avewright/proofwiki-math huggingface dataset:
Theorem: Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal

Let $G$ be a group.
Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$.
Let $K \lhd G / H$.
Let $L = q_H^{-1} \sqbrk K$, where:
:$q_H: G \to G / H$ is the quotient epimorphism from $G$ to the quotient group $G / H$
:$q_H^{-1} \sqbrk K$ is the preimage of $K$ under $q_H$.
Then:
:$L \lhd G$
and there exists a group isomorphism $\phi: \paren {G / H} / K \to G / L$ defined as:
:$\phi \circ q_K \circ q_H = q_L$

Proof:
Let $e$ be the identity element of $G$.
Let $\RR$ be the congruence relation defined by $H$ in $G$.
Let $\SS$ be the congruence relation defined by $K$ in $G / H$.
Let $\TT$ be the relation on $G$ defined as:
:$\forall x, y \in G: x \mathrel \TT y \iff x H \mathrel \SS y H$
From Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence:
:$\TT$ is a congruence relation on $G$
Hence from Congruence Relation on Group induces Normal Subgroup:
:the equivalence class under $\TT$ of $e$, that is $\eqclass e \TT$, is a normal subgroup of $G$.
Then we have:
{{begin-eqn}}
{{eqn | l = L
      | r = q_H^{-1} \sqbrk K
      | c = {{hypothesis}}
}}
{{eqn | r = \set {x \in G: \map {q_H} x \in K}
      | c = {{Defof|Preimage of Subset under Mapping}}
}}
{{eqn | r = \set {x \in G: x H \in K}
      | c = {{Defof|Quotient Mapping}}
}}
{{end-eqn}}
Recall that $H$ is the identity of $G / H$.
Then as $K$ is a subgroup of $G / H$:
:$H \in K$
from Identity of Subgroup.
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \eqclass e \TT
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = \TT
      | r = e
      | c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadstoandfrom
      | l = x H
      | o = \SS
      | r = e H
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x H
      | o = \in
      | r = K
      | c = Definition of $\SS$, and {{apriori}} $H \in K$
}}
{{end-eqn}}
That is:
:$\eqclass e \TT = L$
and so $L$ is a normal subgroup of $G$ {{apriori}}.
{{qed|lemma}}
We can identify:
{{begin-eqn}}
{{eqn | l = q_\RR
      | o = \equiv
      | r = q_H
      | c = as $\RR$ is the congruence relation defined by $H$ in $G$
}}
{{eqn | l = q_\SS
      | o = \equiv
      | r = q_K
      | c = as $\SS$ is the congruence relation defined by $K$ in $G / H$
}}
{{eqn | l = q_\TT
      | o = \equiv
      | r = q_L
      | c = Congruence Relation on Group induces Normal Subgroup
}}
{{end-eqn}}
From Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence we have:
:there exists a unique isomorphism $\phi$ from $\paren {G / H} / K$ to $G / L$ which satisfies:
::$\phi \circ q_\SS \circ q_\RR = q_\TT$
Thus using the above identfications:
:$\phi \circ q_K \circ q_H = q_L$
{{qed}}
Theorem: Trivial Quotient Group is Quotient Group

Let $G$ be a group.
Then the trivial quotient group:
:$G / \set {e_G} \cong G$
where:
:$\cong$ denotes group isomorphism
:$e_G$ denotes the identity element of $G$
is a quotient group.

Proof:
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}}
Theorem: Centralizer of Group Element is Subgroup

Let $\struct {G, \circ}$ be a group and let $a \in G$.
Then $\map {C_G} a$, the centralizer of $a$ in $G$, is a subgroup of $G$.

Proof:
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}}
Theorem: Centralizer of Group Element is Subgroup

Let $\struct {G, \circ}$ be a group and let $a \in G$.
Then $\map {C_G} a$, the centralizer of $a$ in $G$, is a subgroup of $G$.

Proof:
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}}
Theorem: Centralizer in Subgroup is Intersection

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then:
:$\forall x \in G: \map {C_H} x = \map {C_G} x \cap H$
That is, the centralizer of an element in a subgroup is the intersection of that subgroup with the centralizer of the element in the group.

Proof:
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}}
Theorem: Kernel of Inner Automorphism Group is Center

Let the mapping $\kappa: G \to \Inn G$ from a group $G$ to its inner automorphism group $\Inn G$ be defined as:
:$\map \kappa a = \kappa_a$
where $\kappa_a$ is the inner automorphism of $G$ given by $a$.
Then $\kappa$ is a group epimorphism, and its kernel is the center of $G$:
:$\map \ker \kappa = \map Z G$

Proof:
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}}
Theorem: Normalizer is Subgroup

Let $G$ be a group.
The normalizer of a subset $S \subseteq G$ is a subgroup of $G$.
:$S \subseteq G \implies \map {N_G} S \le G$

Proof:
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|Normalizer}}
}}
{{eqn | r = S
      | c = {{Defof|Normalizer}}
}}
{{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}}
Theorem: Subgroup is Normal Subgroup of Normalizer

Let $G$ be a group.
A subgroup $H \le G$ is a normal subgroup of its normalizer:
:$H \le G \implies H \lhd \map {N_G} H$

Proof:
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}}
Theorem: Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then $\map {N_G} H$, the normalizer of $H$ in $G$, is the largest subgroup of $G$ containing $H$ as a normal subgroup.

Proof:
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}}
Theorem: Normal Subgroup iff Normalizer is Group

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then $H$ is normal in $G$ {{iff}} the normalizer of $H$ is equal to $G$:
:$H \lhd G \iff \map {N_G} H = G$

Proof:
=== 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}}
Theorem: Normalizer of Conjugate is Conjugate of Normalizer

Let $G$ be a group.
Let $a \mathop \in G$.
Let $S$ be a subset of $G$.
Let $S^a$ denote the conjugate of $S$ by $a$.
Let $\map {N_G} S$ denote the normalizer of $S$ in $G$.
Then:
:$\map {N_G} {S^a} = \paren {\map {N_G} S}^a$
That is, the normalizer of a conjugate is the conjugate of the normalizer:

Proof:
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}}
Theorem: Normalizer of Center is Group

Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Let $x \in G$.
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Then:
:$\map Z G = \set {x \in G: \map {N_G} x = G}$
That is, the center of a group $G$ is the set of elements $x$ of $G$ such that the normalizer of $x$ is the whole of $G$.
Thus:
:$x \in \map Z G \iff \map {N_G} x = G$
and so:
:$\index G {\map {N_G} x} = 1$
where $\index G {\map {N_G} x}$ is the index of $\map {N_G} x$ in $G$.

Proof:
$\map {N_G} x$ is the normalizer of the set $\set x$.
Thus:
{{begin-eqn}}
{{eqn | l = \map {N_G} x
      | r = G
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall a \in G
      | l = \set x^a
      | r = \set x
      | c = {{Defof|Normalizer}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall a \in G
      | l = a x a^{-1}
      | r = x
      | c = {{Defof|Conjugate of Group Element}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \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}}
Theorem: Quotient Group of Ideal is Coset Space

Let $\struct {R, +, \circ}$ be a ring.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +}$ be the quotient group of $\struct {R, +}$ by $\struct {J, +}$.
Then each element of $\struct {R / J, +}$ is a coset of $J$ in $R$, that is, is of the form $x + J = \set {x + j: j \in J}$ for some $x \in R$.
The rule of addition of these cosets is: $\paren {x + J} + \paren {y + J} = \paren {x + y} + J$.
The identity of $\struct {R / J, +}$ is $J$ and for each $x \in R$, the inverse of $x + J$ is $\paren {-x} + J$.

Proof:
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}}
Theorem: Quotient Ring is Ring/Quotient Ring Product is Well-Defined

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$.
Then $\circ$ is well-defined on $R / J$, that is:
:$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies x_1 \circ y_1 + J = x_2 \circ y_2 + J$

Proof:
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}}
Theorem: Quotient Ring of Commutative Ring is Commutative

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
If $\struct {R, +, \circ}$ is a commutative ring, then so is $\struct {R / J, +, \circ}$.

Proof:
Let $\struct {R, +, \circ}$ be a commutative ring
That means $\circ$ is commutative on $R$.
Thus:
{{begin-eqn}}
{{eqn | q = \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}}
Theorem: Quotient Ring of Ring with Unity is Ring with Unity

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
Then $\struct {R / J, +, \circ}$ is a ring with unity, and its unity is $1_R + J$.

Proof:
Let $\struct {R, +, \circ}$ be a ring with unity.
First, let $J \subsetneq R$.
By {{Corollary|Ideal of Unit is Whole Ring}}:
:$1_R \in J \implies J = R$
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}}
Theorem: Ring Epimorphism with Trivial Kernel is Isomorphism

$\phi$ is an isomorphism {{iff}} $K = \set {0_{R_1} }$.

Proof:
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.
{{qed}}
Theorem: Quotient Epimorphism is Epimorphism/Ring

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
Let $\phi: R \to R / J$ be the quotient (ring) epimorphism from $R$ to $R / J$:
:$x \in R: \map \phi x = x + J$
Then $\phi$ is a ring epimorphism whose kernel is $J$.

Proof:
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 = Quotient Ring Addition is Well-Defined
}}
{{eqn | r = \map \phi x + \map \phi y
      | c = Definition of $\phi$
}}
{{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 = Quotient Ring Product is Well-Defined
}}
{{eqn | r = \map \phi x \, \map \phi y
      | c = Definition of $\phi$
}}
{{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}}
Theorem: Ring Homomorphism Preserves Subrings

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$.

Proof:
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}}
Theorem: Ring Homomorphism Preserves Subrings

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$.

Proof:
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}}
Theorem: Ring Homomorphism Preserves Subrings

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$.

Proof:
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}}
Theorem: Kernel of Ring Homomorphism is Subring

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then the kernel of $\phi$ is a subring of $R_1$.

Proof:
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}}