hcju/proofwikiqa huggingface dataset:
From Subgroup of Elements whose Order Divides Integer, both $H$ and $K$ are subgroups of $G$. Also, because $G$ is abelian, $H$ and $K$ are normal by Subgroup of Abelian Group is Normal. In order to prove $G = H \times K$, by the Internal Direct Product Theorem it suffices to show that $G = H K$ and $H \cap K = \set e$. Since we have $\gcd \set {m, p^n} = 1$, there are integers $s$ and $t$ such that $1 = s m + t p^n$ by Bézout's Lemma.  Let $s$ and $t$ be two such integers. So: :$\forall x \in G: x = x^{s m + t p^n} = x^{s m} x^{t p^n}$ From Element to Power of Group Order is Identity: :$x^{\order G} = e$ Therefore: :$\paren {x^{s m} }^{p^n} = \paren {x^{p^n m} }^s = e^s = e$ and: :$\paren {x^{t p^n} }^m = \paren {x^{p^n m} }^t = e^t = e$ By definition, $x^{s m} \in H$ and $x^{t p^n} \in K$. We conclude that $G = H K$. Now suppose that some $x \in H \cap K$. Then $x^{p^n} = e = x^m$, so from Element to Power of Multiple of Order is Identity $\order x$ divides both $p^n$ and $m$. Since $p$ does not divide $m$, it follows that $\order x = 1$. Therefore, by Identity is Only Group Element of Order 1: :$x = e$ Thus: :$H \cap K = \set e$ It follows that $G = H \times K$. Suppose that $p \divides \order K$. From Cauchy's Lemma (Group Theory), this implies that there is some element (call it $k$) of $K$ with order $p$. But we also have $k^m = e$ from the definition of $K$, so by Element to Power of Multiple of Order is Identity we must have $p \divides m$, a contradiction. We conclude that: :$p$ does not divide $\order K$ It follows that: :$p^n \divides \order H$ {{AimForCont}} a prime $q$ (with $q \ne p$) divides $\order H$. Again from Cauchy's Lemma (Group Theory), this implies that there is some element $h$ of $H$ with order $q$. But since $h^{p^n} = e$ from the definition of $H$, Element to Power of Multiple of Order is Identity gives us $q \divides p^n$, a contradiction. It follows by Proof by Contradiction that $q$ does not divide $\order H$. We conclude $\order H = p^n$, as desired. {{qed}} Category:Abelian Groups 80nvuvd4fl9aabpvi454szqm13gilrl
By definition of abelian group, $x$ and $y$ commute. That is: :$x \circ y = y \circ x$ The result follows from Power of Product of Commutative Elements in Group. {{qed}}
From the definition of mapping, it is clear that $h$ is a relation. Suppose $\tuple {x, y_1}, \tuple {x, y_2} \in h$. Clearly $x \in S_1$ or $x \in S_2$ but as $S_1 \cap S_2 = \O$ it is not in both. If $x \in S_1$ then $y_1 = y_2$ as $\tuple {x, y_1}, \tuple {x, y_2} \in f$, and $f$ is a mapping. Similarly, if $x \in S_2$ then $y_1 = y_2$ as $\tuple {x, y_1}, \tuple {x, y_2} \in g$, and $g$ is a mapping. So $\tuple {x, y_1}, \tuple {x, y_2} \in h \implies y_1 = y_2$. Now suppose $x \in \Dom h$. Either $x \in S_1$ or $x \in S_2$. As both $f$ and $g$ are mappings it follows that either $\exists y \in T_1: \tuple {x, y} \in f$ or $\exists y \in T_2: \tuple {x, y} \in g$. In either case, $\exists y \in T_1 \cup T_2: \tuple {x, y} \in h$. So $h$ is a mapping whose domain is $S_1 \cup S_2$, as we were to show. {{qed}}
Take the definition of natural numbers $\N$ as a naturally ordered semigroup $\struct {\N, +, \le}$. The result follows from Zero is Identity in Naturally Ordered Semigroup. {{qed}}
{{begin-eqn}} {{eqn | l = \tan x + \cot x       | r = \frac {\sin x} {\cos x} + \cot x       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {\sin x} {\cos x} + \frac {\cos x} {\sin x}       | c = Cotangent is Cosine divided by Sine }} {{eqn | r = \frac {\sin^2 x + \cos^2x} {\cos x \sin x}       | c =  }} {{eqn | r = \frac 1 {\cos x \sin x}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \sec x \frac 1 {\sin x}       | c = Secant is Reciprocal of Cosine }} {{eqn | r = \sec x \csc x       | c = Cosecant is Reciprocal of Sine }} {{end-eqn}} {{qed}} Category:Trigonometric Identities jz8w9priubh8wtrir60t5myg7ke5ubr
{{begin-eqn}} {{eqn|l = \tan x \left({\tan x + \cot x}\right)      |r = \tan x \sec x \csc x      |c = Sum of Tangent and Cotangent }} {{eqn|r = \frac {\sin x} {\cos x} \sec x \csc x      |c = Tangent is Sine divided by Cosine }} {{eqn|r = \frac {\sin x} {\cos^2 x} \csc x      |c = Secant is Reciprocal of Cosine }} {{eqn|r = \frac {\sin x} {\cos^2 x \sin x}      |c = Cosecant is Reciprocal of Sine }} {{eqn|r = \frac 1 {\cos^2 x} }} {{eqn|r = \sec^2 x      |c = Secant is Reciprocal of Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan x \paren {\tan x + \cot x}       | r = \frac {\sin x} {\cos x} \paren {\frac {\sin x} {\cos x} + \frac {\cos x} {\sin x} }       | c = {{Defof|Tangent Function}} and {{Defof|Cotangent}} }} {{eqn | r = \frac {\sin x} {\cos x} \paren {\frac {\sin^2 x + \cos^2 x} {\cos x \sin x} } }} {{eqn | r = \frac {\sin x} {\cos x} \paren {\frac 1 {\cos x \sin x} }       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \frac 1 {\cos^2 x} }} {{eqn | r = \sec^2 x       | c = {{Defof|Secant Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l=\sec x - \cos x       | r=\frac 1 {\cos x} - {\cos x}       | c=Secant is Reciprocal of Cosine }} {{eqn | r=\frac {1 - \cos^2 x} {\cos x} }} {{eqn | r=\frac {\sin^2 x} {\cos x}       | c=Sum of Squares of Sine and Cosine }} {{eqn | r=\sin x \tan x       | c=Tangent is Sine divided by Cosine }} {{end-eqn}} {{qed}} Category:Trigonometric Identities ei1gjknt8znlqnk5zyvhatfj1v9yfzf
{{begin-eqn}} {{eqn | l=\tan^2 x - \sin^2 x       | r=\frac {\sin^2 x} {\cos^2x} - \sin^2 x       | c=Tangent is Sine divided by Cosine }} {{eqn | r=\frac {\sin^2 x - \sin^2 x \ \cos^2 x} {\cos^2x} }} {{eqn | r=\frac{\sin^2 x \left({1 - \cos^2 x}\right)} {\cos^2 x} }} {{eqn | r=\tan^2 x \left({1 - \cos^2 x}\right)       | c=Tangent is Sine divided by Cosine }} {{eqn | r=\tan^2 x \ \sin^2 x       | c=Sum of Squares of Sine and Cosine }} {{end-eqn}} {{qed}} Category:Trigonometric Identities 0udy0som302lkqdfshoshg6bzttkw29
{{begin-eqn}} {{eqn | l = \sin^4 x - \cos^4 x       | r = \sin^2 x \left({1 - \cos^2 x}\right) - \cos^2 x \left({1 - \sin^2 x}\right)       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \sin^2 x - \sin^2 x \ \cos^2 x - \cos^2 x + \sin^2 x \ \cos^2 x }} {{eqn | r = \sin^2 x - \cos^2 x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin^2 x - \cos^2 x       | r = \left({\sin^2 x - \cos^2 x}\right) \left({\sin^2 x + \cos^2 x}\right)       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \sin^4 x - \cos^4 x       | c = Difference of Two Squares }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l=\csc x - \sin x       | r=\frac 1 {\sin x} - \sin x       | c=by definition of cosecant }} {{eqn | r=\frac {1 - \sin^2 x} {\sin x} }} {{eqn | r=\frac {\cos^2x} {\sin x}       | c=Sum of Squares of Sine and Cosine }} {{eqn | r=\cos x \ \cot x       | c=by definition of cotangent }} {{end-eqn}} {{qed}} Category:Trigonometric Identities 5bn2pzhm5n2wwhu5hzm0z03edro6f1c
{{begin-eqn}} {{eqn | l = \sec^2 x + \csc^2 x       | r = \frac 1 {\cos^2 x} + \frac 1 {\sin^2 x}       | c = {{Defof|Secant Function}} and {{Defof|Cosecant}} }} {{eqn | r = \frac {\sin^2 x + \cos^2 x} {\cos^2 x \sin^2 x}       | c =  }} {{eqn | r = \frac 1 {\cos^2 x \sin^2 x}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \sec^2 x \csc^2 x       | c = {{Defof|Secant Function}} and {{Defof|Cosecant}} }} {{end-eqn}} {{qed}} Category:Trigonometric Identities f2qk26s0otejvk2vlje2jx2os2b089o
{{begin-eqn}} {{eqn | l=\sec^4 x - \tan^4 x       | r=\sec^4 x - \frac {\sin^4 x} {\cos^4 x}       | c=Tangent is Sine divided by Cosine }} {{eqn | r=\frac 1 {\cos^4 x} - \frac {\sin^4 x} {\cos^4 x}       | c=Secant is Reciprocal of Cosine }} {{eqn | r=\frac {1 - \sin^4 x} {\cos^4x} }} {{eqn | r=\frac {1 - \sin^2 x \left({1 - \cos^2 x}\right)} {\cos^2 x \left({1 - \sin^2 x}\right)}       | c=Sum of Squares of Sine and Cosine }} {{eqn | r=\frac {1 - \sin^2 x + \sin^2 x \ \cos^2 x} {\cos^2 x - \sin^2 x \ \cos^2 x} }} {{eqn | r=\frac {\cos^2 x + \sin^2 x \ \cos^2 x} {\cos^2 x - \sin^2 x \ \cos^2 x}       | c=Sum of Squares of Sine and Cosine }} {{eqn | r=\frac {\cos^2 x \left({1 + \sin^2 x}\right)} {\cos^2 x \left({1 - \sin^2 x}\right)} }} {{eqn | r=\frac {1 + \sin^2 x} {1 - \sin^2 x} }} {{eqn | r=\frac {1 + \sin^2 x} {\cos^2 x}       | c=Sum of Squares of Sine and Cosine }} {{eqn | r=\frac 1 {\cos^2 x} + \frac {\sin^2 x} {\cos^2 x} }} {{eqn | r=\frac 1 {\cos^2 x} x + \tan^2 x       | c=Tangent is Sine divided by Cosine }} {{eqn | r=\sec^2 x + \tan^2 x       | c=Secant is Reciprocal of Cosine }} {{end-eqn}} {{qed}} Category:Trigonometric Identities 5tnn2pdykcwtp537jooy1gy7gtkcecn
{{begin-eqn}} {{eqn | l = \frac 1 {1 - \sin x} + \frac 1 {1 + \sin x}       | r = \frac {1 + \sin x + 1 - \sin x} {1 - \sin^2 x}       | c = Difference of Two Squares }} {{eqn | r = \frac 2 {\cos^2 x}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 2 \sec^2 x       | c = {{Defof|Secant Function}} }} {{end-eqn}} {{qed}} Category:Trigonometric Identities Category:Sine Function eoc9yum0imskau1zik8jiechj2yjo89
{{begin-eqn}} {{eqn | l=\frac 1 {1 - \sin x} - \frac 1 {1 + \sin x}       | r=\frac {1 + \sin x} {1 - \sin^2 x} - \frac {1 - \sin x} {1 - \sin^2 x}       | c=Difference of Two Squares }} {{eqn | r=\frac {2 \sin x} {\cos^2 x}       |c = Sum of Squares of Sine and Cosine }} {{eqn | r=\frac {2 \tan x} {\cos x}       | c=Tangent is Sine divided by Cosine }} {{eqn | r=2 \tan x \ \sec x       | c=Secant is Reciprocal of Cosine }} {{end-eqn}} {{qed}} Category:Trigonometric Identities Category:Sine Function kr8f565ecjdh42ma8fnbv7d6yd46ffh
{{begin-eqn}} {{eqn | l = \sec x + \tan x       | r = \sec x + \frac {\sin x} {\cos x}       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac 1 {\cos x} + \frac {\sin x} {\cos x}       | c = Secant is Reciprocal of Cosine }} {{eqn | r = \frac {1 + \sin x} {\cos x}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {\cos x} {\sec x + \tan x}       | r = \frac {\cos^2 x} {1 + \sin x}       | c = Sum of Secant and Tangent }} {{eqn | r = \frac {1 - \sin^2 x} {1 + \sin x}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \frac {\paren {1 + \sin x} \paren {1 - \sin x} } {1 + \sin x}       | c = Difference of Two Squares }} {{eqn | r = 1 - \sin x }} {{end-eqn}} {{qed}} Category:Trigonometric Identities j1dy19ag3ylw8gkc1yh8rx9voltltmz
{{begin-eqn}} {{eqn | l = \frac {\sec x + 1} {\sec^2 x}       | r = \cos^2 x \paren {\frac 1 {\cos x} + 1}       | c = {{Defof|Secant Function}} }} {{eqn | r = \cos x + \cos^2x }} {{eqn | r = \cos x \paren {1 + \cos x} }} {{eqn | r = \cos x \frac {\paren {1 + \cos x} \paren {1 - \cos x} } {1 - \cos x} }} {{eqn | r = \frac {1 - \cos^2 x} {\frac {1 - \cos x} {\cos x} }       | c = Difference of Two Squares }} {{eqn | r = \frac {\sin^2 x} {\frac 1 {\cos x} - 1}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \frac {\sin^2 x} {\sec x - 1}       | c = {{Defof|Secant Function}} }} {{end-eqn}} {{qed}} Category:Trigonometric Identities gr0g3j3u5vem7n8qr3t81j05zmcwg8m
{{begin-eqn}} {{eqn | l = \paren {\sin x + \cos x} \paren {\tan x + \cot x}       | r = \paren {\sin x + \cos x} \paren {\sec x \csc x}       | c = Sum of Tangent and Cotangent }} {{eqn | r = \frac {\sin x + \cos x} {\sin x \cos x}       | c = {{Defof|Secant Function}} and {{Defof|Cosecant}} }} {{eqn | r = \frac 1 {\cos x} + \frac 1 {\sin x} }} {{eqn | r = \sec x + \csc x       | c = {{Defof|Secant Function}} and {{Defof|Cosecant}} }} {{end-eqn}} {{qed}} Category:Trigonometric Identities mfhi8x6zpj7jjyn9usxkwmft0vebngd
{{begin-eqn}} {{eqn | l=\frac {\tan x} {\sec x + 1}       | r=\frac {\sin x} {\cos x \left({\sec x + 1}\right)}       | c=Tangent is Sine divided by Cosine }} {{eqn | r=\frac {\sin x} {\cos x \left({\frac 1 {\cos x} + 1}\right)}       | c=Secant is Reciprocal of Cosine }} {{eqn | r=\frac {\sin x} {1 + \cos x}       | c=simplifying }} {{eqn | r=\frac {\sin^2 x} {\sin x \left({1 + \cos x}\right)}       | c=Multiply by $1 = \dfrac {\sin x} {\sin x}$ }} {{eqn | r=\frac {1 - \cos^2 x} {\sin x \left({1 + \cos x}\right)}       | c=Sum of Squares of Sine and Cosine }} {{eqn | r=\frac {\left({1 + \cos x}\right) \left({1 - \cos x}\right)} {\sin x \left({1 + \cos x}\right)}       | c=Difference of Two Squares }} {{eqn | r=\frac {1 - \cos x} {\sin x} }} {{eqn | r=\frac {\cos x \left({\frac 1 {\cos x} - 1}\right)} {\sin x} }} {{eqn | r=\frac {\cos x \left({\sec x - 1}\right)} {\sin x}       | c=Secant is Reciprocal of Cosine }} {{eqn | r=\frac {\sec x - 1} {\tan x}       | c=Tangent is Sine divided by Cosine }} {{end-eqn}} {{qed}} Category:Trigonometric Identities mptiv8ej98x641dj9lb50tv2y1p3z3h
{{begin-eqn}} {{eqn | l=\left({a \cos x + b \sin x}\right)^2 + \left({b \cos x - a \sin x}\right)^2       | r=a^2 \cos^2 x + 2 a b \cos x \ \sin x + b^2 \sin^2 x }} {{eqn | o=       | r=\quad + \ b^2 \cos^2 x - 2 a b \sin x \ \cos x + a^2 \sin^2 x }} {{eqn | r=\left({a^2 + b^2}\right) \left({\sin^2 x + \cos^2 x}\right) }} {{eqn | r=a^2 + b^2       | c=Sum of Squares of Sine and Cosine }} {{end-eqn}} {{qed}} Category:Trigonometric Identities p3ven6bi1sv54ounhwlbp1pmwqtvuor
{{begin-eqn}} {{eqn | l = \frac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3}       | r = \frac {1 - \cos^2 x - 2 \cos x - 1} {1 - \cos^2 x + 3 \cos x - 3}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \frac {\cos^2 x - 2 \cos x} {\cos^2 x - 3 \cos x + 2} }} {{eqn | r = \frac {\cos x \paren {\cos x - 2} } {\paren {\cos x - 1} \paren {\cos x - 2} } }} {{eqn | r = \frac {\cos x} {\cos x - 1} }} {{eqn | r = \frac 1 {1 - \frac 1 {\cos x} } }} {{eqn | r = \frac 1 {1 - \sec x}       | c = {{Defof|Secant Function}} }} {{end-eqn}} {{qed}} Category:Trigonometric Identities Category:Reciprocals jw6l8r12fpiz009jcnowmdwb2xhvxx0
{{begin-eqn}} {{eqn | l = \frac {\cos^2 x + 3 \sin x - 1} {\cos^2 x + 2 \sin x + 2}       | r = \frac {1 - \sin^2 x + 3 \sin x - 1} {1 - \sin^2 x + 2 \sin x + 2}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \frac {\sin^2 x - 3 \sin x} {\sin^2 x - 2 \sin x - 3} }} {{eqn | r = \frac {\sin x \paren {\sin x - 3} } {\paren {\sin x - 3} \paren {\sin x + 1} } }} {{eqn | r = \frac {\sin x} {\sin x + 1} }} {{eqn | r = \frac 1 {1 + \dfrac 1 {\sin x} } }} {{eqn | r = \frac 1 {1 + \csc x}       | c = {{Defof|Real Cosecant Function}} }} {{end-eqn}} {{qed}} Category:Trigonometric Identities Category:Reciprocals 5pi4374lgdmak9nbd7rwo0oczh4a6oa
=== Sufficient Condition === Let $\RR \circ \RR^{-1} = I_T$. {{AimForCont}}: :$\exists t \in T: t \notin \Img \RR$ Then: :$t \notin \Img {\RR \circ \RR^{-1} }$ But: :$t \in \Img {I_T}$ by definition of the identity mapping on $T$. Hence: :$\RR \circ \RR^{-1} \ne I_T$ From this contradiction we deduce that: :$\RR \circ \RR^{-1} = I_T \implies T \setminus \Img \RR = \O$ where $T \setminus \Img \RR$ denotes set difference. So from Set Difference with Superset is Empty Set: :$T \subseteq \Img \RR$ But from Image is Subset of Codomain we have: :$T \supseteq \Img \RR$ and so: :$\Img \RR = T$ which means $\RR$ is right-total. {{qed|lemma}} Suppose $\RR$ is not many-to-one. Then: :$\exists s \in S: \exists t_1, t_2 \in T, t_1 \ne t_2: \tuple {s, t_1} \in \RR \land \tuple {s, t_2} \in \RR$ By definition of inverse relation: :$\exists s \in S: \exists t_1, t_2 \in T: \tuple {t_1, s} \in \RR^{-1} \land \tuple {t_2, s} \in \RR^{-1}$ The composite of $\RR^{-1}$ and $\RR$ is defined as: :$\RR \circ \RR^{-1} = \set {\tuple {x, z} \in T \times T: \exists y \in S: \tuple {x, y} \in \RR^{-1} \land \tuple {y, z} \in \RR}$ Thus: {{begin-eqn}} {{eqn | l = \tuple {t_1, s}       | o = \in       | r = \RR^{-1}       | c =  }} {{eqn | l = \tuple {s, t_2}       | o = \in       | r = \RR       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {t_1, t_2}       | o = \in       | r = \RR \circ \RR^{-1}       | c = where $t_1 \ne t_2$ (from above) }} {{end-eqn}} So, by definition of identity mapping: :$\RR \circ \RR^{-1} \ne I_T$ From this contradiction we deduce that $\RR$ must be many-to-one. {{qed|lemma}} So it has been demonstrated that if: :$\RR \circ \RR^{-1} = I_T$ then: :$\RR$ is many-to-one and :$\RR$ is right-total. {{qed|lemma}} === Necessary Condition === Let: :$\RR$ be many-to-one and :$\RR$ be right-total. Let $\tuple {t_1, t_2} \in \RR \circ \RR^{-1}$. The composite of $\RR^{-1}$ and $\RR$ is defined as: :$\RR \circ \RR^{-1} = \set {\tuple {t_1, t_2} \in T \times T: \exists s \in S: \tuple {t_1, s} \in \RR^{-1} \land \tuple {s, t_2} \in \RR}$ By definition of inverse: :$\RR \circ \RR^{-1} = \set {\tuple {t_1, t_2} \in T \times T: \exists s \in S: \tuple {s, t_1} \in \RR \land \tuple {s, t_2} \in \RR}$ But $\RR$ is many-to-one, and so: :$t_1 = t_2$ So: :$\forall \tuple {t_1, t_2} \in \RR \circ \RR^{-1}: t_1 = t_2$ and so: :$\RR \circ \RR^{-1} \subseteq I_T$ Now let $t \in T$. By definition of identity mapping on $T$: :$\tuple {t, t} \in I_T$ As $\RR$ is right-total: :$\Img \RR = T$ and so: :$\exists s \in S: \tuple {s, t} \in \RR$ and so: :$\exists s \in S: \tuple {t, s} \in \RR^{-1}$ Hence by definition of the composite of $\RR^{-1}$ and $\RR$: :$\tuple {t, t} \in \RR \circ \RR^{-1}$ So: :$\RR \circ \RR^{-1} \supseteq I_T$ and so: :$\RR \circ \RR^{-1} = I_T$ {{qed|lemma}} Hence the result. {{qed}}
=== Closure === A relation followed by another relation is another relation, from the definition of composition of relations. As the domain and codomain of two relations are the same, then: : $\forall \mathcal R_1, \mathcal R_2 \subseteq S \times S: \mathcal R_1 \circ \mathcal R_2 \subseteq S \times S$ Therefore composition of relations on a set is closed. {{qed|lemma}} === Associativity === Composition of Relations is Associative. {{qed|lemma}} === Identity === From Identity Mapping is Left Identity and Identity Mapping is Right Identity we have that: : $\forall \mathcal R \subseteq S \times S: \mathcal R \circ I_S = \mathcal R = I_S \circ \mathcal R$ Hence the identity mapping is the identity element of this set of relations. {{qed|lemma}} It is closed and associative, so it is a semigroup. It has an identity element, so it is a monoid. === Inverses === Now if $\mathcal R \in \Bbb E$ were to be invertible, it would need to satisfy: * $\mathcal R^{-1} \circ \mathcal R = I_S$ and * $\mathcal R \circ \mathcal R^{-1} = I_S$. From Inverse Relation is Left and Right Inverse iff Bijection this can only happen if $\mathcal R$ is a bijection on $S$, that is, a permutation, and its inverse is then $\mathcal R^{-1}$. {{qed}}
=== Necessary Condition === Let $\RR \subseteq S \times T$ be such that: :$\RR^{-1} \circ \RR = I_S$ and :$\RR \circ \RR^{-1} = I_T$. From Left and Right Inverse Relations Implies Bijection, it follows that $\RR$ is a bijection. {{qed|lemma}} === Sufficient Condition === Suppose $\RR$ is a bijection. From Bijective Relation has Left and Right Inverse we have that: :$\RR^{-1} \circ \RR = I_S$ and :$\RR \circ \RR^{-1} = I_T$. {{qed|lemma}} Hence the result. {{qed}}
Let $x, y \in n \Z$. Then $\exists p, q \in \Z: x = n p, y = n q$. So $x + y = n p + n q = n \paren {p + q}$ where $p + q \in \Z$. Thus $x + y \in n \Z$ and so $\struct {n \Z, +}$ is closed. {{qed}}
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$. $\boxminus$ is the congruence relation defined on $\N \times \N$ as: :$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested. Integer multiplication is defined as: :$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {a c + b d, a d + b c} {}$ From Integer Multiplication is Closed, we have that $x, y \in \Z \implies x y \in \Z$. From Ring of Integers has no Zero Divisors, we have that $x, y \in \Z: x, y \ne 0 \implies x y \ne 0$. Therefore multiplication on the non-zero integers is closed. {{qed}}
From Element under Right Operation is Left Identity we have that $\left({S, \rightarrow}\right)$ is a semigroup, whatever the nature of $S$. From Element under Left Operation is Right Identity we have that $\left({S, \leftarrow}\right)$ is a semigroup, whatever the nature of $S$. Let $T \in \mathcal P \left({S}\right)$. Then: : From Element under Right Operation is Left Identity, $\left({T, \rightarrow}\right)$ is a semigroup, and therefore a subsemigroup of $\left({S, \rightarrow}\right)$. : From Element under Left Operation is Right Identity, $\left({T, \leftarrow}\right)$ is a semigroup, and therefore a subsemigroup of $\left({S, \leftarrow}\right)$. This applies whatever $S$ is and whatever the subset $T$ is. {{qed}}
This is a reformulation of the Two-Step Subgroup Test in terms of subset product. === Necessary Condition === Let $H$ is a subgroup of $G$. Then $H$ is closed. It follows from Magma Subset Product with Self: :$H H \subseteq H$ Then: :$g \in H^{-1} \implies \exists h \in H: g = h^{-1} \implies g \in H$ so: :$H^{-1} \subseteq H$ {{qed|lemma}} === Sufficient Condition === Let: :$H H \subseteq H$ :$H^{-1} \subseteq H$ From the definition of subset product: :$\forall x, y \in H: x y \in H$ :$\forall x \in H^{-1}: x^{-1} \in H$ So by the Two-Step Subgroup Test, $H$ is a subgroup of $G$. {{qed}}
This is a reformulation of the One-Step Subgroup Test in terms of subset product. === Necessary Condition === Let $H$ be a subgroup of $G$. Let $x, y \in H$. Then by the definition of subset product: :$x y^{-1} \in H H^{-1}$ As $H \le G$, from the One-Step Subgroup Test, $x y^{-1} \in H$. Thus $H H^{-1} \subseteq H$. {{qed|lemma}} === Sufficient Condition === Let: : $H H^{-1} \subseteq H$ From the definition of subset product: : $\forall x, y \in H: x y^{-1} \in H$ So by the One-Step Subgroup Test, $H$ is a subgroup of $G$. {{qed}}
Let $x \circ t \in X \circ \paren {Y \cup Z}$. We have $x \in X, t \in Y \cup Z$ by definition of subset product. By definition of set union, it follows that $t \in Y$ or $t \in Z$. So we also have $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$. That is: :$x \circ t \in \paren {X \circ Y} \cup \paren {X \circ Z}$ and so: :$X \circ \paren {Y \cup Z} \subseteq \paren {X \circ Y} \cup \paren {X \circ Z}$ Now let $x \circ t \in \paren {X \circ Y} \cup \paren {X \circ Z}$. By definition of set union, it follows that $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$. So $x \in X$, and $y \in Y$ or $y \in Z$. That is, $x \in X$, and $y \in Y \cup Z$ by definition of set union. Hence: :$x \circ t \in X \circ \paren {Y \cup Z}$ and so: :$\paren {X \circ Y} \cup \paren {X \circ Z} \subseteq X \circ \paren {Y \cup Z}$ That is: :$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$ The result: :$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$ follows similarly. {{Qed}}
Consider the relation $\mathcal R \subseteq G \times G$ defined as: :$\forall g, h \in G: \tuple {g, h} \in \mathcal R \iff \exists g \in X$ Then: :$\forall S \subseteq G: X \circ S = \map {\mathcal R} S$ Then: {{begin-eqn}} {{eqn | l = X \circ \paren {Y \cup Z}       | r = \map {\mathcal R} {Y \cup Z}       | c =  }} {{eqn | r = \map {\mathcal R} y \cup \map {\mathcal R} Z       | c = Image of Union under Relation }} {{eqn | r = \paren {X \circ Y} \cup \paren {X \circ Z}       | c =  }} {{end-eqn}} Next, consider the relation $\mathcal R \subseteq G \times G$ defined as: :$\forall g, h \in G: \tuple {g, h} \in \mathcal R \iff \exists h \in X$ Then: :$\forall S \subseteq G: S \circ X = \map {\mathcal R} S$ Then: {{begin-eqn}} {{eqn | l = \paren {Y \cup Z} \circ X       | r = \map {\mathcal R} {Y \cup Z}       | c =  }} {{eqn | r = \map {\mathcal R} Y \cup \map {\mathcal R} Z       | c = Image of Union under Relation }} {{eqn | r = \paren {Y \circ X} \cup \paren {Z \circ X}       | c =  }} {{end-eqn}} {{qed}}
Follows directly from: * Quotient Group is Group * Cosets of $G$ by $H$ are subsets of $G$ and therefore elements of $\powerset G$ * The operation $\circ_H$ is defined as the subset product of cosets. {{qed}}