z} = \map \exp {2 \pi i x}$ Then $\phi$ is well-defined. For, if $x + \Z = y + \Z$, then $y = x + n$ for some $n \in \Z$, and: :$\map \exp {2 \pi i \paren {x + n} } = \map \exp {2 \pi i x}$ by Complex Exponential Function has Imaginary Period. Moreover, by Exponential of Sum: :$\map \exp {x + y + \Z} = \map \exp {2 \pi i \paren {x + y} } = \map \exp {2 \pi i x} \, \map \exp {2 \pi i y}$ meaning $\phi$ is a group homomorphism. By Euler's Formula: :$\map \exp {2 \pi i x} = \map \cos {2 \pi i x} + i \, \map \sin {2 \pi i x}$ so that, by Sine and Cosine are Periodic on Reals: :$\map \phi x = 1$ {{iff}} $x + \Z = 0 + \Z$ Hence, by Kernel is Trivial iff Monomorphism, $\phi$ is a monomorphism. From polar form for complex numbers, it follows that all $z \in \C$ with $\cmod z = 1$ are of the form: :$z = \map \exp {2 \pi i x}$ for some $x \in \R$. Hence $\phi$ is also an epimorphism. Thus $\phi: \R / \Z \to K$ is a group isomorphism. {{qed}}
From the definition of integers modulo $m$, we have: :$\Z_m = \dfrac \Z {\RR_m} = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$ It is established that Modulo Addition is Well-Defined: :$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$ The group axioms are fulfilled: : '''{{GroupAxiom|0}}''': Addition modulo $m$ is closed. : '''{{GroupAxiom|1}}''': Addition modulo $m$ is associative. : '''{{GroupAxiom|2}}''': The identity element of $\struct {\Z_m, +_m}$ is $\eqclass 0 m$. : '''{{GroupAxiom|3}}''': The inverse of $\eqclass k m \in \Z_m$ is $-\eqclass k m = \eqclass {-k} m = \eqclass {n - k} m$. : Commutativity: Addition modulo $m$ is commutative. From Integers under Addition form Infinite Cyclic Group and Quotient Group of Cyclic Group, $\struct {\dfrac \Z {\RR_m}, +_m}$ is cyclic order $m$. {{Qed}}
Multiplication modulo $m$ is closed. Multiplication modulo $m$ is associative. Multiplication modulo $m$ has an identity: :$\forall k \in \Z: \eqclass k m \eqclass 1 m = \eqclass k m = \eqclass 1 m \eqclass k m$ This identity is unique. Multiplication modulo $m$ is commutative. Thus all the conditions are fulfilled for this to be a commutative monoid. {{qed}}
First, suppose $k \perp m$. That is: :$\gcd \set {k, m} = 1$ Then, by Bézout's Identity: :$\exists u, v \in \Z: u k + v m = 1$ Thus: :$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$ Thus $\eqclass u m$ is an inverse of $\eqclass k m$. Suppose that: :$\exists u \in \Z: \eqclass u m \eqclass k m = \eqclass {u k} m = 1$. Then: :$u k \equiv 1 \pmod m$ and: :$\exists v \in \Z: u k + v m = 1$ Thus from Bézout's Identity: :$k \perp m$ {{qed}}
From Ring of Integers Modulo m is Ring, $\left({\Z_m, +_m, \times_m}\right)$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity $\left[\!\left[{1}\right]\!\right]_m$]]. Thus by definition $\left({\Z_m, \times_m}\right)$ is a commutative monoid. The result follows from Multiplicative Inverse in Monoid of Integers Modulo m. {{qed}}
By definition of reduced residue system modulo $m$: :$\Z'_m = \set {x \in \Z_m: x \perp m}$ From Subset of Set with Propositional Function: :$\Z'_m \subseteq \Z_m$ We have that: :$\gcd \set {m, 0} = m$ Thus it follows that: :$m > 1 \implies \gcd \set {m, 0} \ne 1$ So: :$\eqclass 0 m \notin \Z'_m$ However: :$\eqclass 0 m \in \Z_m$ so: :$\Z'_m \ne \Z_m$ Thus: :$\Z'_m \subset \Z_m$ Then: :$\eqclass 1 m = 1 \implies 1 \perp m$ So: :$\forall m \in \Z: \eqclass 1 m \in \Z'_m$ Thus: :$\Z'_m \ne \O$ and therefore: :$\O \subset \Z'_m$ {{qed}} Category:Residue Classes Category:Reduced Residue Systems 6472l2hgwcmn9kek2nctjmqg87ke7g6
From Ring of Integers Modulo m is Ring, $\struct {\Z_m, +, \times}$ forms a (commutative) ring with unity. Then we have that the units of a ring with unity form a group. By Multiplicative Inverse in Ring of Integers Modulo m we have that the elements of $\struct {\Z'_m, \times}$ are precisely those that have inverses, and are therefore the units of $\struct {\Z_m, +, \times}$. The fact that $\struct {\Z'_m, \times}$ is abelian follows from Restriction of Commutative Operation is Commutative. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === From Modulo Multiplication on Reduced Residue System is Closed: :$\struct {\Z'_m, \times}$ is closed. {{qed|lemma}} === $\text G 1$: Associativity === We have that Modulo Multiplication is Associative. {{qed|lemma}} === $\text G 2$: Identity === From Modulo Multiplication has Identity, $\eqclass 1 m$ is the identity element of $\struct {\Z'_m, \times_m})$. {{qed|lemma}} === $\text G 3$: Inverses === From Multiplicative Inverse in Monoid of Integers Modulo m, $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ {{iff}} $k$ is coprime to $m$. Thus every element of $\struct {\Z'_m, \times_m}$ has an inverse. {{qed|lemma}} All the group axioms are thus seen to be fulfilled, and so $\struct {\Z'_m, \times_m}$ is a group. {{qed}}
Taking the finite group axioms in turn: === $\text {FG} 0$: Closure === From Modulo Multiplication on Reduced Residue System is Closed: :$\struct {\Z'_m, \times}$ is closed. {{qed|lemma}} === $\text {FG} 1$: Associativity === We have that Modulo Multiplication is Associative. {{qed|lemma}} === $\text {FG} 2$: Finiteness === The order of $\struct {\Z'_m, \times}$ is $\map \phi n$ by definition, where $\map \phi n$ denotes the Euler $\phi$ function. As $\map \phi n < n$ it follows that $\struct {\Z'_m, \times}$ is of finite order. {{qed|lemma}} === $\text {FG} 3$: Cancellability === We have that Modulo Multiplication on Reduced Residue System is Cancellable. {{qed|lemma}} Thus all the finite group axioms are fulfilled, and $\struct {\Z'_m, \times}$ is a group. It remains to note that Modulo Multiplication is Commutative to confirm that $\struct {\Z'_m, \times}$ is abelian. {{qed}}
=== Prime Modulus === $\struct {\Z_m, +, \times}$ is a commutative ring with unity by definition. From Reduced Residue System under Multiplication forms Abelian Group, $\struct {\Z'_m, \times}$ is an abelian group. $\Z'_m$ consists of all the elements of $\Z_m$ coprime to $m$. Now when $m$ is prime, we have, from Reduced Residue System Modulo Prime: :$\Z'_m = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {m - 1} m}$ That is: :$\Z'_m = \Z_m \setminus \set {\eqclass 0 m}$ where $\setminus$ denotes set difference. Hence the result. {{qed|lemma}} === Composite Modulus === Now suppose $m \in \Z: m \ge 2$ is composite. From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain. From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field. {{qed}}
Let $p$ be prime. From Irreducible Elements of Ring of Integers, we have that $p$ is irreducible in the ring of integers $\struct {\Z, +, \times}$. From Ring of Integers is Principal Ideal Domain, $\struct {\Z, +, \times}$ is a principal ideal domain. Thus by Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\ideal p$ is a maximal ideal of $\struct {\Z, +, \times}$. Hence by Maximal Ideal iff Quotient Ring is Field, $\Z / \ideal p$ is a field. But $\Z / \ideal p$ is exactly $\struct {\Z_p, +, \times}$. {{qed|lemma}} Let $p$ be composite. Then $p$ is not irreducible in $\struct {\Z, +, \times}$. Thus by Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\ideal p$ is not a maximal ideal of $\struct {\Z, +, \times}$. Hence by Maximal Ideal iff Quotient Ring is Field, $\Z / \ideal p$ is not a field. {{qed}}
Let $m$ be prime. From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain. From Finite Integral Domain is Galois Field, $\struct {\Z_m, +, \times}$ is a field. {{qed|lemma}} Now suppose $m \in \Z: m \ge 2$ is composite. From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain. From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field. {{qed}}
First we note that, from Integer Multiples under Addition form Infinite Cyclic Group, $\struct {n \Z, +}$ is an infinite cyclic group. From Cyclic Group is Abelian, it follows that $\struct {n \Z, +}$ is an infinite abelian group. Let $H$ be a non-trivial subgroup of $\struct {\Z, +}$. Because $H$ is non-trivial: :$\exists m \in \Z: m \in H: m \ne 0$ Because $H$ is itself a group: :$-m \in H$ So either $m$ or $-m$ is positive and therefore in $\Z_{>0}$. Thus: :$H \cap \Z_{>0} \ne \O$ From the Well-Ordering Principle, $H \cap \Z_{>0}$ has a smallest element, which we can call $n$. It follows from Subgroup of Infinite Cyclic Group is Infinite Cyclic Group that: : $\forall a \in \Z: a n \in H$ Thus: : $n \Z \subseteq H$ {{AimForCont}}: :$\exists m \in \Z: m \in H \setminus n \Z$ Then $m \ne 0$, and also $-m \in H \setminus n \Z$. Assume $m > 0$, otherwise we consider $-m$. By the Division Theorem: :$m = q n + r$ If $r = 0$, then $m = q n \in n \Z$, so $0 \le r < n$. Now this means $r = m - q n \in H$ and $0 \le r < n$. This would mean $n$ was not the smallest element of $H \cap \Z$. Hence, by Proof by Contradiction, there can be no such $m \in H \setminus n \Z$. Thus: :$H \setminus n \Z = \O$ Thus from Set Difference with Superset is Empty Set: :$H \subseteq n \Z$ Thus we have $n \Z \subseteq H$ and $H \subseteq n \Z$. Hence: :$H = n \Z$ {{qed}}
From Integer Multiples under Addition form Subgroup of Integers, $\struct {n \Z, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$. From Integers under Addition form Infinite Cyclic Group, $\struct {\Z, +}$ is a cyclic group. So by Subgroup of Cyclic Group is Cyclic, $\struct {n \Z, +}$ is a cyclic group. The final assertions follow from Subgroup of Infinite Cyclic Group is Infinite Cyclic Group. {{qed}}
From Subgroups of Additive Group of Integers, $\struct {m \Z, +}$ is a subgroup of $\struct {\Z, +}$. From Subgroup of Abelian Group is Normal, $\struct {m \Z, +}$ is normal in $\struct {\Z, +}$. Therefore the quotient group $\dfrac {\struct {\Z, +} } {\struct {m \Z, +} }$ is defined. Now $\Z$ modulo $m \Z$ is Congruence Modulo a Subgroup. This is merely congruence modulo an integer. Thus the quotient set $\Z / m \Z$ is $\Z_m$. The left coset of $k \in \Z$ is denoted $k + m \Z$, which is the same thing as $\eqclass k m$ from the definition of residue class. So $\index \Z {m \Z} = m$ follows from the definition of Subgroup Index. {{qed}}
Let $\eqclass a m$ denote the residue class modulo $m$ of $a$. Since $a \perp m$, it follows by Reduced Residue System under Multiplication forms Abelian Group that $\eqclass a m$ belongs to the abelian group $\struct {\Z'_m, \times}$. Let $k = \order {\eqclass a m}$ where $\order {\, \cdot \,}$ denotes the order of a group element. By Order of Element Divides Order of Finite Group: :$k \divides \order {\Z'_m}$ By the definition of the Euler $\phi$ function: :$\order {\Z'_m} = \map \phi m$ Thus: {{begin-eqn}} {{eqn | l = \eqclass a m^k       | r = \eqclass a m       | c = {{Defof|Order of Group Element}} }} {{eqn | ll= \leadsto       | l = \eqclass a m^{\map \phi m}       | r = \eqclass {a^{\map \phi m} } m       | c = Congruence of Powers }} {{eqn | r = \eqclass 1 m       | c =  }} {{eqn | ll= \leadsto       | l = a^{\map \phi m}       | o = \equiv       | r = 1 \pmod m       | c = {{Defof|Residue Class}} }} {{end-eqn}} {{qed}} {{Namedfor|Leonhard Paul Euler|cat = Euler}}
By definition, a symmetry mapping is a bijection, and hence a permutation. From Symmetric Group is Group, the set of all permutations on $P$ form the symmetric group $\struct {\map \Gamma P, \circ}$ on $P$. Thus $S_P$ is a subset of $\struct {\map \Gamma P, \circ}$. Let $A$ and $B$ be symmetry mappings on $P$. From Composition of Symmetries is Symmetry, $A \circ B$ is also a symmetry mapping on $P$. Also, by the definition of symmetry, $A^{-1}$ is also a symmetry mapping on $P$. Thus we have: :$A, B \in S_P \implies A \circ B \in S_P$ :$A \in S_P \implies A^{-1} \in \S_P$ The result follows from the Two-Step Subgroup Test. {{qed}}
By definition, a square is a regular quadrilateral. From Internal Angles of Regular Polygon, the internal angles of a square measure $\dfrac {180 \degrees \paren {4 - 2} } 4 = 90 \degrees$. The result follows from Measurement of Right Angle. {{qed}} Category:Squares n7tjkuonqy6jbbedb9iv3obmyjyrq54
Let $\Box ABCD$ be a square whose side $AB$ is of length $L$. Let $\Box EFGH$ be a square whose side $EF$ is of length $1$. :300px From the Axioms of Area, the area of $\Box EFGH$ is $1$. By definition, $AB : EF = L : 1$. From Similar Polygons are composed of Similar Triangles, the ratio of the areas of $\Box ABCD$ to $\Box EFGH$ is the duplicate ratio of the ratio of $AB$ to $EF$. Thus by definition of duplicate ratio: : $\Box ABCD : \Box EFGH = \left({AB : EF}\right)^2$ That is: : $\dfrac {\Box ABCD} {\Box EFGH} = \left({\dfrac L 1}\right)^2 = L^2$ That is, the area of $\Box ABCD$ has $L^2$ as many units as $\Box EFGH$. Hence the result. {{qed}}
=== Integer Side Length === In the case where $L = 1$, the statement follows from the definition of area. If $L \in \N, L > 1$, then we can divide the square into smaller squares, each of side length one. Since there will be $L$ squares of side length one on each side, it follows that there will be $L \cdot L = L^2$ squares of side length one. Thus, the area of the square of side length $L$ is $L^2 \cdot 1 = L^2$. {{Qed|lemma}} === Rational Side Length === Let $A$ be the area of the square $S$ with side length $L$. If $L$ is a rational number, then: :$\exists p, q \in \N: L = \dfrac p q$ Create a square of side length $p$. From the integer side length case, its area equals $p^2$. Divide the sides into $q$ equal parts. Thus the square of side length $p$ is divided into $q^2$ small squares. As they all have side length $\dfrac p q$, each of them equals $S$. It follows arithmetically: :$A = \dfrac {p^2} {q^2} = \paren {\dfrac p q}^2 = L^2$ {{Qed|lemma}} === Irrational Side Length === Let $L$ be an irrational number. Then from Rationals are Everywhere Dense in Topological Space of Reals we know that within an arbitrarily small distance $\epsilon$ from $L$, we can find a rational number less than $L$ and a rational number greater than $L$. In formal terms, we have: :$\forall \epsilon > 0: \exists A, B \in \Q_+: A < L < B: \left|{A - L}\right| < \epsilon, \left|{B - L}\right| < \epsilon$ Thus: : $\displaystyle \lim_{\epsilon \to 0^+} A = L$ : $\displaystyle \lim_{\epsilon \to 0^+} B = L$ Since a square of side length $B$ can contain a square of side length $L$, which can in turn contain a square of side length $A$, then: : $\operatorname {area} \Box B \ge \operatorname {area} \Box L \ge \operatorname {area}\Box A$ By the result for rational numbers: : $\operatorname {area}\Box B = B^2$ : $\operatorname {area}\Box A = A^2$ We also note that: : $\displaystyle \lim_{B \to L} B^2 = L^2 = \lim_{A \to L} A^2$ Thus: : $\displaystyle \lim_{B \to L} \operatorname {area} \Box B = \lim_{B \to L} B^2 = L^2$ : $\displaystyle \lim_{A \to L} \operatorname {area} \Box A = \lim_{A \to L} A^2 = L^2$ Finally: : $L^2 \ge \operatorname {area}\Box L \ge L^2$ so: : $\operatorname {area}\Box L = L^2$ {{Qed}}
Let a square have a side length $a \in \R$. This square is equivalent to the area under the graph of $\map f x = a$ from $0$ to $a$. Thus from the geometric interpretation of the definite integral, the area of the square will be the integral: :$\displaystyle A = \int_0^a a \rd l$ Thus: {{begin-eqn}} {{eqn | l = A       | r = \int_0^a a \rd l }} {{eqn | r = \bigintlimits {l \cdot a} 0 a       | c = Integral of Constant }} {{eqn | r = a \cdot a - 0 \cdot a }} {{eqn | r = a^2 }} {{end-eqn}} {{Qed}} === Warning === '''This proof is circular.''' The use of the definite integral to represent area is based on the fact that the area of a rectangle is the product of the rectangle's width and height. That fact is in turn derived from this one. However, this demonstration neatly parallels the integration based proofs of the areas of other figures, for example Area of Circle.
Suppose $\pi^k \left({i}\right) = \pi^r \left({i}\right)$ for some $r > 0$. Then, since $\pi$ has an inverse, $\pi^{k - r} \left({i}\right) = i$. This contradicts the definition of $k$, so $r = 0$. {{qed}} Category:Symmetric Groups s6busb2smrfp9a2bew0qxesh1oun8w9
Let $\pi$ be a permutation, and let $x \in \Fix \pi$. From the definition of a fixed element, $\map \pi x = x$. From the definition of a $k$-cycle, we see that $1$ is the smallest $k \in \Z: k > 0$ such that $\map {\pi^k} x = x$. The result follows. {{qed}} Category:Symmetric Groups 5th1obufnxiwimah16mkfvfzxd09v13
By definition of the identity permutation: : $\forall i \in \N_{>0}: e \left({i}\right) = i$ Thus $e$ fixes all elements of $S_n$. Thus each element moved by a permutation $\pi$ is fixed by $e$. The set of elements moved by $e$ is $\O$, so the converse is true vacuously. {{qed}} Category:Symmetric Groups Category:Identity Mappings ej17rkvuch2qhbeqa3wmxc2042jco6v
Let $\rho$ and $\sigma$ be disjoint permutations. Let $i \in \Fix \rho$. Then: :$\map {\sigma \rho} i = \map \sigma i$ whereas: :$\map {\rho \sigma} i = \map \rho {\map \sigma i}$ {{AimForCont}} $\map \sigma i \notin \Fix \rho$. Then because $\sigma$ and $\rho$ are disjoint it follows that: {{begin-eqn}} {{eqn | l = \map \sigma i       | o = \in       | r = \Fix \sigma }} {{eqn | ll= \leadsto       | l = \map {\sigma^2} i       | r = \map \sigma i }} {{eqn | ll= \leadsto       | l = \map {\sigma^{-1} \sigma^2} i       | r = \map {\sigma^{-1} \sigma} i }} {{eqn | ll= \leadsto       | l = \map \sigma i       | r = i }} {{end-eqn}} But it was previously established that $i \in \Fix \rho$. This is a contradiction. Therefore: :$\map \sigma i \in \Fix \rho$ and so: :$\map {\rho \sigma} i = \map \sigma i = \map {\sigma \rho} i$ Let $i \notin \Fix \rho$. Then: : $i \in \Fix \sigma$ and the same proof can be performed with $\rho$ and $\sigma$ exchanged. {{qed}}
Let $\pi \in S_n$. First we note that, from Element of Finite Group is of Finite Order, every element of a finite group has finite order. Thus $\pi$ has finite order, so: :$\exists r \in \Z: \pi^r = e$ Checking in turn each of the criteria for equivalence: === Reflexive === From above, $\exists r \in \Z: \pi^r = e$. Therefore $\exists k \in \Z: \map {\pi^k} i = i$. So $\mathcal R_\pi$ is reflexive. {{qed|lemma}} === Symmetric === Let $\map {\pi^k} i = j$. Because $\pi^r = e$, we have $\map {\pi^r} j = j$ (from above). Thus $\map {\pi^{r - k} } j = i$. So $\mathcal R_\pi$ is symmetric. {{qed|lemma}} === Transitive === Let $\map {\pi^{s_1} } i = j, \map {\pi^{s_2} } j = k$. Then $\map {\pi^{s_1 + s_2} } i = k$. So $\mathcal R_\pi$ is transitive. {{qed|lemma}} All criteria are met, and so $\mathcal R_\pi$ is an equivalence relation. {{qed}} Category:Symmetric Groups Category:Equivalence Relations rjo3omx45c5ldvuigd6uen9pr0jdghw
A direct application of Power of Product of Commutative Elements in Group, and the fact that Disjoint Permutations Commute. {{qed}}
Suppose $\pi$ is a cycle. Then from Order of Cycle is Length of Cycle, $\order \pi$ is its length. As the LCM of $n \in \Z$ is $n$ itself, the result follows. Let $\pi = \rho_1 \rho_2 \cdots \rho_r$ where: : each $\rho_s$ is of order $k_s$ : $\rho_1$ to $\rho_r$ are mutually disjoint permutations. Let $t = \lcm \set {k_1, k_2, \ldots, k_r}$. From Disjoint Permutations Commute: :$\pi^t = \rho_1^t \rho_2^t \cdots \rho_r^t$ We have that: :$\forall s: 1 \le s \le r: \exists m \in \Z: t = m s$ Hence: :$\forall s: \rho_s^t = e$ Thus $\pi^t = e$ and certainly $\order \pi \divides t$. Let $\pi^u = e$ for some $u \in \Z_{>0}$. We have that $\rho_1, \rho_2, \cdots, \rho_r$ are mutually disjoint permutations. Thus $\rho_1^u, \rho_2^u, \cdots, \rho_r^u$ are also mutually disjoint permutations. Then we have that each $\rho_s^u = e$. Hence $k_s \divides u$. Thus, if $\pi^u = e$, then: :$t \divides u$ Thus, by choosing $u = \order \pi$: :$\order \pi = t = \lcm \set {k_1, k_2, \ldots, k_r}$ {{qed}}
Note that, by Group Action Determines Bijection, $\phi_g \in \map \Gamma X$ for $g \in G$. Let $g, h \in G$. From the definition of group action: :$\forall \tuple {g, x} \in G \times X: \map \phi {g, x} \in X = g \wedge x \in X$ First we show that for all $x \in X$: :$\map {\phi_g \circ \phi_h} x = \map {\phi_{g h} } x$ Thus: {{begin-eqn}}      {{eqn | l = \map {\phi_g \circ \phi_h} x       | r = g \wedge \paren {h \wedge x}       | c = Definition of $\phi_g$, $\phi_h$ }} {{eqn | r = \paren {g h} \wedge x       | c = {{Defof|Group Action}} }} {{eqn | r = \map {\phi_{g h} } x       | c = Definition of $\phi_{g h}$ }} {{end-eqn}} Also, we have: : $e \wedge x = x \implies \map {\phi_e} x = x$ where $e$ is the identity of $G$. Therefore, we have shown that $\tilde \phi: G \to \map \Gamma X: g \mapsto \phi_g$ is a group homomorphism. {{qed}}
=== Proof of Injectivity === Let $x, y \in X$ Then: {{begin-eqn}} {{eqn | l = \map {\phi_g} x       | r = \map {\phi_g} y       | c =  }} {{eqn | ll= \leadsto       | l = g * x       | r = g * y       | c =  }} {{eqn | ll= \leadsto       | l = g^{-1} * \paren {g * x}       | r = g^{-1} * \paren {g * y}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {g^{-1} g} * x       | r = \paren {g^{-1} g} * y       | c =  }} {{eqn | ll= \leadsto       | l = e * x       | r = e * y       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y       | c =  }} {{end-eqn}} Thus $\phi_g$ is an injection. {{qed|lemma}} === Proof of Surjectivity === Let $x \in X$. Then: {{begin-eqn}} {{eqn | l = x       | r = e * x       | c =  }} {{eqn | r = \paren {g g^{-1} } * x       | c =  }} {{eqn | r = g * \paren {g^{-1} * x}       | c =  }} {{eqn | r = \map {\phi_g} y       | c = where $y = g^{-1} * x \in X$ }} {{end-eqn}} Thus a group action is a surjection. {{qed|lemma}} So a group action is an injection and a surjection and therefore a bijection. {{qed}}