hcju/proofwikiqa huggingface dataset:
Suppose $l > m$. Then: :$m = \dfrac m 2 + \dfrac m 2 < \dfrac {l + m} 2 < \dfrac l 2 + \dfrac l 2 = l$ Let $\epsilon = \dfrac {l - m} 2$. Then: :$\epsilon > 0$ We are given that: :$\displaystyle \lim_{n \mathop \to \infty} x_n = l$ By definition of the limit of a real sequence, we can find $N_1$ such that: :$\forall n \ge N_1: \size {x_n - l} < \epsilon$ where $\size {x_n - l}$ denotes the absolute value of $x_n - l$ Suppose $n \ge N_1$ If $x_n \ge l$ then $x_n > \dfrac {l + m} 2$ If $x_n < l$ then: {{begin-eqn}} {{eqn | l = \epsilon       | o = >       | r = \size {x_n - l}       | c = as $n > N_1$ }} {{eqn | r = l - x_n       | c = as $x_n < l$ }} {{eqn | ll= \leadsto       | l = \dfrac {l - m} 2       | o = >       | r = l - x_n       | c = as $\epsilon = \dfrac {l - m} 2$ }} {{eqn | ll= \leadsto       | l = x_n       | o = >       | r = \dfrac {l + m} 2       | c = rearranging terms }} {{end-eqn}} In either case $x_n > \dfrac {l+m} 2$ We are also given that: :$\displaystyle \lim_{n \mathop \to \infty} y_n = m$ Similarly we can find $N_2$ such that: :$\forall n > N_2: \size {y_n - m} < \epsilon$ Suppose $n \ge N_2$ If $y_n \le m$ then $y_n < \dfrac {l + m} 2$ If $y_n > m$ then: {{begin-eqn}} {{eqn | l = \epsilon       | o = >       | r = \size {y_n - l}       | c = as $n > N_2$ }} {{eqn | r = y_n - m       | c = as $y_n > m$ }} {{eqn | ll= \leadsto       | l = \dfrac {l - m} 2       | o = >       | r = y_n - m       | c = as $\epsilon = \dfrac {l - m} 2$ }} {{eqn | ll= \leadsto       | l = \dfrac {l + m} 2       | o = >       | r = y_n       | c = rearranging terms }} {{end-eqn}} In either case $y_n < \dfrac {l + m} 2$ Let $N = \max \set {N_1, N_2}$. Then if $n > N$, both the above inequalities will be true: :$n > N_1$ :$n > N_2$ Thus $\forall n > N$: :$y_n < \dfrac {l + m} 2 < x_n$ It has been shown that: :$l > m \implies \forall n \in \N: \exists m \ge n: x_n > y_n$ Taking the contrapositive: :$\exists N \in \N: \forall n \ge N: x_n \le y_n \implies l \le m$ Hence the result. {{qed}}
{{BeginTableau|p \implies \left({p \lor q}\right)|Instance 2 of the Hilbert-style systems}} {{TableauLine  |n = 1  |f = q \implies \paren{ p \lor q }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A2$ }} {{TableauLine  |n = 2  |f = p \implies \paren{ q \lor p }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep  = 1  |c    = $p \,/\, q, q \,/\, p$ }} {{TableauLine  |n = 3  |f = \paren{ p \lor q } \implies \paren{ q \lor p }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A3$ }} {{TableauLine  |n = 4  |f = \paren{ q \lor p } \implies \paren{ p \lor q }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep  = 3  |c    = $p \,/\, q, q \,/\, p$ }} {{TableauLine  |n = 5  |f = p \implies \paren{ p \lor q }  |rlnk = Hypothetical Syllogism/Formulation 1/Proof 3  |rtxt = Hypothetical Syllogism  |dep  = 2,4 }} {{EndTableau}} {{Qed}}
Let $\pi = \tuple {a_0, a_1, \ldots, a_{k - 1} }$. Observe that: {{begin-eqn}} {{eqn | l = \paren {j + n} \pmod k + 1       | r = \paren {j + n + 1} \pmod k       | c =  }} {{eqn | lll=\leadsto       | l = \map \pi {a_{\paren {j + n} \pmod k} }       | r = a_{\paren {j + n + 1} \pmod k}       | c =  }} {{eqn | lll=\leadsto       | ll= \forall n \in \N, \forall j \in \Z/ k \Z:       | l = \map {\pi^n} {a_j}       | r = a_{\paren {j + n} \pmod k}       | c =  }} {{eqn | lll=\leadsto       | ll= \forall j < k:       | l = \pi^k       | r = I_{S_n}       | c =  }} {{eqn | lo= \land       | l = \pi^j       | o = \ne       | r = I_{S_n}       | c = where $I_{S_n}$ denotes the identity mapping on $S_n$ }} {{end-eqn}} Hence the result, by definition of order of group element. {{qed}}
It is first shown that $\sequence {\norm {x_n} }$ is a real Cauchy sequence in $\R$. Let $\epsilon \in \R_{>0}$ be given. By the definition of Cauchy sequence then: :$\exists N \in \N: \forall n, m > N, \norm {x_n - x_m} < \epsilon$ By the reverse triangle inequality, then: :$\forall n, m > N: \cmod {\norm {x_n} - \norm {x_m} } \le \norm {x_n - x_m} < \epsilon$ From the definition of a real Cauchy sequence it follows that $\sequence {\norm {x_n} }$ is a real Cauchy sequence in $\R$. By Real Sequence is Cauchy iff Convergent the sequence $\sequence {\norm {x_n} }$ has a limit in $\R$. {{qed}}
Let: :$l = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}$ and: :$m = \displaystyle \lim_{n \mathop \to \infty} \norm {y_n}$ By Norm Sequence of Cauchy Sequence has Limit, both of these limits exist. Then: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty } \paren {\norm {x_n} - \norm {y_n} }        | r = l - m       | c = Difference Rule for Real Sequences }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \size {\norm {x_n} - \norm {y_n} }       | r = \size {l - m}       | c = Modulus of Limit }} {{end-eqn}} By Reverse Triangle Inequality, for $n \in \N$: :$l = \size {\norm {x_n} - \norm {y_n} } \le \norm {x_n - y_n} = \to 0$ as $n \to \infty$. By the Squeeze Theorem: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\size {\norm {x_n} - \norm {y_n} } } = 0$ So: :$\size {l - m} = 0$ and therefore: $l = m$ {{qed}} Category:Cauchy Sequences Category:Normed Division Rings o6fk3yrl6ydw69omhl6r33ifdxxbfal
{{BeginTableau|\vdash p \implies \paren{ q \implies \paren{ p \land q } }|Instance 2 of the Hilbert-style systems}} {{TableauLine  |n = 1  |f = \neg p \lor p  |rlnk = Law of Excluded Middle/Sequent Form/Proof 3  |rtxt = Law of Excluded Middle }} {{TableauLine  |n = 2  |f = \paren{ \neg p \lor p } \implies \paren{ p \lor \neg p }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A3$  |c = $\neg p \,/\, p, p \,/\, q$ }} {{TableauLine  |n = 3  |f = p \lor \neg p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep  = 1,2 }} {{TableauLine  |n = 4  |f = \paren{ \neg p \lor \neg q } \lor \neg \paren{ \neg p \lor \neg q }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep  = 3  |c = $\paren{ \neg p \lor \neg q } \,/\, p$ }} {{TableauLine  |n = 5  |f = \paren{ \neg p \lor \neg q } \lor \paren{ p \land q }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 2 (1)$ }} {{TableauLine  |n = 6  |f = \paren{ \paren{ \neg p \lor \neg q } \lor \paren{ p \land q } } \implies \paren{ \neg p \lor \paren{ \neg q \lor \paren{ p \land q } } }  |rlnk = Rule of Association/Disjunction/Formulation 2/Reverse Implication  |rtxt = Rule of Association  |c = $\neg p \,/\, p, \neg q \,/\, q, \paren{ p \land q } \,/\, r$ }} {{TableauLine  |n = 7  |f = \neg p \lor \paren{ \neg q \lor \paren{ p \land q } }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep  = 5,6 }} {{TableauLine  |n = 8  |f = p \implies \paren{ q \implies \paren{ p \land q } }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 2 (2)$  |dep = 7 }} {{EndTableau}} {{qed}}
Recall the statement of $RST \, 4$: :If $\mathbf A$ and $\mathbf B$ are theorems of $\mathscr H_2$, then so is $\mathbf A \land \mathbf B$. Suppose that $\mathbf A$ and $\mathbf B$ are theorems of $\mathscr H_2$. From Rule of Conjunction/Sequent Form/Formulation 2, we have as a theorem: :$p \implies \paren{ q \implies \paren{ p \land q } }$ which can be recursively checked to not have used $RST \, 4$ anywhere in a proof. Applying $RST \, 1$, this becomes: :$\mathbf A \implies \paren{ \mathbf B \implies \paren{ \mathbf A \land \mathbf B } }$ Using $RST \, 3$ twice, the result follows. {{qed}}
{{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r} }} {{Assumption  |1|p \lor \paren {q \lor r} }} {{SequentIntro|2|1|\paren {p \lor q} \lor r|1|Rule of Association: Formulation 1}} {{Implication |3||\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}|1|2}} {{Assumption  |4|\paren {p \lor q} \lor r}} {{SequentIntro|5|4|p \lor \paren {q \lor r}|4|Rule of Association: Formulation 1}} {{Implication |6||\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }|4|5}} {{BiconditionalIntro|7||\paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}|3|6}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}|Instance 2 of the Hilbert-style systems}} {{TableauLine  | n = 1  | f = \paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}  | rlnk = Rule of Association/Disjunction/Formulation 2/Forward Implication  | rtxt = Rule of Association: Forward Implication }} {{TableauLine  | n = 2  | f = \paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }  | rlnk = Rule of Association/Disjunction/Formulation 2/Reverse Implication  | rtxt = Rule of Association: Reverse Implication }} {{TableauLine  | n = 3  | f = \paren {\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r} } \land \paren {\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} } }  | rlnk = Definition:Hilbert Proof System/Instance 2  | rtxt = Rule $RST \, 4$  | dep  = 1,2 }} {{TableauLine  | n = 4  | f = \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}  | rlnk = Definition:Hilbert Proof System/Instance 2  | rtxt = Rule $RST \, 2 (3)$  | dep  = 3 }} {{EndTableau}} {{qed}}
By Subgroup Generated by One Element is Set of Powers: :$\gen a = \set {a^n : n \in \Z}$ The result follows by definition of cyclic group. {{qed}}
{{AimForCont}} $\gen a$ is of finite order. We have that $a \in \gen a$ by definition. From Element of Finite Group is of Finite Order it follows that $a$ is of finite order. From this contradiction it follows that $\gen a$ must be of infinite order after all. {{qed}}
=== $(1)$ implies $(2)$ === Let $G$ be a Cyclic Group by definition 1. Then by definition: : $(1): \quad g \in G$ : $(2): \quad$ every element of $G$ is expressible as a power of $g$. From $(1)$, it follows from Group Axion $G 1$: Closure that: :$\gen g \subseteq G$ From $(2)$ it follows that: :$G \subseteq \gen g$ Thus: :$G = \gen g$ and $G$ is a Cyclic Group by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $G$ be a Cyclic Group by definition 2. Then by definition: :$G = \gen g$ Thus as $g^1 \in \gen g$ :$g \in G$ and by definition of generator: :$\forall h \in \gen g: h = g^n$ for some $n \in \Z$. Thus $G$ is a Cyclic Group by definition 1. {{qed}}
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$. Consider the subgroup $\gen 2$ of $\struct {\R_{\ne 0}, \times}$ generated by $2$. By definition, $\gen 2$ is a cyclic group. Consider the element $4 \in \struct {\R_{\ne 0}, \times}$. We have that $4 = 2^2$. Thus $4 \in \gen 2$. There exists no $n \in \Z$ such that $4^n = 2$. But $2 \in \gen 2$. Thus $\gen 2$ is not generated by $4$. That is, $4$ is not a generator of $\gen 2$. Hence the result, by Proof by Counterexample. {{qed}}
From Additive Function is Linear for Rational Factors: :$f(q) = q f(1)$ for all $q\in\Q$. Without loss of generality, let :$f(q) = q$ for all $q\in\Q$. Since $f$ is not linear, let $\alpha\in\R\setminus\Q$ be such that :$f(\alpha) = \alpha+\delta$ for some $\delta \neq 0$. Consider an arbitrary nonempty circle in the plane. Let its centre be :$(x,y)$ where $x\neq y$ and $x,y\in\Q$ and its radius be $r>0$. We will show how to find a point of the graph of $f$ inside this circle. As $x\neq y$ and $r$ can be arbitrarily small, this will prove the theorem. Since $\delta\neq0$, let :$\beta = \frac{y - x}{\delta}$  Since $x\neq y$, :$\beta\neq0$. As Rationals are Everywhere Dense in Topological Space of Reals, there exists a rational number $b\neq 0$ such that: :$\left\vert  \beta - b  \right\vert < \frac{r}{2 \left\vert\delta\right\vert}$ As Rationals are Everywhere Dense in Topological Space of Reals, there also exists a rational number $a$ such that: :$\left\vert  \alpha - a  \right\vert < \frac{r}{2\left\vert b\right\vert} $ Now put: :$X = x + b (\alpha - a) \ $ :$ Y = f(X)  \ $ Then: :$|X-x| = |b (\alpha - a)| < \frac{r}{2}$ so $X$ is in the circle. Then: {{begin-eqn}} {{eqn|l= Y       |r=f(x + b (\alpha - a))       |c=Definition of $Y$ and $X$  }} {{eqn|r=f(x) + f(b \alpha) - f( b a)       |c=Cauchy functional equation }} {{eqn|r=x + b f(\alpha) - b f(a)      |c=Additive Function is Linear for Rational Factors }} {{eqn|r=y - \delta \beta + b f(\alpha) - b f(a)      |c=Definition of $y$ }} {{eqn|r=y - \delta \beta + b (\alpha + \delta) - b a      |c=Additive Function is Linear for Rational Factors }} {{eqn|r=y + b (\alpha - a) - \delta (\beta - b)}} {{end-eqn}} Therefore  :$|Y-y| = |b (\alpha - a) - \delta (\beta - b)| \le |b (\alpha - a)| + |\delta (\beta - b)| \le r$ so $Y$ is in the circle as well. Hence the point $(X, Y)$ is inside the circle. {{qed}}
From Identity is Only Group Element of Order 1, the identity element $e$ , and only $e$, is of order $1$. From here, we inspect the Cayley table: {{:Quaternion Group/Cayley Table}} It is immediately seen that: :$\paren {a^2}^2 = e$ and so by definition $a^2$ is of order $2$. As can be seen from inspection of the main diagonal, for all other $x \in Q$, we have: :$x^2 = a^2 \ne e$ and so: :$x^4 = \paren {a^2}^2 = e$ There are $6$ such elements. Hence the result by definition of order of group element. {{qed}}
From Reduced Residue System under Multiplication forms Abelian Group it is noted that $\struct {\Z'_7, \times_7}$ is a group. It remains to be shown that $\struct {\Z'_7, \times_7}$ is cyclic. It will be demonstrated that: :$\gen {\eqclass 3 7} = \struct {\Z'_7, \times_7}$ That is, that $\eqclass 3 7$ is a generator of $\struct {\Z'_7, \times_7}$. We note that $\eqclass 1 7$ is the identity element of $\struct {\Z'_7, \times_7}$. Thus successive powers of $\eqclass 3 7$ are taken, until $n \in \Z$ is found such that $\eqclass 3 7^n = \eqclass 1 7$: {{begin-eqn}} {{eqn | l = \eqclass 3 7^2       | r = \eqclass {3 \times 3} 7       | c =  }} {{eqn | r = \eqclass 9 7       | c =  }} {{eqn | r = \eqclass 2 7       | c =  }} {{eqn | l = \eqclass 3 7^3       | r = \eqclass 3 7^2 \times \eqclass 3 7       | c =  }} {{eqn | r = \eqclass {2 \times 3} 7       | c =  }} {{eqn | r = \eqclass 6 7       | c =  }} {{eqn | l = \eqclass 3 7^4       | r = \eqclass 3 7^3 \times \eqclass 3 7       | c =  }} {{eqn | r = \eqclass {6 \times 3} 7       | c =  }} {{eqn | r = \eqclass {18} 7       | c =  }} {{eqn | r = \eqclass 4 7       | c =  }} {{eqn | l = \eqclass 3 7^5       | r = \eqclass 3 7^4 \times \eqclass 3 7       | c =  }} {{eqn | r = \eqclass {4 \times 3} 7       | c =  }} {{eqn | r = \eqclass {12} 7       | c =  }} {{eqn | r = \eqclass 5 7       | c =  }} {{eqn | l = \eqclass 3 7^6       | r = \eqclass 3 7^5 \times \eqclass 3 7       | c =  }} {{eqn | r = \eqclass {5 \times 3} 7       | c =  }} {{eqn | r = \eqclass {15} 7       | c =  }} {{eqn | r = \eqclass 1 7       | c =  }} {{end-eqn}} All elements of $\struct {\Z'_7, \times_7}$ are seen to be in $\gen {\eqclass 3 7}$. Hence the result by definition of cyclic group. {{qed}}
From Reduced Residue System under Multiplication forms Abelian Group it is noted that $\struct {\Z'_5, \times_5}$ is a group. It remains to be shown that $\struct {\Z'_5, \times_5}$ is cyclic. It will be demonstrated that: :$\gen {\eqclass 2 5} = \struct {\Z'_5, \times_5}$ That is, that $\eqclass 2 5$ is a generator of $\struct {\Z'_5, \times_5}$. We note that $\eqclass 1 5$ is the identity element of $\struct {\Z'_5, \times_5}$. Thus successive powers of $\eqclass 2 5$ are taken, until $n \in \Z$ is found such that $\eqclass 2 5^n = \eqclass 1 5$: {{begin-eqn}} {{eqn | l = \eqclass 2 5^2       | r = \eqclass {2 \times 2} 5       | c =  }} {{eqn | r = \eqclass 4 5       | c =  }} {{eqn | l = \eqclass 2 5^3       | r = \eqclass 2 5^2 \times \eqclass 2 5       | c =  }} {{eqn | r = \eqclass {2 \times 4} 5       | c =  }} {{eqn | r = \eqclass 8 5       | c =  }} {{eqn | r = \eqclass 3 5       | c =  }} {{eqn | l = \eqclass 2 5^4       | r = \eqclass 2 5^3 \times \eqclass 2 5       | c =  }} {{eqn | r = \eqclass {3 \times 2} 5       | c =  }} {{eqn | r = \eqclass 6 5       | c =  }} {{eqn | r = \eqclass 1 5       | c =  }} {{end-eqn}} All elements of $\struct {\Z'_5, \times_5}$ are seen to be in $\gen {\eqclass 2 5}$. Hence the result by definition of cyclic group. {{qed}} Category:Multiplicative Groups of Reduced Residues Category:Multiplicative Group of Reduced Residues Modulo 5 Category:Examples of Cyclic Groups Category:Cyclic Group of Order 4 5n9m9g0fku8wbu7y1ytu42a6n0ncayy
From Finite Cyclic Group has Euler Phi Generators, $G$ has $\map \phi {2^n}$ generators. The result follows from corollary to Euler Phi Function of Prime Power. {{qed}}
Using a tableau proof for instance 1 of a Gentzen proof system: {| border="1" |+$\vdash \left({p \lor \left({q \land r}\right)}\right) \implies \left({\left({p \lor q}\right) \land \left({p \lor r}\right)}\right)$ |- ! Line !! ! Pool ! Formula ! Rule ! Depends upon ! Notes |- | 1 || || | $\neg p, p, q$ | Axiom || || |- | 2 || || | $\neg p, p \lor q$ | $\beta$-Rule: $\beta\lor$ | 1 || |- | 3 || || | $\neg p, p, r$ | Axiom || || |- | 4 || || | $\neg p, p \lor r$ | $\beta$-Rule: $\beta\lor$ | 3 || |- | 5 || || | $\neg p, \left({p \lor q}\right) \land \left({p \lor r}\right)$ | $\alpha$-Rule: $\alpha\land$ | 2, 4 || |- | 6 || || | $\neg q, \neg r, p, q$ | Axiom || || |- | 7 || || | $\neg q, \neg r, p \lor q$ | $\beta$-Rule: $\beta\lor$ | 6 || |- | 8 || || | $\neg q, \neg r, p, r$ | Axiom || || |- | 9 || || | $\neg q, \neg r, p \lor r$ | $\beta$-Rule: $\beta\lor$ | 8 || |- | 10 || || | $\neg q, \neg r, \left({p \lor q}\right) \land \left({p \lor r}\right)$ | $\alpha$-Rule: $\alpha\land$ | 7, 9 || |- | 11 || || | $\neg \left({q \land r}\right), \left({p \lor q}\right) \land \left({p \lor r}\right)$ | $\beta$-Rule: $\beta\land$ | 10 || |- | 12 || || | $\neg \left({p \lor \left({q \land r}\right)}\right), \left({p \lor q}\right) \land \left({p \lor r}\right)$ | $\alpha$-Rule: $\alpha\lor$ | 5, 11 || |- | 13 || || | $\left({p \lor \left({q \land r}\right)}\right) \implies \left({ \left({p \lor q}\right) \land \left({p \lor r}\right)}\right)$ | $\beta$-Rule: $\beta\implies$ | 12 || |} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \equiv^l       | r = y \pmod H }} {{eqn | ll= \iff       | l = x^{-1} y       | o = \in       | r = H       | c = {{Defof|Left Congruence Modulo Subgroup}} }} {{eqn | ll= \iff       | l = \exists h \in H: x^{-1} y       | r = h       | c = {{Defof|Element}} of $H$ }} {{eqn | ll= \iff       | l = \exists h \in H: y       | r = x h       | c = Division Laws for Groups }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \equiv^r       | r = y \pmod H }} {{eqn | ll= \leadstoandfrom       | l = x y^{-1}       | o = \in       | r = H       | c = {{Defof|Right Congruence Modulo Subgroup}} }} {{eqn | ll= \leadstoandfrom       | l = \exists h \in H: x y^{-1}       | r = h       | c = {{Defof|Element}} of $H$ }} {{eqn | ll= \leadstoandfrom       | l = \exists h \in H: x       | r = h y       | c = Division Laws for Groups }} {{end-eqn}} {{qed}}
Let $e$ be the identity of $G$. From Identity of Subgroup, we have $e \in H$. From the behaviour of the identity, we have: :$x = x e$ That is: : $\exists h \in H: x h = x$ The result follows by definition of left coset. {{Qed}}
Let $e$ be the identity of $G$. From Identity of Subgroup, we have $e \in H$. From the behaviour of the identity, we have: :$x = e x$ That is: : $\exists h \in H: h x = x$ The result follows by definition of right coset. {{Qed}}
Follows directly from: * Left Congruence Modulo Subgroup is Equivalence Relation * Element in its own Equivalence Class. {{qed}}
Follows directly from: * Right  Congruence Modulo Subgroup is Equivalence Relation * Element in its own Equivalence Class. {{qed}}
=== $(1)$ implies $(2)$ === Let $x$ be an infinite order element of $G$ by definition 1. Then by definition there exists no $k \in \Z_{>0}$ such that $x^k = e_G$. {{AimForCont}} not all $x^r$ are distinct for all $r \in \Z_{>0}$. Then $x^m = x^n$ for some $m, n \in \Z$ where $m \ne n$. {{WLOG}}, let $m > n$. Then: : $x^{m - n} = e^G$ As $m > n$ we have that $m - n \in \Z_{>0}$. Let $k = m - n$. and so: :$\exists k \in \Z_{>0}$ such that $x^k = e_G$. But this contradicts the statement that no such $k$ exists. Hence no such distinct $m$ and $n$ exist. Thus $x$ is an infinite order element of $G$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $x$ be an infinite order element of $G$ by definition 2. Then by definition the powers $x^r$ of $x$ are distinct for all $r \in \Z_{>0}$. That is, there exist no $x^m = x^n$ for some $m, n \in \Z$ where $m \ne n$. {{AimForCont}} there exists $k \in \Z_{>0}$ such that $x^k = e^G$. Consider some $n \in \Z_{>0}$ such that $m = n + k$. {{begin-eqn}} {{eqn | l = x^m       | r = x^{n + k}       | c =  }} {{eqn | r = x^n x^k       | c =  }} {{eqn | r = x^n e^G       | c =  }} {{eqn | r = x^n       | c =  }} {{end-eqn}} But this contradicts the statement that no such $m, n \in \Z_{>0}$ exist. Thus there can be no $k \in \Z_{>0}$ such that $x^k = e_G$. Thus $x$ is an infinite order element of $G$ by definition 1. {{qed|lemma}} === $(2)$ implies $(3)$ === Let $x$ be an infinite order element of $G$ by definition 2. Then by definition the powers $x^r$ of $x$ are distinct for all $r \in \Z_{>0}$. {{AimForCont}} not all $\gen x = r$ for some $r \in \Z_{>0}$. Let $s > r + 1$. Then by the Pigeonhole Principle at least two of the elements of $\gen x$ are such that $x^m = x^n$ for some $m, n \in \closedint 0 r$ But this contradicts the statement that all $x^k$ are different for all $k \in \Z_{>0}$. Thus $x$ is an infinite order element of $G$ by definition 3. {{qed|lemma}} === $(3)$ implies $(1)$ === Let $x$ be an infinite order element of $G$ by definition 3. Then by definition the group $\gen x$ generated by $x$ is of infinite order: :$\order {\gen x} = \infty$ {{AimForCont}} there exists $k \in \Z_{>0}$ such that $x^k = e^G$. Then by Order of Cyclic Group equals Order of Generator, $\gen x$ has order $k$ But this contradicts the statement that $\order {\gen x} = \infty$. Thus there can be no $k \in \Z_{>0}$ such that $x^k = e_G$. Thus $x$ is an infinite order element of $G$ by definition 1. {{qed}} Category:Order of Group Elements n1x6apzrqazf04kf1at8kzdmhtaw3a2
=== $(1)$ implies $(2)$ === Let $x$ be a finite order element of $G$ by definition 1. Then by definition there exists $k \in \Z_{>0}$ such that $x^k = e_G$. Consider some $m, n \in \Z_{>0}$ such that $m = n + k$. {{begin-eqn}} {{eqn | l = x^m       | r = x^{n + k}       | c =  }} {{eqn | r = x^n x^k       | c =  }} {{eqn | r = x^n e^G       | c =  }} {{eqn | r = x^n       | c =  }} {{end-eqn}} Thus $x$ is a finite order element of $G$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $x$ be a finite order element of $G$ by definition 2. That is, there exists $m, n \in \Z_{>0}$ such that $x^m = x^n$ but $m \ne n$. {{WLOG}}, suppose that $m > n$. Let $k = m - n$. From $x^m = x^n$ it follows from Powers of Group Elements that: : $x^k = x^{m - n} = x^m x^{-n} = x^n x^{-n} = e_G$ Thus there exists $k \in \Z_{>0}$ such that $x^k = e_G$. Thus $x$ is a finite order element of $G$ by definition 1. {{qed}} Category:Order of Group Elements 1k644uuqbcryr87lijipa3qkq30t988
Let $\gen x$ be the subgroup of $G$ generated by $x$. By definition, $\gen x \le G$. All the elements of $\gen x$ are powers of $x$. As $x \in H$ it follows by group axiom $G0$: closure that all the powers of $x$ are elements of $H$. That is: :$\gen x \le H$ By Order of Cyclic Group equals Order of Generator: :$\order x = \order {\gen x}$ But $\gen x$ is the same cyclic group whether it is considered as embedded in $H$ or $G$. Hence the result. {{qed}}
Let $g_1 g_2 = g_2 g_1$. We have: :$\paren {g_1 g_2}^{n_1 n_2} = e$ Thus: :$\order {g_1 g_2} \le n_1 n_2$ Suppose $\order {g_1 g_2}^r = e$. Then: {{begin-eqn}} {{eqn | l = {g_1}^r       | r = {g_2}^{-r}       | c =  }} {{eqn | o = \in       | r = \gen {g_1} \cap \gen {g_2}       | c =  }} {{eqn | ll= \leadsto       | l = {g_1}^r = {g_2}^r       | r = e       | c =  }} {{end-eqn}} Thus $r$ is divisible by both $m$ and $n$. The result follows. {{qed}}