hcju/proofwikiqa huggingface dataset:
{{AimForCont}} that $\struct {\Q, \times}$ is a group. By the definition of the number $0 \in \Q$: :$\forall x \in \Q: x \times 0 = 0 = 0 \times x$ Thus $0$ is a zero in the abstract algebraic sense. From Group with Zero Element is Trivial, $\struct {\Q, \times}$ is the trivial group. But $\Q$ contains other elements besides $0$. From this contradiction it follows that $\struct {\Q, \times}$ is not a group. {{qed}}
{{AimForCont}} that $\struct {\R, \times}$ is a group. By the definition of the number $0 \in \R$: :$\forall x \in \R: x \times 0 = 0 = 0 \times x$ Thus $0$ is a zero in the abstract algebraic sense. From Group with Zero Element is Trivial, $\struct {\R, \times}$ is the trivial group. But $\R$ contains other elements besides $0$. From this contradiction it follows that $\struct {\R, \times}$ is not a group. {{qed}}
{{AimForCont}} that $\struct {\C, \times}$ is a group. By the definition of the number $0 \in \C$: :$\forall x \in \C: x \times 0 = 0 = 0 \times x$ Thus $0$ is a zero in the abstract algebraic sense. From Group with Zero Element is Trivial, $\struct {\C, \times}$ is the trivial group. But $\C$ contains other elements besides $0$. From this contradiction it follows that $\struct {\C, \times}$ is not a group. {{qed}}
Let us refer to this group as $D_4$. Taking the group axioms in turn: === G0: Closure === From the Cayley table it is seen directly that $D_4$ is closed. {{qed|lemma}} === G1: Associativity === Composition of Mappings is Associative. {{qed|lemma}} === G2: Identity === The identity is $e$ as defined. {{qed|lemma}} === G3: Inverses === Each element can be seen to have an inverse: :$r^{-1} = r^3$ and so $\left({r^3}\right)^{-1} = r$ :$r^2$, $t_{AC}$, $t_{BD}$, $t_x$ and $t_y$ are all self-inverse. {{qed|lemma}} Thus $D_4$ is seen to be a group. Note that from the Cayley table it can be observed directly that: :$r \circ t_x = t_{BD}$ :$t_x \circ r = t_{AC}$ thus illustrating by counterexample that $D_4$ is not abelian. {{qed}}
By Symmetric Group is Group, $\struct {\Gamma \paren S, \circ}$ is a group. Let $\phi \in \Gamma \paren S$ be a permutation on $S$. As a permutation is a self-map, it follows that $\phi \in S^S$. Thus by definition $\Gamma \paren S$ is a subset of $S^S$. So by definition, $\Gamma \paren S$, is a subgroup of $\struct {S^S, \circ}$. {{qed}} Category:Symmetric Groups Category:Permutation Theory 7qa78uey4kkvp6bapsfmznyb8xgrmxv
=== Necessary Condition === Let $a \in S$ be cancellable for $\circ$. Suppose there exists $x \in S$ which appears twice in a row in $\mathcal T$ headed by $a$. Thus by definition of the structure of a Cayley table: :$\exists y_1, y_2 \in S: a \circ y_1 = x = a \circ y_2$ such that $y_1 \ne y_2$. That contradicts the stipulation that $a$ is cancellable for $\circ$. So no element of $S$ is repeated in $\mathcal T$ in the row headed by $a$. Similarly, suppose there exists $x \in S$ which appears twice in a column in $\mathcal T$ headed by $a$. Thus by definition of the structure of a Cayley table: :$\exists y_1, y_2 \in S: y_1 \circ a = x = y_2 \circ a$ such that $y_1 \ne y_2$. That contradicts the stipulation that $a$ is cancellable for $\circ$. So no element of $S$ is repeated in $\mathcal T$ in the column headed by $a$. {{qed|lemma}} === Sufficient Condition === Let $a$ be such that: :$(1): \quad$ no element of $S$ is repeated in $\mathcal T$ in the row headed by $a$ and: :$(2): \quad$ no element of $S$ is repeated in $\mathcal T$ in the column headed by $a$. Thus from $(1)$: :$\forall y_1, y_2 \in S: y_1 \ne y_2 \implies a \circ y_1 \ne a \circ y_2$ So by the Rule of Transposition: :$\forall y_1, y_2 \in S: a \circ y_1 = a \circ y_2 \implies y_1 = y_2$ Also from $(2)$: :$\forall y_1, y_2 \in S: y_1 \ne y_2 \implies y_1 \circ a \ne y_2 \circ a$ So by the Rule of Transposition: :$\forall y_1, y_2 \in S: y_1 \circ a = y_2 \circ a \implies y_1 = y_2$ Thus it follows that $a$ is cancellable for $\circ$. {{qed}}
Let $\struct {\Z, -}$ be the magma which is the set of integers under the operation of subtraction. We have that the natural numbers $\N$ are a subset of the integers. Consider $\struct {\N, -}$, the natural numbers under subtraction. We have that Natural Number Subtraction is not Closed. For example: : $1 - 2 = -1 \notin \N$ Thus $\struct {\N, -}$ is not closed. So $\struct {\N, -}$ is not a submagma of $\struct {\Z, -}$ Hence it is not true to write $\struct {\N, -} \subseteq \struct {\Z, -}$, despite the fact that $\N \subseteq \Z$. Thus $\struct {\N, -}$ is not a submagma of $\struct {\Z, -}$. {{qed}}
Recall that a submagma of an algebraic structure $\mathcal S$ is a subsets of $\mathcal S$ which is closed. Let $\mathcal X$ be the set of all submagmas of $\mathcal S$. From Empty Set is Submagma of Magma: :$\O \in \mathcal X$ From Magma is Submagma of Itself: :$\mathcal S \in \mathcal X$ From Idempotent Magma Element forms Singleton Submagma: :$\set e \in \mathcal X$ Let us refer to the Cayley table: === Cayley Table of Symmetry Group of Square === {{:Symmetry Group of Square/Cayley Table}} Taking each of the elements of $\mathcal S$ in order: {{begin-eqn}} {{eqn | l = t_x \circ t^x       | r = e }} {{eqn | l = t_y \circ t^y       | r = e }} {{eqn | l = t_{AC} \circ t^{AC}       | r = e }} {{eqn | l = t_{BD} \circ t^{BD}       | r = e }} {{eqn | l = r^2 \circ r^2       | r = e }} {{eqn | l = r \circ r       | r = r^2 }} {{eqn | l = r \circ r^2       | r = r^3 }} {{eqn | l = r \circ r^3       | r = e }} {{eqn | l = r^3 \circ r^3       | r = r^2 }} {{eqn | l = r^3 \circ r^2       | r = r }} {{eqn | l = r^3 \circ r       | r = e }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = \set {e, r^2}       | o = \in       | r = \mathcal X }} {{eqn | l = \set {e, t_x}       | o = \in       | r = \mathcal X }} {{eqn | l = \set {e, t_y}       | o = \in       | r = \mathcal X }} {{eqn | l = \set {e, t_{AC} }       | o = \in       | r = \mathcal X }} {{eqn | l = \set {e, t_{BD} }       | o = \in       | r = \mathcal X }} {{eqn | l = \set {e, r, r^2, r^3}       | o = \in       | r = \mathcal X }} {{end-eqn}} Next note by inspection that: :$\set {e, r^2, t_x, t_y} \in \mathcal X$ and: :$\set {e, r^2, t_{AC}, t_{BD} } \in \mathcal X$ {{begin-eqn}} {{eqn | l = \set {e, r^2, t_x, t_y}       | o = \in       | r = \mathcal X }} {{eqn | l = \set {e, r^2, t_{AC}, t_{BD} }       | o = \in       | r = \mathcal X }} {{end-eqn}} Finally note by inspection that: : any closed subset of $\mathcal S$ which contains both $r$ and any of the reflections contains all the elements of $\mathcal S$ : any closed subset of $\mathcal S$ which contains both $r^3$ and any of the reflections contains all the elements of $\mathcal S$. Thus there are no more proper subsets of $\mathcal S$ which are submagmas of $\mathcal S$. {{qed}}
Consider the function: :$f \left({x}\right) = \begin{cases} 0 & : x = a \\ \dfrac 1 {x - a} & : x \in \left({a \,.\,.\, b}\right] \end{cases}$ Since $\dfrac 1 {x - a}$ is continuous on $\left({a \,.\,.\, b}\right)$, $f$ is continuous on $\left({a \,.\,.\, b}\right)$. Therefore, $f$ is a piecewise continuous function for the (finite) subdivision $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$. We now consider whether the improper integral $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ exists. Let $c$ be a point in $\left({a \,.\,.\, b}\right)$. From the definition of improper integral, the existence of $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ requires that $\displaystyle \lim_{\gamma \mathop \to a^+} \int_\gamma^c \dfrac 1 {x - a} \rd x$ exists. We have {{begin-eqn}} {{eqn | l = \int_\gamma^c \frac 1 {x - a} \rd x       | r = \ln \left({x - a}\right) \vert _\gamma^c }} {{eqn | r = \ln \left({c - a}\right) - \ln \left({\gamma - a}\right) }} {{eqn | r = \ln \left({c - a}\right) + \ln \left({\frac 1 {\gamma - a} }\right) }} {{end-eqn}} The last {{RHS}} approaches $\infty$ as $\gamma$ approaches $a$ from above. So $\displaystyle \int_{a^+}^c \dfrac 1 {x - a} \rd x$ does not exist. Therefore, $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ does not exist either. Accordingly, $f$ is not a piecewise continuous function with improper integrals. {{qed}} Category:Piecewise Continuous Functions 7t74sablucz0x8v3jocui9occt1lzso
Let $x \in T$ be right cancellable in $S$. That is: : $\forall a, b \in S: a \circ x = b \circ x \implies a = b$ Therefore: : $\forall c, d \in T: c \circ x = d \circ x \implies c = d$ Thus $x$ is right cancellable in $T$. {{qed}} Category:Abstract Algebra hzjs6p195emmj6ujwsi0gz9wpdsg06q
Let $x \in T$ be left cancellable in $S$. That is: : $\forall a, b \in S: x \circ a = x \circ b \implies a = b$ Therefore: : $\forall c, d \in T: x \circ c = x \circ d \implies c = d$ Thus $x$ is left cancellable in $T$. {{qed}} Category:Abstract Algebra msfv3rbqh4ysjgyrf5b5uzs8ty9ymm3
Let $C_\lambda$ be the set of left cancellable elements of $\struct {S, \circ}$: :$C_\lambda = \set {x \in S: \forall a, b \in S: x \circ a = x \circ b \implies a = b}$ Let $x, y \in C_\lambda$. Then: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ a       | r = \paren {x \circ y} \circ b       | c = }} {{eqn | ll= \leadsto       | l = x \circ \paren {y \circ a}       | r = x \circ \paren {y \circ b}       | c = by associativity of $\circ$ }} {{eqn | ll= \leadsto       | l = x \circ a       | r = x \circ b       | c = as $y \in C_\lambda$ }} {{eqn | ll= \leadsto       | l = a       | r = b       | c = as $x \in C_\lambda$ }} {{eqn | ll= \leadsto       | l = x \circ y       | o = \in       | r = C_\lambda       | c =  }} {{end-eqn}} Thus $\struct {C_\lambda, \circ}$ is closed. Therefore by the Subsemigroup Closure Test $\struct {C_\lambda, \circ}$ is a subsemigroup of $\struct {S, \circ}$. {{qed}} Category:Subsemigroups Category:Cancellability b9eqfcllj7pbuvrl4sw72qy0mq7828r
Let $C_\rho$ be the set of right cancellable elements of $\struct {S, \circ}$: :$C_\rho = \set {x \in S: \forall a, b \in S: a \circ x = b \circ x \implies a = b}$ Let $x, y \in C_\rho$. Then: {{begin-eqn}} {{eqn | l = a \circ \paren {x \circ y}       | r = b \circ \paren {x \circ y}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a \circ x} \circ y       | r = \paren {b \circ x} \circ y       | c = by associativity of $\circ$ }} {{eqn | ll= \leadsto       | l = a \circ x       | r = b \circ x       | c = as $y \in C_\rho$ }} {{eqn | ll= \leadsto       | l = a       | r = b       | c = as $x \in C_\rho$ }} {{eqn | ll= \leadsto       | l = x \circ y       | o = \in       | r = C_\rho       | c =  }} {{end-eqn}} Thus $\struct {C_\rho, \circ}$ is closed. Therefore by the Subsemigroup Closure Test $\struct {C_\rho, \circ}$ is a subsemigroup of $\struct {S, \circ}$. {{qed}} Category:Subsemigroups Category:Cancellability mv9ktdrtnrrawipd59a65a8rjr13i30
Let $f: X \to Y$ be surjective. Let $h_1: Y \to Z, h_2: Y \to Z: h_1 \circ f = h_2 \circ f$. As $f$ is a surjection: :$\Img f = Y$ by definition. But in order for $h_1 \circ f$ to be defined, it is necessary that $Y = \Dom {h_1}$. Similarly, for $h_2 \circ f$ to be defined, it is necessary that $Y = \Dom {h_2}$. So it follows that the domains of $h_1$ and $h_2$ are the same. Also: :The codomain of $h_1$ equals the codomain of $h_1 \circ f$ :The codomain of $h_2$ equals the codomain of $h_2 \circ f$ again by definition of composition of mappings. Now, we have shown that the domains and codomains of $h_1$ and $h_2$ are the same. All we need to do now to prove that $h_1 = h_2$, and therefore that $f$ is right cancellable, is to show that: :$\forall y \in Y: h_1 \paren y = h_2 \paren y$. So, let $y \in Y$. As $f$ is surjective: : $\exists x \in X: y = f \paren x$ Thus: {{begin-eqn}} {{eqn | l = h_1 \paren y       | r = h_1 \paren {f \paren x}       | c = Definition of $y$ }} {{eqn | r = h_1 \circ f \paren x       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = h_2 \circ f \paren x       | c = By Hypothesis }} {{eqn | r = h_2 \paren {f \paren x}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = h_2 \paren y       | c = Definition of $y$ }} {{end-eqn}} Thus $h_1 \paren y = h_2 \paren y$ and thus $f$ is right cancellable.
Let $f: X \to Y$ be surjective. Then from Surjection iff Right Inverse: :$\exists g: Y \to X: f \circ g = I_Y$ Suppose $h \circ f = k \circ f$ for two mappings $h: Y \to Z$ and $k: Y \to Z$. Then: {{begin-eqn}} {{eqn | l = h       | r = h \circ I_Y       | c =  }} {{eqn | r = h \circ \paren {f \circ g}       | c =  }} {{eqn | r = \paren {h \circ f} \circ g       | c = Composition of Mappings is Associative }} {{eqn | r = \paren {k \circ f} \circ g       | c = by hypothesis }} {{eqn | r = k \circ \paren {f \circ g}       | c = Composition of Mappings is Associative }} {{eqn | r = k \circ I_Y       | c =  }} {{eqn | r = k       | c =  }} {{end-eqn}} Thus $f$ is right cancellable. So surjectivity implies right cancellability.
Suppose $f$ is a mapping which is not surjective. Then: : $\exists y_1 \in Y: \neg \exists x \in X: f \paren x = y_1$ Let $Z = \set {a, b}$. Let $h_1$ and $h_2$ be defined as follows.  : $h_1 \paren y = a: y \in Y$ : $h_2 \paren y = \begin{cases} a & : y \ne y_1 \\ b & : y = y_1 \end{cases}$ Thus we have $h_1 \ne h_2$ such that $h_1 \circ f = h_2 \circ f$. Therefore $f$ is not right cancellable. It follows from the Rule of Transposition that if $f$ is right cancellable, then $f$ must be surjective.
Let $f: X \to Y$ be a right cancellable mapping. Let $Y$ contain exactly one element. Then by definition $Y$ is a singleton. From Mapping to Singleton is Surjection it follows that $f$ is a surjection. So let $Y$ contain at least two elements. Call those two elements $a$ and $b$, and we note that $a \ne b$. We define the two mappings $h, k$ as follows: :$h: Y \to Y: \forall x \in Y: \map h x = \begin{cases} x & : x \in \Img f \\ a & : x \notin \Img f \end{cases}$ :$k: Y \to Y: \forall x \in Y: \map k x = \begin{cases} x & : x \in \Img f \\ b & : x \notin \Img f \end{cases}$ It is clear that: :$\forall y \in X: \map h {\map f y} = \map f y = \map k {\map f y}$ and so: :$h \circ f = k \circ f$ But by hypothesis, $f$ is right cancellable. Thus $h = k$. {{AimForCont}} $Y \ne \Img f$. Then: :$\Img f \subsetneq Y$ That is: :$\exists x \in Y: x \notin \Img f$ It follows that: :$a = \map h x = \map k x = b$ But we posited that $a \ne b$. From this contradiction we conclude that: :$Y = \Img f$ So, by definition, $f$ must be a surjection.
$\mathcal R$ is shown to be an equivalence relation thus: === Reflexivity === : $\forall x \in \Z: \left\vert{x}\right\vert = \left\vert{x}\right\vert$ Thus $\mathcal R$ is reflexive. {{qed|lemma}} === Symmetry === : $\forall x, y \in \Z: \left\vert{x}\right\vert = \left\vert{y}\right\vert \implies \left\vert{y}\right\vert = \left\vert{x}\right\vert$ Thus $\mathcal R$ is symmetric. {{qed|lemma}} === Transitive === : $\forall x, y, z \in \Z: \left\vert{x}\right\vert = \left\vert{y}\right\vert \land \left\vert{y}\right\vert = \left\vert{z}\right\vert \implies \left\vert{x}\right\vert = \left\vert{z}\right\vert$ Thus $\mathcal R$ is transitive. {{qed|lemma}} Thus, by definition, $\mathcal R$ is an equivalence relation. {{qed}}
From Absolute Value Function on Integers induces Equivalence Relation, $\RR$ is an equivalence relation. However, consider that: {{begin-eqn}} {{eqn | l = \size {-1} = \size 1       | o = \leadsto       | r = -1 \mathop \RR 1       | c =  }} {{eqn | l = \size 2 = \size 2       | o = \leadsto       | r = 2 \mathop \RR 2       | c =  }} {{end-eqn}} By conventional integer addition: :$-1 + 2 = 1$ while: :$1 + 2 = 3$ But it does not hold that: :$\size 1 = \size 3$ Therefore $\RR$ is not a congruence relation for integer addition. Hence the result, by Proof by Counterexample. {{qed}}
Let $\mathcal R$ be an equivalence relation on the structure $\left({S, \rightarrow}\right)$. Then: : $x_1 \rightarrow y_1 = y_1$ : $x_2 \rightarrow y_2 = y_2$ Suppose $x_1 \mathop {\mathcal R} x_2 \land y_1 \mathop {\mathcal R} y_2$. It follows directly that: : $\left({x_1 \rightarrow y_1}\right) \mathop {\mathcal R} \left({x_2 \rightarrow y_2}\right)$ {{Qed}}
Let $\mathcal R$ be an equivalence relation on the structure $\left({S, \leftarrow}\right)$. Then: : $x_1 \leftarrow y_1 = x_1$ : $x_2 \leftarrow y_2 = x_2$ Suppose $x_1 \mathop {\mathcal R} x_2 \land y_1 \mathop {\mathcal R} y_2$. It follows directly that: : $\left({x_1 \leftarrow y_1}\right) \mathop {\mathcal R} \left({x_2 \leftarrow y_2}\right)$ {{Qed}}
Follows directly from: : Left Congruence Modulo Subgroup is Equivalence Relation : Relation Partitions Set iff Equivalence {{qed}}
Follows directly from: : Right Congruence Modulo Subgroup is Equivalence Relation : Relation Partitions Set iff Equivalence. {{qed}}
{{begin-eqn}} {{eqn | l = H x       | r = H y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \equiv^r       | r = y \bmod H       | c = Right Coset Space forms Partition }} {{eqn | ll= \leadstoandfrom       | l = x y^{-1}       | o = \in       | r = H       | c = Equivalent Statements for Congruence Modulo Subgroup }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l = x H       | r = y H       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \equiv^l       | r = y \pmod H       | c = Left Coset Space forms Partition }} {{eqn | ll= \leadstoandfrom       | l = x^{-1} y       | o = \in       | r = H       | c = Equivalent Statements for Congruence Modulo Subgroup }} {{end-eqn}} {{Qed}}
Let $x \in \eqclass g {\mathcal R^l_H}$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \eqclass g {\mathcal R^l_H}       | c =  }} {{eqn | ll= \leadsto       | lo= \exists h \in H:       | l = g^{-1} x       | r = h       | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}} }} {{eqn | ll= \leadsto       | lo= \exists h \in H:       | l = x       | r = g h       | c = Group properties }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = g H       | c = {{Defof|Left Coset}} }} {{eqn | ll= \leadsto       | l = \eqclass g {\mathcal R^l_H}       | o = \subseteq       | r = g H       | c = {{Defof|Subgroup}} }} {{end-eqn}} Now let $x \in g H$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = g H       | c =  }} {{eqn | ll= \leadsto       | lo= \exists h \in H:       | l = x       | r = g h       | c = {{Defof|Left Coset}} }} {{eqn | ll= \leadsto       | l = g^{-1} x       | r = h \in H       | c = Group properties }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \eqclass g {\mathcal R^l_H}       | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}} }} {{eqn | ll= \leadsto       | r = g H \subseteq \eqclass g {\mathcal R^l_H}       | c = {{Defof|Subgroup}} }} {{end-eqn}} Thus: :$\eqclass g {\mathcal R^l_H} = g H$ that is, the equivalence class $\eqclass g {\mathcal R^l_H}$ of an element $g \in G$ equals the left coset $g H$. {{Qed}}
Let $x \in \eqclass g {\mathcal R^r_H}$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \eqclass g {\mathcal R^r_H}       | c =  }} {{eqn | ll= \leadsto       | lo= \exists h \in H:       | l = x g^{-1}       | r = h       | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}} }} {{eqn | ll= \leadsto       | lo= \exists h \in H:       | l = x       | r = h g       | c = Group Properties }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = H g       | c = {{Defof|Right Coset}} }} {{eqn | ll= \leadsto       | l = \eqclass g {\mathcal R^r_H}       | o = \subseteq       | r = H g       | c = {{Defof|Subgroup}} }} {{end-eqn}} Now let $x \in g H$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = H g       | c =  }} {{eqn | ll= \leadsto       | lo= \exists h \in H:       | l = x       | r = h g       | c = {{Defof|Right Coset}} }} {{eqn | ll= \leadsto       | l = x g^{-1}       | r = h \in H       | c = Group Properties }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \eqclass g {\mathcal R^r_H}       | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}} }} {{eqn | ll= \leadsto       | l = H g       | o = \subseteq       | r = \eqclass g {\mathcal R^r_H}       | c = {{Defof|Subgroup}} }} {{end-eqn}} Thus: :$\eqclass g {\mathcal R^r_H} = H g$ That is, the equivalence class $\eqclass g {\mathcal R^r_H}$ of an element $g \in G$ equals the right coset $g H$. {{Qed}}
Let $G$ be a group whose identity is $e$. Let $H$ be a subgroup of $G$. For clarity of expression, we will use the notation: :$\tuple {x, y} \in \mathcal R^l_H$ for: :$x \equiv^l y \pmod H$ From the definition of left congruence modulo a subgroup, we have: :$\mathcal R^l_H = \set {\tuple {x, y} \in G \times G: x^{-1} y \in H}$ We show that $\mathcal R^l_H$ is an equivalence: === Reflexive === We have that $H$ is a subgroup of $G$. From Identity of Subgroup: :$e \in H$ Hence: :$\forall x \in G: x^{-1} x = e \in H \implies \tuple {x, x} \in \mathcal R^l_H$ and so $\mathcal R^l_H$ is reflexive. {{qed|lemma}} === Symmetric === {{begin-eqn}} {{eqn | l = \tuple {x, y}       | o = \in       | r = \mathcal R^l_H       | c =  }} {{eqn | ll= \leadsto       | l = x^{-1} y       | o = \in       | r = H       | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}} }} {{eqn | ll= \leadsto       | l = \tuple {x^{-1} y}^{-1}       | o = \in       | r = H       | c = {{GroupAxiom|0}} }} {{end-eqn}} But then: : $\tuple {x^{-1} y}^{-1} = y^{-1} x \implies \tuple {y, x} \in \mathcal R^l_H$ Thus $\mathcal R^l_H$ is symmetric. {{qed|lemma}} === Transitive === {{begin-eqn}} {{eqn | l = \tuple {x, y}, \tuple {y, z}       | o = \in       | r = \mathcal R^l_H       | c =  }} {{eqn | ll= \leadsto       | l = x^{-1} y       | o = \in       | r = H       | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}} }} {{eqn | lo= \land       | l = y^{-1} z       | o = \in       | r = H       | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}} }} {{eqn | ll= \leadsto       | l = \tuple {x^{-1} y} \tuple {y^{-1} z} = x^{-1} z       | o = \in       | r = H       | c = Group Properties }} {{eqn | ll= \leadsto       | l = \tuple {x, z}       | o = \in       | r = R^l_H       | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}} }} {{end-eqn}} Thus $\mathcal R^l_H$ is transitive. {{qed|lemma}} So $\mathcal R^l_H$ is an equivalence relation. {{qed}}
Let $G$ be a group whose identity is $e$. Let $H$ be a subgroup of $G$. For clarity of expression, we will use the notation: :$\tuple {x, y} \in \mathcal R^l_H$ for: :$x \equiv^l y \pmod H$ From the definition of left congruence modulo a subgroup, we have: :$\mathcal R^l_H = \set {\tuple {x, y} \in G \times G: x^{-1} y \in H}$ We show that $\mathcal R^l_H$ is an equivalence: === Reflexive === We have that $H$ is a subgroup of $G$. From Identity of Subgroup: :$e \in H$ Hence: :$\forall x \in G: x x^{-1} = e \in H \implies \tuple {x, x} \in \mathcal R^r_H$ and so $\mathcal R^r_H$ is reflexive. {{qed|lemma}} === Symmetric === {{begin-eqn}} {{eqn | l = \tuple {x, y}       | o = \in       | r = \mathcal R^r_H       | c =  }} {{eqn | ll= \leadsto       | l = x y^{-1}       | o = \in       | r = H       | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}} }} {{eqn | ll= \leadsto       | l = \tuple {x y^{-1} }^{-1}       | o = \in       | r = H       | c = {{GroupAxiom|0}} }} {{end-eqn}} But then: : $\tuple {x y^{-1} }^{-1} = y x^{-1} \implies \tuple {y, x} \in \mathcal R^r_H$ Thus $\mathcal R^r_H$ is symmetric. {{qed|lemma}} === Transitive === {{begin-eqn}} {{eqn | l = \tuple {x, y}, \tuple {y, z}       | o = \in       | r = \mathcal R^r_H       | c =  }} {{eqn | ll= \leadsto       | l = x y^{-1}       | o = \in       | r = H       | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}} }} {{eqn | lo= \land       | l = y z^{-1}       | o = \in       | r = H       | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}} }} {{eqn | ll= \leadsto       | l = \tuple {x y^{-1} } \tuple {y z^{-1} } = x z^{-1}       | o = \in       | r = H       | c = Group Properties }} {{eqn | ll= \leadsto       | l = \tuple {x, z}       | o = \in       | r = R^r_H       | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}} }} {{end-eqn}} Thus $\mathcal R^r_H$ is transitive. {{qed|lemma}} So $\mathcal R^r_H$ is an equivalence relation. {{qed}}