hcju/proofwikiqa huggingface dataset:
By Quotient Mapping on Structure is Canonical Epimorphism, both $q_K$ and $q_H$ are epimorphisms. From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \left({G / H}\right) / K$ is also an epimorphism. Now: :$\forall x \in G: x \in \ker \left({q_K \circ q_H}\right) \iff q_K \left({q_H \left({x}\right)}\right) = K = e_{G/H}$ This means the same as: :$q_H \left({x}\right) \in \ker \left({q_K}\right) = K$ But: :$q_H \left({x}\right) \in K \iff x \in q_H^{-1} \left({K}\right) = L$ Thus: :$L = \ker \left({q_K \circ q_H}\right)$ By Kernel is Normal Subgroup of Domain: : $L \lhd G$ {{qed}}
From Trivial Subgroup is Normal: :$\set {e_G} \lhd G$ Let $x \in G$. Then: :$x \set {e_G} = \set {x e_G} = \set x$ So each (left) coset of $G$ modulo $\set {e_G}$ has one element. Now we set up the quotient epimorphism $\psi: G \to G / \set {e_G}$: :$\forall x \in G: \map \phi x = x \set {e_G}$ which is of course a surjection. We now need to establish that it is an injection. Let $p, q \in G$. {{begin-eqn}} {{eqn | l = \map \phi p       | r = \map \phi q       | c =  }} {{eqn | ll= \leadsto       | l = p \set {e_G}       | r = q \set {e_G}       | c = Definition of $\phi$ }} {{eqn | ll= \leadsto       | l = \set p       | r = \set q       | c = from above }} {{eqn | ll= \leadsto       | l = p       | r = q       | c = {{Defof|Set Equality}} }} {{end-eqn}} So $\psi$ is a group isomorphism and therefore: :$G / \set {e_G} \cong G$ {{qed}}
Let $\struct {G, \circ}$ be a group. We have that: : $\forall a \in G: e \circ a = a \circ e \implies e \in C_G \paren a$ Thus $C_G \paren a \ne \O$. Let $x, y \in C_G \paren a$. Then: {{begin-eqn}} {{eqn | l = x \circ a       | r = a \circ x       | c =  }} {{eqn | l = y \circ a       | r = a \circ y       | c =  }} {{eqn | ll= \leadsto       | l = x \circ y \circ a       | r = x \circ a \circ y       | c =  }} {{eqn | r = a \circ x \circ y       | c =  }} {{eqn | ll= \leadsto       | l = x \circ y       | o = \in       | r = C_G \paren a       | c =  }} {{end-eqn}} Thus $C_G \paren a$ is closed under $\circ$. Let $x \in C_G \paren a$. Then: {{begin-eqn}} {{eqn | l = x \circ a       | r = a \circ x       | c =  }} {{eqn | ll= \leadsto       | l = x^{-1} \circ x \circ a \circ x^{-1}       | r = x^{-1} \circ a \circ x \circ x^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = a \circ x^{-1}       | r = x^{-1} \circ a       | c =  }} {{end-eqn}} So: : $x \in C_G \paren a \implies x^{-1} \in C_G \paren a$ Thus, by the Two-Step Subgroup Test, the result follows. {{qed}}
Let $\struct {G, \circ}$ be a group. We have that: :$\forall a \in G: e \circ a = a \circ e \implies e \in \map {C_G} a$ Thus $\map {C_G} a \ne \O$. Let $x, y \in \map {C_G} a$. Then from Commutation with Group Elements implies Commuation with Product with Inverse: :$a \circ x \circ y^{-1} = x \circ y^{-1} \circ a$ so: :$x \circ y^{-1} \in\map {C_G} a$ The result follows by the One-Step Subgroup Test, the result follows. {{qed}}
It is clear that: :$g \in \map {C_H} x \iff g \in \map {C_G} x \land g \in H$ The result follows by definition of set intersection. {{qed}}
Let $\kappa: G \to \Aut G$ be a mapping defined by $\map \kappa x = \kappa_x$. It is clear that $\Img \kappa = \Inn G$. It is also clear that $\kappa$ is a homomorphism: {{begin-eqn}} {{eqn | l = \map \kappa x \map \kappa y       | r = \kappa_x \circ \kappa_y       | c =  }} {{eqn | r = \kappa_{x y}       | c = Inner Automorphism is Automorphism }} {{eqn | r = \map \kappa {x y}       | c =  }} {{end-eqn}} Note that $\forall \kappa_x \in \Inn G: \exists x \in G: \map \kappa x = \kappa_x$. Thus $\kappa: G \to \Inn G$ is a surjection and therefore an group epimorphism. Now we investigate the kernel of $\kappa$: {{begin-eqn}} {{eqn | l = \map \ker \kappa       | r = \set {x \in G: \kappa_x = I_G}       | c = {{Defof|Kernel of Group Homomorphism}} }} {{eqn | r = \set {x \in G: \forall g \in G: \map {\kappa_x} g = \map {I_G} g}       | c = Equality of Mappings }} {{eqn | r = \set {x \in G: \forall g \in G: x g x^{-1} = g}       | c = Definition of $\kappa_x$ }} {{eqn | r = \set {x \in G: \forall g \in G: x g = g x}       | c =  }} {{eqn | r = \map Z G       | c = {{Defof|Center of Group}} }} {{end-eqn}} So the kernel of $\kappa$ is the center of $G$. {{Qed}}
=== Necessary Condition === Let $G$ be abelian. Then: :$\forall a \in G: \forall x \in G: a x = x a$ Thus: :$\forall a \in G: a \in \map Z G = G$ {{qed|lemma}} === Sufficient Condition === Let $\map Z G = G$. Then by the definition of center: :$\forall a \in G: \forall x \in G: a x = x a$ and thus $G$ is abelian by definition. {{qed}}
Let $a, b \in \map {N_G} S$. Then: {{begin-eqn}} {{eqn | l = S^{a b}       | r = \paren {S^b}^a       | c = Conjugate of Set by Group Product }} {{eqn | r = S^a       | c = {{Defof|Normal Subgroup}} }} {{eqn | r = S       | c = {{Defof|Normal Subgroup}} }} {{end-eqn}} Therefore $a b \in \map {N_G} S$. Now let $a \in \map {N_G} S$: :$a \in \map {N_G} S \implies S^{a^{-1} } = \paren {S^a}^{a^{-1} } = S^{a^{-1} a} = S$ Therefore $a^{-1} \in \map {N_G} S$. Thus, by the Two-Step Subgroup Test, $\map {N_G} S \le G$. {{qed}}
From Subgroup is Subgroup of Normalizer we have that $H \le \map {N_G} H$. It remains to show that $H$ is normal in $\map {N_G} H$. Let $a \in H$ and $b \in \map {N_G} H$. By the definition of normalizer: : $b a b^{-1} \in H$ Thus $H$ is normal in $\map {N_G} H$. {{qed}}
From Subgroup is Subgroup of Normalizer, we have that $H \le \map {N_G} H$. Now we need to show that $H \lhd \map {N_G} H$. For $a \in \map {N_G} H$, the conjugate of $H$ by $a$ in $\map {N_G} H$ is: {{begin-eqn}} {{eqn | l = H^a       | r = \set {x \in \map {N_G} H: a x a^{-1} \in H}       | c = {{Defof|Conjugate of Group Subset}} }} {{eqn | r = H^a \cap \map {N_G} H       | c = {{Defof|Set Intersection}} }} {{eqn | r = H \cap \map {N_G} H       | c = {{Defof|Normalizer}} }} {{eqn | r = H       | c = Intersection with Subset is Subset }} {{end-eqn}} so: :$\forall a \in \map {N_G} H: H^a = H$ and so by definition of normal subgroup: :$H \lhd \map {N_G} H$ Now we need to show that $\map {N_G} H$ is the largest subgroup of $G$ containing $H$ such that $H \lhd \map {N_G} H$. That is, to show that any subgroup of $G$ in which $H$ is normal is also a subset of $\map {N_G} H$. Take any $N$ such that $H \lhd N \le G$. In $N$, the conjugate of $H$ by $a \in N$ is $N \cap H^a = H$. Therefore: :$H \subseteq H^a$ Similarly, $H \subseteq H^{a^{-1} }$, so: :$H^a \subseteq \paren {H^a}^{a^{-1} } = H$ Thus: :$\forall a \in N: H^a = H, a \in \map {N_G} H$ That is: :$N \subseteq \map {N_G} H$ So what we have shown is that any subgroup of $G$ in which $H$ is normal is a subset of $\map {N_G} H$, which is another way of saying that $\map {N_G} H$ is the largest such subgroup. {{Qed}}
=== Sufficient Condition === Let $H$ be normal in $G$. Then $G$ is trivially the largest subgroup of $G$ in which $H$ is normal. Thus from Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup: :$\map {N_G} H = G$ {{qed|lemma}} === Necessary Condition === Let $\map {N_G} H = G$. From Subgroup is Normal Subgroup of Normalizer, $H$ is normal in $\map {N_G} H$. Hence $H$ is normal in $G$. {{qed}}
From the definition of conjugate: :$S^a = \set {y \in G: \exists x \in S: y = a x a^{-1} } = a S a^{-1}$ From the definition of normalizer: :$\map {N_G} S = \set {x \in G: S^x = S}$ Thus: {{begin-eqn}} {{eqn | l = \map {N_G} {S^a}       | r = \set {x \in G: \paren {S^a}^x = S^a}       | c =  }} {{eqn | r = \set {x \in G: x a S a^{-1} x^{-1} = a S a^{-1} }       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {\map {N_G} S}^a       | r = \set {x \in G: S^x = S}^a }} {{eqn | r = \set {x \in G: x S x^{-1} = S}^a }} {{eqn | r = \set {y \in G: \exists z \in \set {x \in G: x S x^{-1} = S}: y = a z a^{-1} } }} {{end-eqn}} Suppose that $x \in \map {N_G} S$. It is to be shown that: : $a x a^{-1} \in \map {N_G} {S^a} = \map {N_G} {a S a^{-1} }$ To this end, compute: {{begin-eqn}} {{eqn | l = \paren {a x a^{-1} } a S a^{-1} \paren {a x a^{-1} }^{-1}       | r = \paren {a x a^{-1} } a S a^{-1} \paren {a x^{-1} a^{-1} }       | c = Power of Conjugate equals Conjugate of Power }} {{eqn | r = a x S x^{-1} a^{-1} }} {{eqn | r = a S a^{-1}       | c = as $x \in \map {N_G} S$ }} {{end-eqn}} Hence $a x a^{-1} \in \map {N_G} {S^a}$, and it follows that: :$z \in \paren {\map {N_G} S}^a \implies z \in \map {N_G} {S^a}$ Conversely, let $x \in \map {N_G} {S^a}$. That is, let $x \in G$ such that $x a S a^{-1} x^{-1} = a S a^{-1}$. Now if we can show that $a^{-1} x a \in \map {N_G} S$, then: :$x = a \left({a^{-1} x a}\right) a^{-1} \in \paren {\map {N_G} S}^a$ establishing the remaining inclusion. Thus, we compute: {{begin-eqn}} {{eqn | l = \paren {a^{-1} x a} S \paren {a^{-1} x a}^{-1}       | r = a^{-1} x a S a^{-1} x^{-1} a       | c = Power of Conjugate equals Conjugate of Power }} {{eqn | r = a^{-1} a S a^{-1} a       | c = as $x \in \map {N_G} {S^a}$ }} {{eqn | r = S }} {{end-eqn}} Combined with the above observation, this establishes that: :$z \in \map {N_G} {S^a} \implies z \in \paren {\map {N_G} S}^a$ Hence the result, by definition of set equality. {{qed}}
$\map {N_G} x$ is the normalizer of the set $\set x$. Thus: {{begin-eqn}} {{eqn | l = \map {N_G} x       | r = G       | c =  }} {{eqn | ll= \leadstoandfrom       | lo= \forall a \in G:       | l = \set x^a       | r = \set x       | c = {{Defof|Normalizer}} }} {{eqn | ll= \leadstoandfrom       | lo= \forall a \in G:       | l = a x a^{-1}       | r = x       | c = {{Defof|Conjugate of Group Element}} }} {{eqn | ll= \leadstoandfrom       | lo= \forall a \in G:       | l = a x       | r = x a       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \map Z G       | c = {{Defof|Center of Group}} }} {{end-eqn}} {{qed}}
From Ideal is Additive Normal Subgroup that $J$ is a normal subgroup of $R$ under $+$. From Quotient Ring Addition is Well-Defined, $+$ is a well-defined operation. The rest follows directly from the definition of quotient group. {{qed}}
From Left Cosets are Equal iff Product with Inverse in Subgroup, we have: {{begin-eqn}} {{eqn | l = x_1 + J       | r = x_2 + J       | c =  }} {{eqn | ll= \leadsto       | l = x_1 + \paren {-x_2}       | o = \in       | r = J       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y_1 + J       | r = y_2 + J       | c =  }} {{eqn | ll= \leadsto       | l = y_1 + \paren {-y_2}       | o = \in       | r = J       | c =  }} {{end-eqn}} Hence from the definition of ideal: {{begin-eqn}} {{eqn | l = \paren {x_1 + \paren {-x_2} } \circ y_1       | o = \in       | r = J }} {{eqn | l = x_2 \circ \paren {y_1 + \paren {-y_2} }       | o = \in       | r = J }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \paren {x_1 + \paren {-x_2} } \circ y_1 + x_2 \circ \paren {y_1 + \paren {-y_2} }       | o = \in       | r = J       | c = as $\struct {J, +}$ is a group }} {{eqn | ll= \leadsto       | l = x_1 \circ y_1 + \paren {-\paren {x_2 \circ y_2} }       | o = \in       | r = J       | c = Various ring properties }} {{eqn | ll= \leadsto       | l = x_1 \circ y_1 + J       | r = x_2 \circ y_2 + J       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{end-eqn}} <includeonly>{{qed|lemma}}</includeonly>
Let $\struct {R, +, \circ}$ be a commutative ring That means $\circ$ is commutative on $R$. Thus: {{begin-eqn}} {{eqn | ll= \forall x, y \in R:       | l = \paren {x + J} \circ \paren {y + J}       | r = x \circ y + J       | c = Definition of $\circ$ in $R / J$ }} {{eqn | r = y \circ x + J       | c = Commutativity of $\circ$ }} {{eqn | r = \paren {y + J} \circ \paren {x + J}       | c = Definition of $\circ$ in $R / J$ }} {{end-eqn}} {{Qed}}
Let $\struct {R, +, \circ}$ be a ring with unity. First, let $J \subsetneq R$. By Ideal of Unit is Whole Ring: Corollary: :$1_R \in J \implies J = R$ So $1_R \notin J$. Thus $1_R + J \ne J$, so $1_R + J \ne 0_{R/J}$. Now let $x \in R$. {{begin-eqn}} {{eqn | l = \paren {1_R + J} \circ \paren {x + J}       | r = 1_R \circ x + J       | c = Definition of $\circ$ in $R / J$ }} {{eqn | r = x + J       | c = {{Defof|Unity of Ring}} }} {{eqn | r = x \circ 1_R + J       | c = {{Defof|Unity of Ring}} }} {{eqn | r = \paren {x + J} \circ \paren {1_R + J}       | c = Definition of $\circ$ in $R / J$ }} {{end-eqn}} Thus $R / J$ has a unity, and that unity is $1_R + J$. Now suppose $J = R$. Then $1_R + J = J$ and therefore $1_R = 0_R$. The only ring to have $1_R = 0_R$ is the null ring. This is appropriate, because: :$R / J = R / R = \set {0_{R / R} }$ which is the null ring. {{Qed}}
Follows directly from Kernel is Trivial iff Monomorphism and the definition of a ring epimorphism. {{qed}} Category:Ring Epimorphisms Category:Ring Isomorphisms 7eicjqz42mvwfnab4mh5x2fqlui7ox6
Let $x, y \in R$. Then: {{begin-eqn}} {{eqn | l = \map \phi {x + y}       | r = \paren {x + y} + J       | c = Definition of $\phi$ }} {{eqn | r = \paren {x + J} + \paren {y + J}       | c = {{Defof|Quotient Ring Addition}} }} {{eqn | r = \map \phi x + \map \phi y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map \phi {x \circ y}       | r = \paren {x \circ y} + J       | c = Definition of $\phi$ }} {{eqn | r = \paren {x + J} \circ \paren {y + J}       | c = {{Defof|Quotient Ring Product}} }} {{eqn | r = \map \phi x \, \map \phi y       | c =  }} {{end-eqn}} Thus $\phi$ is a homomorphism. $\phi$ is surjective because: :$\forall x + J \in R / J: x + J = \map \phi x$ Therefore $\phi$ is an epimorphism. Let $x \in \map \ker \phi$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map \ker \phi       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map \phi x       | r = 0_{R/J}       | c = {{Defof|Kernel of Ring Homomorphism}} }} {{eqn | ll= \leadstoandfrom       | l = x + J       | r = J       | c = $J$ is the zero of $\struct {R / J, +, \circ}$ }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = J       | c = Left Coset Equals Subgroup iff Element in Subgroup }} {{end-eqn}} Thus: :$\map \ker \phi = J$ {{Qed}}
Since $S \ne \O$, $\phi \sqbrk S \ne \O$. From Group Homomorphism Preserves Subgroups, $\struct {\phi \sqbrk S, +_2}$ is a subgroup of $\struct {R_2, +_2}$. From Homomorphism Preserves Subsemigroups, $\struct {\phi \sqbrk S, \circ_2}$ is a subsemigroup of $\struct {R_2, \circ_2}$. Thus, as $\struct {R_2, +_2}$ is a group and $\struct {R_2, \circ_2}$ is a semigroup, the result follows. {{qed}}
From Morphism Property Preserves Closure, $\phi \sqbrk {R_1}$ is a closed algebraic structure. From Epimorphism Preserves Rings, $\phi \sqbrk S$ is a ring. Hence the result, from the definition of subring. {{qed}}
Let $S$ be a subring of $R_1$. Since $S \ne \O$ it follows that $\phi \sqbrk S \ne \O$. Let $x, y \in \phi \sqbrk S$. Then: :$\exists s, t \in S: x = \map \phi s, y = \map \phi t$ So: {{begin-eqn}} {{eqn | l = x +_2 \paren {-y}       | r = \map \phi s +_2 \paren {-\map \phi t}       | c =  }} {{eqn | r = \map \phi {s +_1 \paren {-t} }       | c = as $\phi$ is a homomorphism }} {{end-eqn}} As $S$ is a subring of $R_1$, it is closed under $+_1$ and the taking of negatives. Thus $s +_1 \paren {-t} \in S$ and so $x +_2 \paren {-y} \in \phi \sqbrk S$. Similarly: {{begin-eqn}} {{eqn | l = x \circ_2 y       | r = \map \phi s \circ_2 \map \phi t       | c =  }} {{eqn | r = \map \phi {s \circ_1 t}       | c = as $\phi$ is a homomorphism }} {{end-eqn}} Because $S$ is a subring of $R_1$, it is closed under $\circ_1$. Thus $s \circ_1 t \in S$ and so $x \circ_2 y \in \phi \sqbrk S$. The result follows from Subring Test. {{qed}}
From Ring Homomorphism of Addition is Group Homomorphism and Kernel of Group Homomorphism is Subgroup: :$\struct {\map \ker \phi, +_1} \le \struct {R_1, +_1}$ where $\le$ denotes subgroup. Let $x, y \in \map \ker \phi$. {{begin-eqn}} {{eqn | l = \map \phi {x \circ_1 y}       | r = \map \phi x \circ_2 \map \phi y       | c = Morphism Property }} {{eqn | r = 0_{R_2} \circ_2 0_{R_2}       | c = Definition of Kernel }} {{eqn | r = 0_{R_2}       | c =  }} {{end-eqn}} Thus $x \circ_1 y \in \map \ker \phi$. Thus the conditions for Subring Test are fulfilled, and $\map \ker \phi$ is a subring of $R_1$. {{qed}}
By Kernel of Ring Homomorphism is Subring, $\map \ker \phi$ is a subring of $R_1$. Let $s \in \map \ker \phi$, so $\map \phi s = 0_{R_2}$. Suppose $x \in R_1$. Then: {{begin-eqn}} {{eqn | l = \map \phi {x \circ_1 s}       | r = \map \phi x \circ_2 \map \phi s       | c = {{Defof|Morphism Property}} }} {{eqn | r = \map \phi x \circ_2 0_{R_2}       | c = as $s \in \map \ker \phi$ }} {{eqn | r = 0_{R_2}       | c = Properties of $0_{R_2}$ }} {{end-eqn}} and similarly for $\map \phi {s \circ_1 x}$. The result follows. {{Qed}}
From Null Ring is Ideal, $\set {0_R}$ is an ideal of $\struct {R, +, \circ}$, as $\struct {R, +, \circ}$, being a division ring, is also a ring. By definition, every non-zero element of a division ring is a unit. So $S \ne \set {0_R} \implies \exists x \in S: x \ne 0_R$ such that $r$ is a unit of $R$. The result follows from Ideal of Unit is Whole Ring. {{qed}}
From the Quotient Theorem for Epimorphisms, there is one and only one isomorphism that satisfies the conditions for each of the operations on $R_1$. So the first statement follows directly. From Kernel is Trivial iff Monomorphism, $\phi$ is a ring monomorphism {{iff}} $K = \set {0_{R_1} }$. As $\phi$ is also an epimorphism, the result follows.
$J$ is an ideal of $R_1$, so it is also a subring of $R_1$. From Ring Homomorphism Preserves Subrings, it follows that $\phi \sqbrk J$ is a subring of $R_2$. Now suppose $u \in \phi \sqbrk J$. Let $v \in R_2$. Then $\exists x \in J, y \in R_1$ such that $\map \phi x = u, \map \phi y = v$. Thus, by the morphism property: :$u \circ_2 v = \map \phi x \circ_2 \map \phi y = \map \phi {x \circ_1 y}$ So $u \circ_2 v \in \phi \sqbrk J$ because $x \circ_1 y \in J$. Similarly $v \circ_2 u \in \phi \sqbrk J$ also. The result follows. {{qed}}
Let $x \in \phi^{-1} \left[{\phi \left[{J}\right]}\right]$. Then: {{begin-eqn}} {{eqn | l=x       | o=\in       | r=\phi^{-1} \left[{\phi \left[{J}\right]}\right]       | c= }} {{eqn | ll=\implies       | l=\phi \left({x}\right)       | o=\in       | r=\phi \left[{J}\right]       | c=Definition of preimage }} {{eqn | ll=\implies       | l=\exists b \in J: \phi \left({x}\right)       | r=\phi \left({b}\right)       | c=Definition of image }} {{eqn | ll=\implies       | l=\phi \left({x}\right) + \left({- \phi \left({b}\right)}\right)       | r=0_{R_2}       | c=Definition of ring negative on $R_2$ }} {{eqn | ll=\implies       | l=\phi \left({x + \left({-b}\right)}\right)       | r=0_{R_2}       | c=as $\phi$ is a homomorphism }} {{eqn | ll=\implies       | l=x + \left({-b}\right)       | o=\in       | r=K       | c=Definition of kernel }} {{eqn | ll=\implies       | l=x       | o=\in       | r=J + K       | c=as $x = b + \left({x + \left({-b}\right)}\right)$ }} {{end-eqn}} So we have shown that: :$\phi^{-1} \left[{\phi \left[{J}\right]}\right] \subseteq J + K$ Now suppose that $x \in J + K$. Then: {{begin-eqn}} {{eqn | l=x       | o=\in       | r=J + K       | c= }} {{eqn | ll=\implies       | l=\exists b \in J, a \in K: x       | r=b + a       | c= }} {{eqn | ll=\implies       | l=\phi \left({x}\right)       | r=\phi \left({b}\right) + \phi \left({a}\right)       | c= }} {{eqn | r=\phi \left({b}\right) + 0_{R_2}       | c=as $a \in K$ }} {{eqn | r=\phi \left({b}\right)       | c= }} {{eqn | ll=\implies       | l=\phi \left({x}\right)       | o=\in       | r=\phi \left[{J}\right]       | c=as $b \in J$ }} {{eqn | ll=\implies       | l=x       | o=\in       | r=\phi^{-1} \left[{\phi \left[{J}\right]}\right]       | c=Definition of preimage }} {{end-eqn}} So we have shown that: :$J + K \subseteq \phi^{-1} \left[{\phi \left[{J}\right]}\right]$ Hence the result by definition of set equality. {{Qed}}
Let $K = \ker \left({\phi}\right)$ be the kernel of $R_1$. We have that $0_{R_2} \in S_2$ and so $\left\{{0_{R_2}}\right\} \subseteq S_2$. From Subset Maps to Subset: :$\phi^{-1} \left[{\left\{{0_{R_2}}\right\}}\right] \subseteq \phi^{-1} \left[{S_2}\right] = S_1$ But by definition, $K = \phi^{-1} \left[{\left\{{0_{R_2}}\right\}}\right]$ and so $S_1$ is a subset of $R_1$ containing $K$, that is: :$K \subseteq S_1 \subseteq R_1$ Now we need to show that $S_1$ is a subring of $R_1$. Let $r, r' \in S_1$. Then $\phi \left({r}\right), \phi \left({r'}\right) \in S_2$. Hence: {{begin-eqn}} {{eqn | l=\phi \left({r +_1 r'}\right)       | r=\phi \left({r}\right) +_2 \phi \left({r'}\right)       | c=as $\phi$ is a ring homomorphism }} {{eqn | o=\in       | r=S_2       | c=because $S_2$ is a subring }} {{end-eqn}} So: :$r + r' \in S_1$ Then: {{begin-eqn}} {{eqn | l=\phi \left({-r}\right)       | r=-\phi \left({r}\right)       | c=Group Homomorphism Preserves Inverses }} {{eqn | o=\in       | r=S_2       | c=because $S_2$ is a subring }} {{end-eqn}} So: :$-r \in \phi^{-1} S_1$ Finally: {{begin-eqn}} {{eqn | l=\phi \left({r \circ_1 r'}\right)       | r=\phi \left({r}\right) \circ_2 \phi \left({r'}\right)       | c=as $\phi$ is a ring homomorphism }} {{eqn | o=\in       | r=S_2       | c=because $S_2$ is a subring }} {{end-eqn}} So: :$r \circ_1 r' \in S_1$ So from the Subring Test we have that $\phi^{-1} \left[{S_2}\right]$ is a subring of $R$ containing $K$. {{qed}}
From Preimage of Subring under Ring Homomorphism is Subring we have that $S_1 = \phi^{-1} \left[{S_2}\right]$ is a subring of $R_1$ such that $\ker \left({\phi}\right) \subseteq S_1$. We now need to show that $S_1$ is an ideal of $R_1$. Let $s_1 \in S_1, r_1 \in R_1$. Then: {{begin-eqn}} {{eqn | l = \phi \left({r_1 \circ_1 s_1}\right)       | r = \phi \left({r_1}\right) \circ_2 \phi \left({s_1}\right)       | c = as $\phi$ is a homomorphism }} {{eqn | r = S_2       | c = as $S_2$ is an ideal of $R_2$ }} {{end-eqn}} Thus: :$r_1 \circ_1 s_1 \in \phi^{-1} \left[{S_2}\right]= S_1$ Similarly for $s_1 \circ_1 r_1$. So $S_1$ is an ideal of $R_1$. {{qed}}
As $\phi$ is an epimorphism, it is a surjection, and so $\Img \phi = R_2$. So $S_2 \subseteq \Img {R_1}$. The result then follows from Image of Preimage under Mapping. {{qed}}
We have that the diagonal relation is an equivalence relation. Let $\left({S, \circ}\right)$ be any algebraic structure. {{begin-eqn}} {{eqn | o=       | r=x_1 \Delta_S x_2 \land y_1 \Delta_S y_2       | c= }} {{eqn | o=\implies       | r=x_1 = x_2 \land y_1 = y_2       | c=Definition of Diagonal Relation }} {{eqn | o=\implies       | r=x_1 \circ y_1 = x_2 \circ y_2       | c=(consequence of equality) }} {{eqn | o=\implies       | r=\left({x_1 \circ y_1}\right) \Delta_S \left({x_2 \circ y_2}\right)       | c=Definition of Diagonal Relation }} {{end-eqn}} {{qed}} It can therefore be described as universally congruent. {{Qed}}